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Inside Collection:

Collection by: Rupinder Sekhon. E-mail the author

# Linear Equations

Module by: Rupinder Sekhon. E-mail the author

Summary: This chapter covers principles of linear equations. After completing this chapter students should be able to: graph a linear equation; find the slope of a line; determine an equation of a line; solve linear systems; and complete application problems using linear equations.

## Chapter Overview

In this chapter, you will learn to:

1. Graph a linear equation.
2. Find the slope of a line.
3. Determine an equation of a line.
4. Solve linear systems.
5. Do application problems using linear equations.

## Graphing a Linear Equation

Equations whose graphs are straight lines are called linear equations. The following are some examples of linear equations:

2x3y=62x3y=6 size 12{2x - 3y=6} {}, 3x=4y73x=4y7 size 12{3x=4y - 7} {}, y=2x5y=2x5 size 12{y=2x - 5} {}, 2y=32y=3 size 12{2y=3} {}, and x2=0x2=0 size 12{x - 2=0} {}.

A line is completely determined by two points, therefore, to graph a linear equation, we need to find the coordinates of two points. This can be accomplished by choosing an arbitrary value for xx size 12{x} {} or yy size 12{y} {} and then solving for the other variable.

### Example 1

#### Problem 1

Graph the line: y=3x+2y=3x+2 size 12{y=3x+2} {}

### Example 2

#### Problem 1

Graph the line: 2x+y=42x+y=4 size 12{2x+y=4} {}

The points at which a line crosses the coordinate axes are called the intercepts. When graphing a line, intercepts are preferred because they are easy to find. In order to find the x-intercept, we let y=0y=0 size 12{y=0} {}, and to find the y-intercept, we let x=0x=0 size 12{x=0} {}.

### Example 3

#### Problem 1

Find the intercepts of the line: 2x3y=62x3y=6 size 12{2x - 3y=6} {}, and graph.

### Example 4

#### Problem 1

Graph the line given by the parametric equations: x=3+2tx=3+2t size 12{x=3+2t} {}, y=1+ty=1+t size 12{y=1+t} {}

### Horizontal and Vertical Lines

When an equation of a line has only one variable, the resulting graph is a horizontal or a vertical line.

The graph of the line x=ax=a size 12{x=a} {}, where aa size 12{a} {} is a constant, is a vertical line that passes through the point ( aa size 12{a} {}, 0). Every point on this line has the x-coordinate aa size 12{a} {}, regardless of the y-coordinate.

The graph of the line y=by=b size 12{y=b} {}, where bb size 12{b} {} is a constant, is a horizontal line that passes through the point (0, bb size 12{b} {}). Every point on this line has the y-coordinate bb size 12{b} {}, regardless of the x-coordinate.

#### Example 5

##### Problem 1

Graph the lines: x=2x=2 size 12{x= - 2} {} , and y=3y=3 size 12{y=3} {}.

## Slope of a Line

### Section Overview

In this section, you will learn to:

1. Find the slope of a line if two points are given.
2. Graph the line if a point and the slope are given.
3. Find the slope of the line that is written in the form y=mx+by=mx+b size 12{y= ital "mx"+b} {}.
4. Find the slope of the line that is written in the form Ax+By=cAx+By=c size 12{ ital "Ax"+ ital "By"=c} {}.

In the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the "steepness" of the line is known. The precise number that refers to the steepness or inclination of a line is called the slope of the line.

From previous math courses, many of you remember slope as the "rise over run," or "the vertical change over the horizontal change" and have often seen it expressed as:

riserun, vertical changehorizontal change, ΔyΔxetc.riserun size 12{ { {"rise"} over {"run"} } } {}, vertical changehorizontal change size 12{ { {"vertical change"} over {"horizontal change"} } } {}, ΔyΔx size 12{ { {Δy} over {Δx} } } {}etc.
(5)

We give a precise definition.

Definition 1:

If ( x1x1 size 12{x rSub { size 8{1} } } {}, y1y1 size 12{y rSub { size 8{1} } } {}) and ( x2x2 size 12{x rSub { size 8{2} } } {}, y2y2 size 12{y rSub { size 8{2} } } {}) are two different points on a line, then the slope of the line is

Slope = m = y 2 y 1 x 2 x 1 Slope = m = y 2 y 1 x 2 x 1 size 12{"Slope"=m= { {y rSub { size 8{2} } - y rSub { size 8{1} } } over {x rSub { size 8{2} } - x rSub { size 8{1} } } } } {}

### Example 6

#### Problem 1

Find the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.

### Example 7

#### Problem 1

Find the slope of the line that passes through the points (2, 3) and (2, -1), and graph.

### Example 8

#### Problem 1

Graph the line that passes through the point (1, 2) and has slope 3434 size 12{ - { {3} over {4} } } {} .

### Example 9

#### Problem 1

Find the slope of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.

### Example 10

#### Problem 1

Find the slope of the line y=3x+2y=3x+2 size 12{y=3x+2} {}.

### Example 11

#### Problem 1

Determine the slope and y-intercept of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.

## Determining the Equation of a Line

### Section Overview

In this section, you will learn to:

1. Find an equation of a line if a point and the slope are given.
2. Find an equation of a line if two points are given.

So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on it, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.

An equation of a line can be written in two forms, the slope-intercept form or the standard form.

The Slope-Intercept Form of a Line: y = mx + b y = mx + b size 12{y= ital "mx"+b} {}

A line is completely determined by two points, or a point and slope. So it makes sense to ask to find the equation of a line if one of these two situations is given.

### Example 12

#### Problem 1

Find an equation of a line whose slope is 5, and y-intercept is 3.

### Example 13

#### Problem 1

Find the equation of the line that passes through the point (2, 7) and has slope 3.

### Example 14

#### Problem 1

Find an equation of the line that passes through the points (–1, 2), and (1, 8).

### Example 15

#### Problem 1

Find an equation of the line that has x-intercept 3, and y-intercept 4.

The Standard form of a Line: Ax + By = C Ax + By = C size 12{ ital "Ax"+ ital "By"=C} {}

Another useful form of the equation of a line is the Standard form.

Let LL size 12{L} {} be a line with slope mm size 12{m} {}, and containing a point (x1,y1)(x1,y1) size 12{ $$x rSub { size 8{1} } ,y rSub { size 8{1} }$$ } {}. If (x,y)(x,y) size 12{ $$x,y$$ } {} is any other point on the line LL size 12{L} {}, then by the definition of a slope, we get

m = y y 1 x x 1 m = y y 1 x x 1 size 12{m= { {y - y rSub { size 8{1} } } over {x - x rSub { size 8{1} } } } } {}
(21)
y y 1 = m ( x x 1 ) y y 1 = m ( x x 1 ) size 12{y - y rSub { size 8{1} } =m $$x - x rSub { size 8{1} }$$ } {}
(22)

The last result is referred to as the point-slope form or point-slope formula. If we simplify this formula, we get the equation of the line in the standard form, Ax+By=CAx+By=C size 12{ ital "Ax"+ ital "By"=C} {}.

### Example 16

#### Problem 1

Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope –3/5.

### Example 17

#### Problem 1

Find the standard form of the line that passes through the points (1, -2), and (4, 0).

### Example 18

#### Problem 1

Write the equation y=2/3x+3y=2/3x+3 size 12{y= - 2/3x+3} {} in the standard form.

### Example 19

#### Problem 1

Write the equation 3x4y=103x4y=10 size 12{3x - 4y="10"} {} in the slope-intercept form.

Finally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.

By solving for yy size 12{y} {}, it can easily be shown that the slope of the line Ax+By=CAx+By=C size 12{ ital "Ax"+ ital "By"=C} {} is A/BA/B size 12{ - A/B} {}. The reader should verify.

### Example 20

#### Problem 1

Find the slope of the following lines, by inspection.

1. 3x5y=103x5y=10 size 12{3x - 5y="10"} {}
2. 2x+7y=202x+7y=20 size 12{2x+7y="20"} {}
3. 4x3y=84x3y=8 size 12{4x - 3y=8} {}

Now that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be very easy.

### Example 21

#### Problem 1

Find an equation of the line that passes through (2, 3) and has slope 4/54/5 size 12{ - 4/5} {}.

If you use this method often enough, you can do these problems very quickly.

## Applications

Now that we have learned to determine equations of lines, we get to apply these ideas in real-life equations.

### Example 22

#### Problem 1

A taxi service charges $0.50 per mile plus a$5 flat fee. What will be the cost of traveling 20 miles? What will be cost of traveling xx size 12{x} {} miles?

### Example 23

#### Problem 1

The variable cost to manufacture a product is $10 and the fixed cost$2500. If xx size 12{x} {} represents the number of items manufactured and yy size 12{y} {} the total cost, write the cost function.

### Example 24

#### Problem 1

It costs $750 to manufacture 25 items, and$1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this function to predict the cost of 100 items.

### Example 25

#### Problem 1

The freezing temperature of water in Celsius is 0 degrees and in Fahrenheit 32 degrees. And the boiling temperatures of water in Celsius, and Fahrenheit are 100 degrees, and 212 degrees, respectively. Write a conversion equation from Celsius to Fahrenheit and use this equation to convert 30 degrees Celsius into Fahrenheit.

### Example 26

#### Problem 1

The population of Canada in the year 1970 was 18 million, and in 1986 it was 26 million. Assuming the population growth is linear, and x represents the year and y the population, write the function that gives a relationship between the time and the population. Use this equation to predict the population of Canada in 2010.

## More Applications

### Section Overview

In this section, you will learn to:

1. Solve a linear system in two variables.
2. Find the equilibrium point when a demand and a supply equation are given.
3. Find the break-even point when the revenue and the cost functions are given.

In this section, we will do application problems that involve the intersection of lines. Therefore, before we proceed any further, we will first learn how to find the intersection of two lines.

### Example 27

#### Problem 1

Find the intersection of the line y=3x1y=3x1 size 12{y=3x - 1} {} and the line y=x+7y=x+7 size 12{y= - x+7} {}.

### Example 28

#### Problem 1

Find the intersection of the lines 2x+y=72x+y=7 size 12{2x+y=7} {} and 3xy=33xy=3 size 12{3x - y=3} {} by the elimination method.

### Example 29

#### Problem 1

Solve the system of equations x+2y=3x+2y=3 size 12{x+2y=3} {} and 2x+3y=42x+3y=4 size 12{2x+3y=4} {} by the elimination method.

### Example 30

#### Problem 1

Solve the system of equations 3x4y=53x4y=5 size 12{3x - 4y=5} {} and 4x5y=64x5y=6 size 12{4x - 5y=6} {}.

### Supply, Demand and the Equilibrium Market Price

In a free market economy the supply curve for a commodity is the number of items of a product that can be made available at different prices, and the demand curve is the number of items the consumer will buy at different prices. As the price of a product increases, its demand decreases and supply increases. On the other hand, as the price decreases the demand increases and supply decreases. The equilibrium price is reached when the demand equals the supply.

### Example 31

#### Problem 1

The supply curve for a product is y=1.5x+10y=1.5x+10 size 12{y=1 "." 5x+"10"} {} and the demand curve for the same product is y=2.5x+34y=2.5x+34 size 12{y= - 2 "." 5x+"34"} {}, where xx size 12{x} {} is the price and y the number of items produced. Find the following.

1. How many items will be supplied at a price of $10? 2. How many items will be demanded at a price of$10?
3. Determine the equilibrium price.
4. How many items will be produced at the equilibrium price?

### Break-Even Point

In a business, the profit is generated by selling products. If a company sells xx size 12{x} {} number of items at a price PP size 12{P} {}, then the revenue RR size 12{R} {} is PP size 12{P} {} times xx size 12{x} {} , i.e., R=PxR=Px size 12{R=P cdot x} {}. The production costs are the sum of the variable costs and the fixed costs, and are often written as C=mx+bC=mx+b size 12{C= ital "mx"+b} {}, where xx size 12{x} {} is the number of items manufactured.

A company makes a profit if the revenue is greater than the cost, and it shows a loss if the cost is greater than the revenue. The point on the graph where the revenue equals the cost is called the Break-even point.

### Example 32

#### Problem 1

If the revenue function of a product is R=5xR=5x size 12{R=5x} {} and the cost function is y=3x+12y=3x+12 size 12{y=3x+"12"} {}, find the following.

1. If 4 items are produced, what will the revenue be?
2. What is the cost of producing 4 items?
3. How many items should be produced to break-even?
4. What will be the revenue and the cost at the break-even point?

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# Product of concave functions and harmonic mean

I discovered something interesting, and I would like to know whether it is a known result or not. Say that a function $$f: \Omega \subset \mathbb{R} \rightarrow \mathbb{R_+^*}$$ is $$\alpha$$-concave if $$f^\alpha$$ is concave.

Let $$f$$ a $$\alpha$$-concave function, $$g$$ a $$\beta$$-concave function. Let $$\gamma$$ be the half of the harmonic mean of $$\alpha$$ and $$\beta$$, ie. $$\begin{equation*}\frac{1}{\gamma} = \frac{1}{\alpha} + \frac{1}{\beta}.\end{equation*}$$ Then $$fg$$ is a $$\gamma$$-concave function.

This can be proved by computing the hessian of $$fg$$. Have you already seen this result in the literature?

• A small remark: $\gamma$ is not the harmonic mean, but half of it. – Ivan Izmestiev Nov 24 '18 at 6:59
• You're perfectly right, thanks! I've just corrected it. – LacXav Nov 26 '18 at 15:51

We want to prove that $$(f(ax+by)g(ax+by))^\g \ge a(f(x)g(x))^\g + b (f(y)g(y))^\g$$ for every $$x, y$$ and $$a+b = 1$$. Since we know that $$f(ax+by)^\a \ge af(x)^\a + bf(y)^\a$$ and $$g(ax+by)^\b \ge ag(x)^\b + bg(y)^\b$$ it is enough to prove that

$$(af(x)^\a + bf(y)^\a)^{\frac{\g}{\a}} (ag(x)^\b + bg(y)^\b)^{\frac{\g}{\b}}\ge af(x)^\g g(x)^\g + b f(y)^\g g(y)^\g.$$

Consider set $$\{x, y\}$$ as measurable space with $$m(x) = a$$, $$m(y) = b$$. Then our desired inequality (after rising to the power $$\frac{1}{\g}$$) is nothing but Holder inequality with parameters $$\a, \b$$.

Unfortunatly, I too do not have any reference for this cute result, but I'm sure it is known.

This is nice, and I do not know a reference for this.

The notion of $$p$$-concavity is mentioned, for example, in Section 9 of

Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008.

Note that a natural interpretation of $$0$$-concavity is log-concavity. Your result generalizes the fact that the product of log-concave functions is log-concave.

Also your result is equivalent to the following.

Theorem. If functions $$F$$ and $$G$$ are concave, then so is $$F^tG^{1-t}$$ for all $$t \in [0,1]$$.

Indeed, substitute $$f^\alpha = F$$, $$g^\beta = G$$, $$t = \frac{\beta}{\alpha+\beta}$$.

The concavity of $$F^tG^{1-t}$$ can be proved as follows. For every $$\lambda \in (0,1)$$ and $$\mu = 1-\lambda$$ the concavity of $$F$$ and $$G$$ imply $$\begin{multline*} F^tG^{1-t}(\lambda x + \mu y) = F^t(\lambda x + \mu y) G^{1-t}(\lambda x + \mu y)\\ \ge (\lambda F(x) + \mu F(y))^t (\lambda G(x) + \mu G(y))^{1-t}\\ \ge (\lambda F(x))^t(\lambda G(x))^{1-t} + (\mu F(y))^t(\mu G(y))^{1-t}\\ = \lambda F^tG^{1-t}(x) + \mu F^tG^{1-t}(y) \end{multline*}$$ (In the middle we have used the inequality $$(1+u)^t(1+v)^{1-t} \ge 1+u^t v^{1-t}$$, for which I have only a nasty proof.)

One can prove the theorem also by computing the second derivative of $$F^tG^{1-t}$$ (which is fun), but the above argument is more general because it does not require differentiability.

I think these results should be known, but have no reference.

EDIT: As Alexei Kulikov pointed out, one can prove the Theorem by first proving the special case $$t=\frac12$$ as a lemma (in this case the nasty inequality becomes $$\sqrt{(1+u)(1+v)} \ge 1 + \sqrt{uv}$$, which follows from Cauchy-Schwarz). From the special case one infers by induction all $$t = p/2^n$$ cases, which implies the general case by a limit argument.

EDIT: A reformulation of the theorem: the space of exp-concave functions is convex.

• It seems to me that your inequality(and the initial problem as well) can be redused to the case $t=\frac{1}{2}$ via mimicking proof of the fact that mid-point concavity implies concavity for continuous functions. And for $t=\frac{1}{2}$ it is just C-S. – Aleksei Kulikov Nov 23 '18 at 23:33
• And moreover it is actually Holder inequality on $2$-point measurable space with appropriate functions. – Aleksei Kulikov Nov 24 '18 at 0:37
• @AlekseiKulikov: I will add the $t=1/2$ approach to my answer. What do you mean by Hoelder inequality on $2$-point measurable space? Could you write this up as an answer? – Ivan Izmestiev Nov 24 '18 at 6:52
• Thanks for the reference, and having put in perspective. – LacXav Nov 26 '18 at 16:35


# Permutations expressed as product of transpositions

There is a theorem that states that all permutations can be expressed as a product of transpositions. I have a couple of questions about this theorem:

1. Does the product which is equal to the permutation always start from the identity permutation?

In the proof for this theorem our professor has argued that every permutation can be transformed into the identity permutation by applying a certain number of transpositions, e.g. if $\sigma$ is a permutation not equal to the identity permutation then you can apply, say l transpositions, so that you get: $\tau_l \circ \tau_{l-1} \circ .... \circ \tau_1 \circ \sigma = id \Rightarrow \tau_1^{-1} \circ \tau_2^{-1} \circ ... \circ \tau_l^{-1}=\tau_1 \circ \tau_2 \circ .... \circ \tau_l$.

Is this product of transpositions always unique, or can you start from any arbitrary permutation and perform the required number of transpositions to get your permutation?

2 If I form the composition of two permutations, say $\sigma_1$ and $\sigma_2$ given by: $$\sigma_1 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 4 &5 &6 &1 &2 \\ \end{pmatrix}$$

$$\sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 4 &6 &1 &5 &3 \\ \end{pmatrix}$$

I can express them in terms of the following transpositions $$\sigma_1 = (1,5)\circ (2,6) \circ (3,5) \circ (4,6)$$ $$\sigma_2 = (1,4) \circ (2,4) \circ (3,6)$$

When I form the composition $\sigma_1 \circ \sigma_2$ I get:

$$\sigma_1 \circ \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 6 &2 &3 &1 &5 \\ \end{pmatrix}$$

But since the product of the transpositions is equal to the permutations, I should get the same result when I use:

$$\sigma_1 \circ \sigma_2 = (1,5)\circ (2,6) \circ (3,5) \circ (4,6) \circ (1,4) \circ (2,4) \circ (3,6)$$

but I get:

$$\sigma_1 \circ \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 &5 &3 &2 &4 \\ \end{pmatrix}$$

Why doesn't this work?

• I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\sigma_1\circ\sigma_2$ is that $\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\mapsto 4\mapsto 6\mapsto 2$ etc. – mdp Apr 22 '14 at 13:49
• Thanks for your comment, but I did start from the right side: $\begin{pmatrix} 1 & 2 & 3&4&5&6\end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 2 & 6&4&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 4 & 6&2&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 6&1&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 6&3&5&1\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 5&3&6&1\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 1 & 5&3&6&4\end{pmatrix} \Rightarrow \begin{pmatrix} 6 & 1 & 5&3&2&4\end{pmatrix}$ I'm not sure what you mean by $1 \rightarrow 4 \rightarrow 6 \rightarrow 2...$ – eager2learn Apr 22 '14 at 13:58
• Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$). – mdp Apr 22 '14 at 14:04
• Yes that was the mistake I made. Thanks a lot for your help. – eager2learn Apr 22 '14 at 14:33

Note that the product of transpositions that you used to express $\sigma_1$, when composed, does not again yield $\sigma_1$; I.e., you did not correctly express $\sigma_1$ as a product of transpositions.
Rather, $\sigma_1$ can be written $(1,5)\circ (1, 3)\circ (2, 6)\circ (2, 4)$, or $(1, 3)\circ (3, 5)\circ (2, 4)\circ(4, 6)$...
Similarly, $\sigma_2$ is incorrectly decomposed. Two correct decompositions include $(1, 4)\circ (1, 2)\circ (3, 6)$ and $(1, 2)\circ (2, 4) \circ (3, 6)$...
...which answers your question about uniqueness. When writing a permutation as a product of transpositions, there are many such ways to do this. What does not vary is the parity: an "odd" permutation is one that can only be decomposed to a product of an odd number of transpositions, and "even" permutations can only be decomposed into a product of an even number of transpositions. So, for example, $\sigma_1$ is even, and $\sigma_2$ is odd.
• I don't understand this. If I apply $(1,5)\circ (1,3) \circ (2,6) \circ (2,4)$ on the id permutation I get the permutation $\begin{pmatrix} 5 &6&1&2&3&4 \end{pmatrix}$ Why is this a transposition for $\sigma_1$? I think this goes back to my original question, from which permutation do I start when applying those transpositions? – eager2learn Apr 22 '14 at 14:26
• You start from the rightmost permutation. Where does it send $1$?. If it sends $1$ to $a$, then you move to the next transposition to its left to see where it sends $a$. Etc. – Namaste Apr 22 '14 at 14:30


By default, glmnet will do two things that you should be aware of: Since regularized methods apply a penalty to the coefficients, we need to ensure our coefficients are on a common scale. Shows the effect of collinearity in the coefficients of an estimator. Ridge regression is a method by which we add a degree of bias to the regression estimates. The second line fits the model to the training data. Keep in mind, ridge is a regression … 11. Ridge, LASSO and Elastic net algorithms work on same principle. In scikit-learn, a ridge regression model is constructed by using the Ridge class. Lasso is great for feature selection, but when building regression models, Ridge regression should be your first choice. In R, the glmnet package contains all you need to implement ridge regression. Active 2 years, 8 months ago. Let us first implement it on our above problem and check our results that whether it performs better than our linear regression model. They all try to penalize the Beta coefficients so that we can get the important variables (all in case of Ridge and few in case of LASSO). Ridge Regression is the estimator used in this example. Linear regression is the standard algorithm for regression that assumes a linear relationship between inputs and the target variable. The value of alpha is 0.5 in our case. Ridge Regression. Elastic net regression combines the properties of ridge and lasso regression. Ridge Regression is a neat little way to ensure you don't overfit your training data - essentially, you are desensitizing your model to the training data. Use the below code for the same. This is also known as $$L1$$ regularization because the regularization term is the $$L1$$ norm of the coefficients. Ridge regression - introduction¶. Yes simply it is because they are good biased. Important things to know: Rather than accepting a formula and data frame, it requires a vector input and matrix of predictors. regression_model = LinearRegression() regression_model.fit(X_train, y_train) ridge = Ridge(alpha=.3) For example, to conduct ridge regression you may use the sklearn.linear_model.Ridge regression model. scikit-learn provides regression models that have regularization built-in. It works by penalizing the model using both the 1l2-norm1 and the 1l1-norm1. Generally speaking, alpha increases the affect of regularization, e.g. The Alpha Selection Visualizer demonstrates how different values of alpha influence model selection during the regularization of linear models. Ridge Regression is a technique for analyzing multiple regression data that suffer from multicollinearity. There are two methods namely fit() and score() used to fit this model and calculate the score respectively. Here, we are using Ridge Regression as a Machine Learning model to use GridSearchCV. One commonly used method for determining a proper Γ \boldsymbol{\Gamma} Γ value is cross validation. Note that scikit-learn models call the regularization parameter alpha instead of $$\lambda$$. The alpha parameter tells glmnet to perform a ridge (alpha = 0), lasso (alpha = 1), or elastic net (0 < alpha < 1) model. ridgeReg = Ridge(alpha=0.05, normalize=True) ridgeReg.fit(x_train,y_train) pred = ridgeReg.predict(x_cv) calculating mse When multicollinearity occurs, least squares estimates are unbiased, but their variances are large so they may be far from the true value. This notebook is the first of a series exploring regularization for linear regression, and in particular ridge and lasso regression.. We will focus here on ridge regression with some notes on the background theory and mathematical derivations that are useful to understand the concepts.. Then, the algorithm is implemented in Python numpy

## ridge regression alpha

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# Lecture 17: Rapidly Decreasing Singular Values

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## Description

Professor Alex Townsend gives this guest lecture answering the question “Why are there so many low rank matrices that appear in computational math?” Working effectively with low rank matrices is critical in image compression applications.

## Summary

Professor Alex Townsend's lecture
Why do so many matrices have low effective rank?
Sylvester test for rapid decay of singular values
Image compression: Rank $$k$$ needs only $$2kn$$ numbers.
Flags give many examples / diagonal lines give high rank.

Related section in textbook: III.3

Instructor: Prof. Alex Townsend

GILBERT STRANG: So let me use the mic to introduce Alex Townsend, who taught here at MIT-- taught Linear Algebra 18.06 very successfully.

And then now he's at Cornell on the faculty, still teaching very successfully. And he was invited here yesterday for a big event over in Engineering.

And he agreed to give a talk about a section of the book-- section 4.3-- which, if you look at it, you'll see is all about his work. And now you get to hear from the creator himself. OK.

ALEX TOWNSEND: OK. Thanks. Thank you, Gil. Thank you for inviting me here. I hope you're enjoying the course.

Today I want to tell you a little about why there so many matrices that are low rank in the world. So as computational mathematicians-- Gil and myself-- we come across low-rank matrices all the time. And we started wondering, as a community, why?

What is it about the problems that we are looking at? What makes low-rank matrices appear? And today I want to give you that story-- or at least an overview of that story.

So for this class, x is going to be n by n real matrix. So nice and square. And you already know, or are very comfortable with, the singular values of a matrix.

So the singular values of a matrix, as you know, are a sequence of numbers that are monotonically non-increasing that tell us all kinds of things about the matrix x.

For example, the number of nonzero singular values tell us the rank of the matrix x. And they also, you probably know, tell us how well a matrix x can be approximated by a low-rank matrix.

So let me just write two facts down that you already are familiar with. So here's a fact-- that, if I look at the number of non-zero singular values in x-- so I'm imagining there's going to be k non-zero singular values-- then we can say a few things about x.

For example, the rank of x, as we know, is k-- the number of non-zero singular values. But we also know from the SVD that we can decompose x into a sum of rank 1 matrices-- in fact, the sum of k of them.

So because x is rank k, we can write down a low-rank representation for x, and it involves k terms, like this.

Each one of these vectors here is a column vector. So if I draw this pictorially, this guy looks like this, right? And we have k of them. So because x is rank k, we can write x as a sum of k rank 1 matrices.

And we also have an initial fact that we already know-- that the dimension of the column space of x is equal to k, and the same with the row space. So the column space of x equals the row space of x-- the dimension-- and they all equal k.

And so there are three facts we can determine from looking at this sequence of singular values of a matrix x. Of course, the singular value sequence is unique. X defines its own singular values.

What we're interested in here is, what makes x? What are the properties of x that make sure that the singular values have a lot of zeros in that sequence? Can we try to understand what kind of x makes that happen?

And we really like matrices that have a lot of zeros here, for the following reason-- we say x is low rank if the following holds, right? Because if we wanted to send x to our friend-- we're imagining x as picture where each entry is a pixel of that image.

If that matrix-- that image-- was low rank, we could send the picture to our friend in two ways. We could send one every single entry of x. And for us to do that, we would have to send n squared pieces of information, because we'd have to send every entry.

But if x is sufficiently low rank, we could also send our friend the vectors-- u, u1, v1, uk, up to vk. And how much pieces of data would we have to send our friend to get x to them if we sent in the low-rank form?

Well, there's 2n here, 2n here numbers. There's k of them. So we'd have to send 2kn numbers. And we strictly say a matrix is low rank if it's more efficient to send x to our friend in low-rank form then in full-rank form.

So this, of course, by a little calculation, just shows us that, provided the rank is less than half the size of the matrix, we are calling the matrix low rank.

Now, often, in practice, we demand more. We demand that k is much smaller than this number, so that it's far more efficient to send our friend the matrix x in low-rank form than in full-rank form. So the colloquial use of the word low rank is kind of this situation. But this is the strict definition of it.

So what do low-rank matrices look like? And to do that, I have some pictures for you. I have some flags-- the world flags. So these are all matrices x-- these examples-- because their flags happen to not be square. I hope you can all see this.

But the top row here are all matrices that are extremely low rank. For example, the Austria flag-- if you want to send that to your friend, that matrix is of rank 1. So all you have to do is send your friend two vectors. You have to tell your friend the column space and the row space. And there's only the dimensions of one of both.

For the English flag, you need to send them two column vectors and two row vectors-- u1, v1, u2 and v2. And as we go down this row, they get slowly fuller and fuller rank.

So the Japanese flag, for example, is low rank but not that small. The Scottish flag is essentially full rank. So it's very inefficient to send your friend the Scottish flag in low-rank form. You're better off sending almost every single entry.

So what do low-rank matrices look like? Well, if the matrix is extremely low rank, like rank 1, then when you look at that matrix-- like here, like the flag-- it's highly aligned with the coordinates-- with the rows and columns.

So if it's rank 1, the matrix is highly aligned-- like the Austria flag. And of course, as we add in more and more rank here, the situation gets a bit blurry. For example, once we get into the medium rank situation, which is a circle, it's very hard to see that the circle is actually, in fact, low rank.

But what I'm going to do was try to understand why the Scottish flag or diagonal patterns-- particularly a bad example for low rank. So I'm going to take the triangular flag to examine that more carefully. So the triangular flag looks like-- I'll take a square matrix and I'll color in the bottom half.

So this matrix is the matrix of ones below the diagonal. And I'm interested in this matrix and, in particular, its singular values, to try to understand why diagonal patterns are not particularly useful for low-rank compression. And this matrix of all ones has a really nice property that, if I take its inverse, it looks a lot like-- getting close to Gil's favorite matrix.

So if I take the inverse of this matrix-- it has an inverse because it's got ones on the diagonal-- then its inverse is the following matrix, which people familiar with finite difference schemes will notice the familiarity between that and the first order finite difference approximation.

In particular, if I go a bit further and times two of these together, and do this, then this is essentially Gil's favorite matrix, except one entry happens to be different-- ends up being this matrix, which is very close to the second order, central, finite difference matrix.

And people have very well studied that matrix and know its eigenvalues, its singular values-- they know everything about that matrix. And you'll remember that if we know the eigenvalues of a matrix, like x transpose x, we know the singular values of x. So this allows us to show, by the fact that we know that, that the singular values of this matrix are not very amenable to low rank. They're all non-zero, and they don't even decay.

So I'm getting this from-- I rang up Gil, and Gil tells me these numbers. That allows us to work out exactly what the singular values of this matrix are, from the connection to finite differences.

And so we can understand why this is not good by looking at the singular values. So the first singular value of x from this expression is going to be approximately 2n over pi. And from this expression, again, for the last guy-- the last singular value of x is going to be approximately a half.

So these singular values are all large. They're not getting close to zero. If I plotted these singular values on a graph-- so here's the first singular value, the second, and the n-th-- then what would the graph look like?

Well, plot these numbers. Divide by this guy so that they all are bounded between 1 and 0 because of the normalization, because I divided by sigma 1 of x.

And so we can plot them, and they will look like this kind of thing. This number happens to be here where they come to be pi over 4n, which is me dividing this number by this number, approximately.

So triangular patterns are extremely bad for low rank. We need things-- or we at least intuitively think that we need things-- aligned with the rows and columns, but the circle case happens to also be low rank. And so what happened to the Japanese flag? Why is the Japanese flag convenient for low rank?

Well it's the fact that it's a circle, and there's lots of symmetry in a circle. So if I try to look at the rank of a circle, the Japanese flag, then I can bound this rank by decomposing the Japanese flag into two things. So this is going to be less than or equal to the rank of sum of two matrices, and I'll do it so that the decomposition works out.

I have the circle. I'm going to cut out a rank one piece that lives in the middle of this circle. OK? And I'm going to cut out a square from the interior of that circle. OK? And I can figure out-- of course the rank is just bounded by the sum of those two ranks. This guy is bounded by rank one because it's highly aligned with the grid. So this guy is bounded by rank one. So this thing here plus 1.

And now I have to try to understand the rank of this piece. Now this piece has lots of symmetry. For example, we know that the rank of that matrix is the dimension of the column space and the dimension of the row space. So when we look at this matrix, because of symmetry, if I divide this matrix in half along the columns, all the columns on the left appear on the right. So for example, the rank of this matrix is the same as the rank of that matrix because I didn't change the column space. OK?

Now I go again and divide along the rows, and now the row dimension of this matrix is the same as the top half, because as I wipe out those, I didn't change the dimension of the row space because the rows are the same top-bottom. And so this becomes the rank of that tiny little matrix there. And because it's small, it won't have too large a rank. So this is definitely less than-- if I divide that up, a little guy here looks like that plus the other guy that looks like that plus 1.

And so of course the row space of this matrix cannot be very high because this is a very thin matrix. There's lots of zeros in that matrix, only a few ones. And so you can go along and do a bit of trig to try to figure out how many rows are non-zero in this matrix.

And a bit of trig tells you-- well it depends on the radius of this original circle. So if I make the original radius r of this Japanese flag, then the bound that you end up getting will be, for this matrix, r 1 minus square root 2 over 2 for this guy. That's a bit of trig. I've got to make sure that's an integer. And then again, here it's the same but for the column space. So this is me just doing trig. OK?

And that's bound on the rank. It happens to be extremely good. And if you work out what that rank is and try to look back, you will find it's extremely efficient to send the Japanese flag to your friend in low rank form, because it's not full rank because these numbers are so small. So this comes out to be, like, approximately 1/2 r plus 1. So much smaller than what you would expect, because remember, a circle is almost the anti-version version of a line with the grid, but yet, it's still low rank. OK.

Now most matrices that we come up with in computational math are not exactly of finite rank. They are of numerical rank. And so I'll just define that. So the numerical rank of a matrix is very similar to the rank, except we allow ourselves a little bit of wiggle room when we define it, and so that amount of wiggle room will be of parameter called tol called epsilon. That's a tolerance. I'm thinking of epsilon as a tolerance. That's the amount of wiggle room I'm going to give myself. OK.

And we say that the numerical rank-- I'll put an epsilon there to denote numerical rank-- is k. k is the first singular value, or the last singular value, above epsilon. In the following sense, I'm copying the definition above but with epsilons instead of zeros. If this singular value is less than epsilon, relatively, and the kth one was not below. So k plus 1 is the first singular value below epsilon in this relative sense.

So of course the rank of 0x, if that was defined, is the same as the rank of x. OK? So this is just allowing ourselves some wiggle room. But this is actually what we're interested more in practice. All right? I don't want to necessarily send my friend the flag to exact precision. I would actually be happy to send my friend the flag up to 16 digits of precision, for example. They're not going to tell the difference between those two flags. And if I can get away with compressing the matrix a lot more once I have a little bit of wiggle room, that would be a good thing.

So we know from the Eckart and Young that the singular values tell us how well we can approximate x by a low-rank matrix. In particular, we know that the k plus 1 singular value of x tells us how well x can be approximated by a rank k matrix. OK? For example, when the rank was exactly k, the sigma k plus 1 was 0, and then this came out to be 0 and we found that x was exactly a rank k matrix.

Here, because we have the wiggle room, the epsilon, we get an approximation, not an exact. So this is telling us how well we can approximate x by a rank k matrix. OK? That's what the singular values are telling us. And so this allows us to try our best to compress matrices but use low-rank approximation rather than doing things exactly.

And of course, on a computer, when we're using floating point arithmetic, or on a computer because we always round numbers to the nearest 16-digit number, if epsilon was 16 digits, your computer wouldn't be able to tell the difference between x or x the rank k approximation if this number satisfied this expression. Your computer would think of x and xk as the same matrix because it would inevitably round both to epsilon, within epsilon. OK.

So what kind of matrices are numerically of low rank? Of course all low-rank matrices are numerically of low rank because the wiggle room can only help you but it's far more than that. There are many full-rank matrices-- matrices that don't have any singular values that are zero-- but the singular values decay rapidly to zero. That are full-rank matrices with low numerical rank because of the wiggle room.

So for example, here is the classic matrix that fits this regime. If I give you this, this is called the Hilbert matrix. This is a matrix that happens to have extremely low numerical rank but it's actually full rank, which means that I can approximate H by a rank k matrix where k is quite small very well, provided you give me some wiggle room, but it's not a low-rank matrix in the sense that if epsilon was zero here, you didn't allow me the wriggle room, all the singular values of this matrix are positive. So it's of low numerical rank but it's not a low-rank matrix.

The other classical example which motivated a lot of the research in this area was the Vandermonde matrix. So here is the Vandermonde matrix. An n by n version of it. Think of the xi's as real. And this is Vandermonde. This is the matrix that comes up when you try to do polynomial interpolation at real points.

This is an extremely bad matrix to deal with because it's numerically low rank, and often, you actually want to solve a linear system with this matrix. And numerical low rank implies that it's extremely hard to invert, so numerical low rank is not always good for you. OK? Often, we want the inverse, which exists, but it's difficult because V has low numerical rank.

OK. So people have been trying to understand why these matrices are numerically of low rank for a number of years, and the classic reason why there are so many low-rank matrices is because the world is smooth, as people say. They say, the world is smooth. That's why matrices are of numerical low rank. And to illustrate that point, I will do an example.

So this is classically understood by a man called Reade in 1983, and this is what his reason was. I have a picture of John Reade. He's not very famous, so I try to make sure his picture gets around. He's playing the piano. It's, like, one of the only pictures I could find of him.

So what is in this reason? Why do people say this? Well here's an example that illustrates it. If I take a polynomial in two variables and I-- for example, this is a polynomial of two variables-- and my x matrix comes from sampling that polynomial integers-- for example, this matrix-- then that matrix happens to be of low rank-- mathematically of low rank, with epsilon equals zero.

Why is that? Well if I write down x in terms of matrices, you could easily see it. So this is made up of a matrix of all ones plus a matrix of j-- so that's 1, 2, up to n, 1, 2, up to n, because every entry of that matrix just depends on the row index. And then this guy depends on both j and k. So this is a multiplication table, right? So this is n, 2, 4, up to 2n, n, 2n, n squared. OK.

Clearly, the matrix of all ones is a rank one matrix. The same with this guy. The column space is just of dimension one. And the last guy also happens to be of rank one because I can write this matrix in rank one form, which is a column vector times a row vector. OK. So this matrix x is of rank three. I guess at lowest rank three is what I've actually shown. OK.

Now of course this hasn't got to numerical low rank yet, so let's get ourselves there. So Reade knew this, and he said to himself, OK, well if I can approximate-- if x is actually coming from sampling a function, and I approximate that function by polynomial, then I'm going to get myself a low-rank approximation and get a bound on the numerical rank.

So in general, if I give you a polynomial of two variables, which can be written down-- it's degree n in both x and y. Let's just keep these indexes away from the matrix index. I give you this such polynomial, and I go away and I sample it and make a matrix X, then X, by looking at each term individually like I did there, will have low rank mathematically, with epsilon equals zero. This will have, at most, m squared rank, and if m is 3 or 4 or 10, it possibly could be low because this X could be a large matrix. OK.

So what Reade did for the Hilbert matrix was said, OK, well look at that guy. That guy looks like it's sampling a function. It looks like it's sampling the function 1 over x plus y minus 1. So he said to himself, well, that x, if I look at the Hilbert matrix, then that is sampling a function. It happens to not be a polynomial. It happens to be this function.

But that's OK because sampling polynomials, integers, gives me low rank exactly. Maybe sampling smooth functions, functions like this, can be well approximated by polynomials and therefore have low numerical rank. And that's what he did in this case. So he tried to find a p, a polynomial approximation to f. In particular, he looked at exactly this kind of approximation. So he has some numbers here so that things get dissolved later. And he tried to find a p that did this kind of approximation. So this approximates f.

And then he would develop a low-rank approximation to X by sampling p. So he would say, OK, well if I let y be a sampling of p, then from the fact that f is a good approximation to p, y is a good approximation to X. And so this has finite rank. He wrote down that this must hold. And the epsilon comes out here because these factors were chosen just right. The divide by n was chosen so that the epsilon came out just there. OK?

So, for many years, that was kind of the canonical reason that people would give, that, well, if the matrix X is sampled from a smooth function, then we can approximate our function by a polynomial and get polynomial rank approximations. And therefore, the matrix X will be of low numerical rank. There's an issue with this reasoning, especially for the Hilbert matrix, that it doesn't actually work that well.

So for example, if I take the 1,000 by 1,000 Hilbert matrix and I look at its rank-- OK, well I've already told you this is full rank. You'll get 1,000. All the singular values are positive. If I look at the numerical rank of this 1,000 by 1,000 Hilbert matrix and I compute it, I compute the SVD and I look at how many are above epsilon where epsilon is 10 to the minus 15, so that means I can approximate the 1,000 by 1,000 Hilbert matrix by a rank 28 matrix and only give up 15-- there will be exact 15 digits, which is a huge amount. So this is what we get in practice, but Reade's argument here shows that the rank of this matrix, the numerical rank, is at most. So it doesn't do a very good job on the Hilbert matrix for bounding the rank, right?

So Reade comes along, takes this function. He tries to find a polynomial that does this, where epsilon is 10 to the minus 15. He finds that the number of terms that he needs in this expression here is around 719, and therefore, that's the rank that he gets. The bound on the numerical rank. The trouble is that 719 tells us that this is not of low numerical rank, but we know it is, so it's an unsatisfactory reason.

So there's been several people trying to come up with more appropriate reasons that explain the 28 here. And so one reason that I've started to use is another slightly different way of looking at things, which is to say the world is Sylvester. Now Sylvester, what does that mean? What does the word "Sylvester" mean in this case? It means that the matrices satisfy a certain type of equation called the Sylvester equation, and so the reason is really, many of these matrices satisfy a Sylvester equation, and that takes the form-- for sum A, B, and C.

OK. So X is your matrix of interest. You want to show X is of numerical low rank. And the task at hand is to find an A, B, and C so that X satisfies that equation. OK. For example, the two matrices I've had on the board satisfy a Sylvester equation-- a Sylvester matrix equation. There is an A, a B, and a C for which they do this.

For example, remember the Hilbert matrix, which we have there still, but I'll write it down again. Has these entries. So all we need to do is to try to figure out an A, a B, and then a C so that we can make it fit a Sylvester equation. There's many different ways of doing this. The one that I like is the following, where if I put 1/2 here and 3/2 here, all the way down to n minus 1/2, times this matrix-- so this is timesing the top of this matrix by 1/2 and then 3/2 and then 5/2. So we're basically timesing each entry of this matrix by j minus 1/2.

And then I do something on the right here, which I'm allowed to do because I've got the B freedom, and I choose this to be the same up to a minus sign. Then when you think about this, what is it doing? It's timing the jk entry-- this is-- by j minus 1/2. That's what this is doing. And what's this doing is timesing the jk entry by k minus 1/2. So this is, in total, timesing the jk entry by j plus k minus 1/2 minus 1/2, which is minus 1, so this is timesing the jk entry by j plus k minus 1. So it knocks out the denominator. And what we get from this equation is a bunch of ones. So in this case, A and B are diagonal, and C is the matrix of all ones. OK?

We can also do this for Vandermonde. So Vandermonde, you'll remember, looks like this. And then over here, we have this guy, the matrix that appears with polynomial interpolation. OK. So if I think about this, I could also come up with an A, B, and C, and for example, here's one that works. I can stick the x's on the diagonal.

So if you imagine what that matrix on the left is doing, it's timesing each column by the vector x. OK? So the first column of this matrix becomes x, the vector x. The second becomes the vector x squared, where squared is done entry-wise. And then the third entry is now x cubed, and when we get to the last, it's x to the n. OK? So that's like, multiply each column by the vector x.

So if I want to try to come up with a matrix-- so what's left is of low rank, is like of this form. What I can do is shift the columns. So I've noticed that this product here, this diagonal matrix, has made the first column x. So if I want to kill off that column, I can take the second column and permute it to the first column. I could take the third column and permute it to the second, the last column and permute it to the penultimate column here. And that will actually kill off a lot of what I've created in this matrix right here.

So let me write that down. This is a circumshift matrix. This does that permutation. I've put a minus 1 there. I could have put any number there. It doesn't make any difference. But this is the one that works out extremely nicely. Now this zeros out lots of things because of the way I've done the multiplication by x and the circumshift of the columns.

And so the first column is zero because this first column is x, this first column is x, so I've got x minus x. This column was x squared minus x squared, so I got zero, and I just keep going along until that last column. That last column is a problem because the last column of this guy is x to the n, whereas I don't have x to the n in V, so there are some numbers here. OK.

You'll notice that C in both cases happens to be a low-rank matrix. In these cases, it happens to be of rank one. And so people were wondering, maybe it's something to do with satisfying these kind of equations that makes these matrices that appear in practice numerically of low rank. And after a lot of work in this area, people have come up with a bound that demonstrates that these kind of equations are key to understanding numerical low rank.

So if X satisfies a Sylvester equation, like this, and A is normal, B is normal-- I don't really want to concentrate on those two conditions. It's a little bit academic. Then-- people have found a bound on the singular values of any matrix that satisfies this kind of expression, and they found this following bound. OK, so here, the rank of C is r. So that goes there. So in our cases, the two examples we have, r is 1, so we can forget about r.

This nasty guy here is called the Zolotarev number. E is a set that contains the eigenvalues of A, and F is a set that contains the eigenvalues of B. OK. Now it looks like we have gained absolutely nothing by this bound, because I've just told you singular values are bound by Zolotarev numbers. That doesn't mean anything to anyone. It means a little bit to me but not that much.

So the key to this bound-- the reason this is useful-- is that so many people have worked out what these Zolotarev numbers actually mean. OK? So these are two key people that worked out what this bound means. And we have gained a lot because people have been studying this number. This is, like, a number that people cared about from 1870 onwards to the present day, and people have studied this number extremely well. So we've gained something by turning it into a more abstract problem that people have thought about previously, and now we can go to the literature on Zolotarev numbers, whatever they are, and discover this whole literature of work on this Zolotarev number.

And the key part-- I'll just tell you the key-- is that the sets E and F are separated. So for example, in the Hilbert matrix, the eigenvalues of A can be read off the diagonal. What are they? They are between minus 1/2 and n minus 1/2. And the eigenvalues of B lie in the set minus 1/2 minus n plus 1/2. And the key reason why the Hilbert matrix is of low numerical rank is the fact that these two sets are separated, and that makes this Zolotarev number gets small extremely quickly with k.

Now you might wonder why there is a question mark on Penzl's name. There is an unofficial curse that's been going on for a while. Both these men died while working on the Zolotarev problem. They both died at the age of 31. One died by being hit by a train, Zolotarev. It's unclear whether he was suicidal or it was accidental.

Penzl died at the age of 31 in the Canadian mountains by an avalanche. I am currently not yet 31 but going to be 31 very soon, and I'm scared that I may join this list. OK. But for the Hilbert matrix, what you get from this analysis, based on these two peoples' work, is a bound on the numerical rank. And the rank that you get is, let's say, a world record bound. For the Hilbert matrix is 34, which is not quite 28, not yet, but it's far more descriptive of 28 than 719.

And so this technique of bounding singular values by using these Zolotarev numbers is starting to gain popularity because we can finally answer to ourselves why there are so many low-rank matrices that appear in computational math. And it's all based on two 31-year-olds that died. And so if you ever wonder when you're doing computational science when a low rank appears and the smoothness argument does not work for you, you might like to think about Zolotarev and the curse. OK, thank you very much.

[APPLAUSE]

GILBERT STRANG: Thank you [INAUDIBLE] Excellent.

ALEX TOWNSEND: How does it work now?

GILBERT STRANG: We're good. Yeah.

ALEX TOWNSEND: I'm happy to take questions if we have a minute, if you have any questions.

GILBERT STRANG: How near of 31 are you?

ALEX TOWNSEND: [INAUDIBLE] I get a spotlight. I'm 31 in December.

GILBERT STRANG: Wow. OK.

ALEX TOWNSEND: So they died at the age of 31, so you know, next year is the scary year for me. So I'm not driving anywhere. I'm not leaving my house until I become 32.

GILBERT STRANG: Well, thank you [INAUDIBLE]

ALEX TOWNSEND: OK, thanks.

[APPLAUSE]


# How many trailing zeroes in $52!$? [duplicate]

Problem

How many trailing zeroes are there in $52!$ ?

My thoughts

I believe I correctly solved it, but I'm not happy with the scalability of my method.

I figure the number of trailing zeroes is equal to the number of times $52!$ is divisible by 10. I wrote out every integer from 1-52 that is divisible by 2 or 5. The idea being that the number of 2 AND 5-factors equals the number of 10-factors.

I quickly noted that the number of 2-factors is greater than the number of 5-factors, so I figured finding the number of 5-factors will do. There were 12.

I'm not very happy with this, because if they now ask me to do the same for $152!$, I'll have to tell them to shove it. I'm not doing this again.

Question

Is there a better way to do this? Perhaps a method that scales better?

## marked as duplicate by Jyrki LahtonenAug 27 '17 at 16:35

• Your method looks good and should be quick now you have the basics. Undertaking the same task for $152!$ should rapidly give you some insight into how to solve the problem quicker. Look out for $125$. – Joffan Aug 27 '17 at 16:39
You get your $12$ by counting $\left\lfloor\frac{52}{5}\right\rfloor+\left\lfloor\frac{52}{25}\right\rfloor$. This counts each multiple of $5$ in the product $52!$, then adds on a count for each multiple of $25$. This can generalize to $152$. You'll need to go up to $125$ with divisors.
In general, the number of trailing $0$s in $n!$ will be $$\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor$$ where for any $n$, only the first few terms in that infinite sum will be nonzero.


# Which transformation needed to make variance independent of population parameter?

Suppose $$s^2$$ is the sample variance of a sample$$(\text{of size }n)$$ from a normal population with mean $$\mu$$ and variance $$\sigma^2. \text{Here }s^2=\frac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}$$
As we know $$\frac{(n-1)s^2}{\sigma^2}\sim \chi^2_{(n-1)}$$ here . Let $$x\sim \chi^2_{(n-1)}$$ then $$E[x]=n-1$$ and $$Var[x]=2(n-1)$$ \begin{align*} E\left[\frac{(n-1)s^2}{\sigma^2}\right]=E[x]&=n-1\\ \implies \frac{(n-1)}{\sigma^2}E[s^2] &= n-1\\ \implies E[s^2]&=\sigma^2 \end{align*} Similarly, \begin{align*} Var\left[\frac{(n-1)s^2}{\sigma^2}\right]=Var[x]&=2(n-1)\\ \implies \frac{(n-1)^2}{\sigma^4}E[s^2] &=2(n-1)\\ \implies Var[s^2]&=\color{red}{\frac{2\sigma^4}{(n-1)} } \end{align*} Now we need a function of $$s^2$$ whose variance will be independent of $$\sigma^2.$$ Let $$f(s^2)$$ be the required transformation.

The required transformation is $$f(s^2)\stackrel{?}{=}\ln{s^2}$$

Question: How they get that transformation $$?$$
Similar things happen with Poisson variate and Binomial proportion with square root and $$\sin^{-1}$$ transformation. So I need a general approach to get that transformation which will make variance independent of population parameter.

• Search for Variance-stabilizing_transformation. – StubbornAtom Nov 6 '19 at 14:33
• "The aim behind the choice of a variance-stabilizing transformation is to find a simple function f to apply to values x in a data set to create new values y=f(x) such that the variability of the values y is not related to their mean value." But I need the variability of the values y is not related to their variance(population) value? @StubbornAtom Sir – emonhossain Nov 6 '19 at 14:45
• The wiki article might not be entirely clear. I gave you the keyword to look for the topic. (And I am no 'sir'.) – StubbornAtom Nov 6 '19 at 14:51

Suppose $$X_1,X_2,\ldots,X_n$$ are i.i.d $$N(\mu,\sigma^2)$$ and define the sample variance as $$S^2=\frac{1}{n}\sum\limits_{i=1}^n (X_i-\overline X)^2$$

Here we are concerned with the large-sample behaviour of $$S^2$$, so it does not matter if we take $$n$$ as the divisor above instead of $$n-1$$.

By CLT it can be argued that $$\sqrt n(S^2-\sigma^2)\stackrel{L}\longrightarrow N(0,2\sigma^4)\tag{1}$$

And by 'Delta-method', for a real-valued function $$g$$ such that $$g'(\sigma^2)$$ exists and $$g'(\sigma^2)\ne 0$$, it follows from $$(1)$$ that

$$\sqrt n(g(S^2)-g(\sigma^2))\stackrel{L}\longrightarrow N(0,2\sigma^4[g'(\sigma^2)]^2)\tag{2}$$

You have to solve for a $$g$$ such that asymptotic variance of $$g(S^2)$$ is free of $$\sigma^2$$.

So for some non-zero constant $$c$$, set $$\sqrt 2\sigma^2 g'(\sigma^2)=c$$

Therefore, $$\int g'(\sigma^2)\,d\sigma^2=\frac{c}{\sqrt 2}\int\frac{d\sigma^2}{\sigma^2}$$

Choosing $$c=\sqrt 2$$ and taking constant of integration to be zero, we have the required transformation $$g(\sigma^2)=\ln\sigma^2\quad,\text{ i.e. }\quad \color{blue}{g(S^2)=\ln S^2}$$

From $$(2)$$ you end up with $$\sqrt n(\ln(S^2)-\ln(\sigma^2))\stackrel{L}\longrightarrow N(0,2)$$

This is the application of variance stabilizing transformation on the sample variance from a normal population. The other transformations that you mention for the Poisson mean and Binomial proportion are derived similarly.

Result $$(1)$$ and many more can be found in A Course in Large Sample Theory by Thomas Ferguson. The relevant part is on page 46 of the first edition (1996):

The main result used in this section is Cramer's Theorem on the asymptotic normality of functions of sample moments, studied through a Taylor-series expansion to one term.

• what I know is CLT tells us $\color{red}{\sqrt{n}(\overline{x}-\mu)\text{ converge in distribution to a normal }N(0,\sigma^2)}$ And I know how to prove this but $\sqrt n(s^2-\sigma^2)\stackrel{L}\longrightarrow N(0,2\sigma^4)$ is looks new to me. So everything will be meaningful to me if you provide a source where I can get that proof $?$ I will accept your answer then. Thanks again @StubbornAtom – emonhossain Nov 6 '19 at 16:07
• @emonhossain Updated. – StubbornAtom Nov 6 '19 at 18:07
• Really you help me a lot. Thanks @StubbornAtom – emonhossain Nov 6 '19 at 18:33


# Putnam Problem of the Day

• Aug 12th 2010, 10:20 PM
simplependulum
Putnam Problem of the Day

Find the smallest natural number with $6$ as the last digit , such that if the final $6$ is moved to the front of the number it is multiplied by $4$ .

Spoiler:
153846
?
• Aug 13th 2010, 03:26 AM
Unbeatable0
Let the number be $a_1a_2...a_n6$.

Let $x=0.\overline{a_1a_2...a_n6}$.

Then $4x = 0.\overline{6a_1a_2...a_n} \Rightarrow 40x = 6.\overline{a_1a_2...a_n6} = 6+x$.

Therefore $x = \frac{6}{39} = \frac{2}{13} = 0.\overline{153846}$.

By backwards argument it's easy to see that all the cycles in $\frac{2}{13}$ are solutions.

Thus we have found all the solutions, the smallest of which is $153846$.

This technique can be applied to this kind of problems in many cases.
• Aug 13th 2010, 02:03 PM
Soroban
Hello, simplependulum!

Just write out the multiplcation,
. . and the digits will appear sequentially.

Quote:

Find the smallest natural number with $6$ as the last digit,
such that if the final $6$ is moved to the front of the number it is multiplied by $4$.

We have: . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&C&D&E&6 \\ \times &&&&& 4\\ \hline 6&A&B&C&D&E \end{array}$

In column-6: . $6 \!\times\! 4 \,=\,24 \quad\Rightarrow\quad E = 4$

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&C&D&4&6 \\
\times &&&&& 4\\ \hline
6&A&B&C&D&4 \end{array}$

In column-5: . $4 \!\times\! 4 + 2 \:=\:18 \quad\Rightarrow\quad D = 8$

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&C&8&4&6 \\
\times &&&&& 4\\ \hline
6&A&B&C&8&4 \end{array}$

In column-4: . $8 \!\times\! 4 + 1 \:=\:33 \quad\Rightarrow\quad C = 3$

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&B&3&8&4&6 \\
\times &&&&& 4\\ \hline
6&A&B&3&8&4 \end{array}$

In column-3: . $3\!\times\!4 + 3 \:=\:15 \quad\Rightarrow\quad B = 5$

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
A&5&3&8&4&6 \\
\times &&&&& 4\\ \hline
6&A&5&3&8&4 \end{array}$

In column-2: . $5\!\times\!4 + 1 \:=\:21 \quad\Rightarrow\quad A = 1$

Therefore: . $\begin{array}{cccccc}
1&5&3&8&4&6 \\
\times &&&&& 4\\ \hline
6&1&5&3&8&4 \end{array}$

• Aug 13th 2010, 05:19 PM
Wilmer
Soroban, the fact that it is a 6digit number is NOT a given.
• Aug 13th 2010, 06:16 PM
Soroban
Quote:

Originally Posted by Wilmer
Soroban, the fact that it is a 6-digit number is NOT a given. . Right!

I originally set up the problem like this:

. . $\begin{array}{ccccccccccc}
A & B & C & D & E & F & G & \hdots & M & N & 6 \\
\times &&&&&&&&&& 4 \\ \hline
6 & A & B & C & D & E & F & G & \hdots & M & N
\end{array}$

And found that "it came out even" after six digits
. . which gave me the smallest number: $153,\!846.$

The next is the 12-digit number: $153,\!846,\!153,\!846.$

• Aug 13th 2010, 07:10 PM
undefined
Fyi, there is a name for these, n-parasitic number, where here n=4 and k=6.
• Aug 13th 2010, 09:32 PM
Wilmer
Let n = number of digits

{20[10^(n-1) - 4] + 78} / 13

Gives integer solution for n=6k where k = any integer > 0


# Circle questions -- Secants, chords and tangents

• Apr 3rd 2014, 05:03 PM
StonerPenguin
Circle questions -- Secants, chords and tangents
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;

"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).
http://i1137.photobucket.com/albums/...mexam10Q32.png
So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)

Did I do that right?
• Apr 3rd 2014, 07:13 PM
JeffM
Re: Circle questions -- Secants, chords and tangents
Quote:

Originally Posted by StonerPenguin
Hello again, just got a few questions pertaining to circles. I'll just post one to start with;

"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).
http://i1137.photobucket.com/albums/...mexam10Q32.png
So solving this.. according to the segments of secants and tangents theorem...
x2 = (2r + 353)(353)
x2 = (12742 + 353)(353)
x2 = 4622535
x = $\sqrt{4622535}$ = 2150km (approx.)

Did I do that right?

Computation looks right to me. But simply memorizing theorems does not promote understanding.

Let's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.

$Let\ r = length\ of\ radius\ of\ given\ circle,$

$u + r = length\ from\ the\ given\ circle's\ center\ to\ a\ given\ point\ outside\ the\ circle,$

$x = length\ of\ the\ line\ that\ is\ tangent\ to\ the\ given\ circle\ and\ runs\ through\ the\ given\ point.$

By the Pythagorean Theorem, we have:

$(u + r)^2 = x^2 + r^2 \implies u^2 + 2ru + r^2 = x^2 + r^2 \implies x^2 = u^2 + 2ru = u(u + 2r) \implies x = \sqrt{u(u + 2r)}.$
• Apr 6th 2014, 06:17 PM
bjhopper
Re: Circle questions -- Secants, chords and tangents
ISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg
cos 71.35 =d/6724
d km to horizon =2150 km
• Apr 6th 2014, 09:56 PM
Soroban
Re: Circle questions -- Secants, chords and tangents
Hello, StonerPenguin!

Quote:

The radius of the earth is approximately 6371 km.
If the international space station (ISS) is orbiting 353 km above the earth,
find the distance from the ISS to the horizon (x).

Code:

                o                 |\                 | \             352 |  \                 |  \ x                 |    \               * * *  \           *    |    *\         *      |      o       *    6371|    *  *                 |  *6371       *        | *      *       *        o        *       *                  *         *                *         *              *           *          *               * * *
Note the right triangle.

The equation is: . $x^2 + 6371^2 \:=\:6723^2$
• Apr 7th 2014, 05:16 PM
StonerPenguin
Re: Circle questions -- Secants, chords and tangents
Thank you JeffM, bjhopper and Soroban! It's nice to see different perspectives :D and the image drawn in code is really cool.
Here's another question I've had trouble with:
http://i1137.photobucket.com/albums/...mexam10Q15.png
"Explain how you know $\overline{AB}$ $\overline{CD}$ given E is the center of the circle. (Include theorem numbers.)"

And here's some pertinent theorems;
Quote:

Theorem 10.4
If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
Quote:

Theorem 10.5
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
Quote:

Theorem 10.6
In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
Obviously from the diagram $\overline{AB}$ $\overline{CD}$ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/
• Apr 8th 2014, 04:24 PM
bjhopper
Re: Circle questions -- Secants, chords and tangents
t is a triangle a is an angle
t AED congruent to BEC isosceles t's same legs and equal altitudes
a DEC = 180- 2* 1 / 2 *AED
a AEB 180-2*1/2 * BEC a AED = aBEC
a AEB = a DEC
AB =DC equal arcs = equal chords


# Tag Info

7

Since $fg\geq 1$, you have $\sqrt{f(x)\cdot g(x)} \geq 1$ for all $x$, and so $$1 \leq \int_X \sqrt{f}\cdot \sqrt{g} d\mu \leq \sqrt{\int_X f d\mu }\cdot \sqrt{\int_X g d\mu }$$ where the second inequality is Cauchy–Schwarz; squaring both sides gives the inequality.

6

It is sufficient to show that $m(E \cap [a, b]) = 0$ for every $a$ and $b$ with $a < b$. Let $t = \inf_{x \in [a, b]}{e^{-x^2/2}}$. Then $t > 0$ and we have $$0 = \int_{E}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}e^{-x^2/2}dx \ge \int_{E\cap [a, b]}tdx = m(E \cap [a, b])t$$ and as $t > 0$, this implies $m(E \cap [a, b]) = 0$.

5

A measure space must specify three things: An underlying set; A $\sigma$-algebra of subsets which are measurable in this measure space; A measure (a function on this $\sigma$-algebra). When we say "a measurable subset of the real numbers" we typically mean "a subset of the real numbers measurable with respect to the Lebesgue measure", ...

5

The questions, despite looking as a representation problem in functional analysis, are much deeper as they bring out the history of the topic involved, notably $BV$-functions and the reasons why the customary definition adopted for the variation of a multivariate function is definition 2 above. And as thus the answers below needs to indulge a bit on this ...

5

$\bigcup_{i\ge 1}A_i$ is by definition simply $$\left\{x:\exists i\in\Bbb Z^+\,(x\in A_i)\right\}\,,$$ which clearly does not depend on the order in which the sets are enumerated. You can simply let $\mathscr{A}=\{A_i:i\in\Bbb Z^+\}$; then $$\bigcup_{i\ge 1}A_i=\bigcup\mathscr{A}=\{x:\exists A\in\mathscr{A}\,(x\in A)\}\,,$$ with no reference to the ...

5

Hint: Consider the sets $[-r,r] \subseteq \mathbb{R}$. Can you show $r \mapsto \mu(E \cup [-r,r] \cap F)$ is continuous? Can you show for any $r$ the set we're measuring is compact? What does the intermediate value theorem buy us? I hope this helps ^_^

5

This is false. For an easy counterexample, let $X_n\geq 0$ be any sequence of random variables with $\liminf_n X_n = 0$ which converges to $X = 1$ in probability. Then let $\mathcal{Q}$ be large enough so that the $X_n$ are $\mathcal{Q}$-measurable. The inequality in question then becomes $X \leq \liminf_n X_n,$ which is false almost surely. For instance, ...

5

It's true that the definition of absolute continuity is the same either we state it in the form of the finite and disjoint intervals or we state it in the form of countable and disjoint intervals. In detail, suppose we have the following definitions of absolute continuity: Definition 1 Let $f:\mathbb{R}\to \mathbb{R}$ be a function. We say that $f$ is ...

4

By assumption, as $Z \notin L^{\infty}$, we have $$\mathbb{P}(Z > n^2) > 0, \quad n \in \mathbb{N}.$$ Therefore, we may define $$X = \sum_{n=1}^{\infty}\frac{1}{n^2} \frac{\mathbb{1}_{[Z > n^2]}}{\mathbb{P}(Z>n^2)} \in L^1(\mathbb{P}).$$ However, it follows that \begin{align*}\mathbb{E}[ZX] = \sum_{n=1}^{\infty}\mathbb{E}\left[\frac{1}{n^2} \frac{...

4

Cannot be true in general: consider $k=1$ (or any odd $k$), take $X_i$'s to be i.i.d normal with mean $0$ and variance $1$, we show below that $Y_n$ is not UI. Note that $$\mathbb{E}\left(|Y_n| 1\{|Y_n|\geq M\}\right) \geq M\mathbb{P}\left(|Y_n|\geq M\right) = M\mathbb{P}\left(|\bar{X}_n|\leq \frac{1}{M}\right) \to M,$$ as $n\to \infty$ by laws of large ...

4

The assertion “not every subset of $\Bbb R$ is measurable” is valid for the Lebesgue measure. Anyway, here Axler is talking about $\sigma$-algebras, not about measures. And, yes, $\mathcal P(\Bbb R)$ is a $\sigma$-algebra. And, on this $\sigma$-algebra you can define the measure $m(X)=\#X$; with respect to this measure, $\Bbb R$ is measurable.

4

If $p > p_0$ and $f \in L^{p_0} \cap L^\infty$ then $|f|^p = |f|^{p - p_0}|f|^{p_0} \le \|f\|_\infty^{p - p_0}|f|^{p_0}$ almost everywhere.

4

The claim is true: Let $S = \sum_{i=1}^\infty \chi_{E_i}$. Suppose for the sake of contradiction that $S < \infty$ a.e. Then, there exists an $M$ such that $\Pr(S\le M) \ge 1 - \frac\varepsilon 2$. If $A$ is the event that $S\le M$, then $\mu(A\cap E_i) \ge \frac\varepsilon 2$ for all $i$. But this would imply $\int_A S\,d\mu = \sum_{i=1}^\infty \mu(A\... 4 Well,$[0,1]=C\hspace{1mm}\cup C^{c} $(I discourage you to use$\mathbb{C}$for the cantor set) and you can write f as: $$f(x)=0\cdot\chi_{C}(x)+x^{2}\chi_{C^{c}}$$ where by$\chi$I mean the characteristic function of the set in subscript. The sum of measurable functions is measurable as well as the product, and polynomial functions are measurable. 4 The family of measurable sets form a$\sigma$-algebra but not all sets belonging in any$\sigma$-algebra are measurable. So to fully define a measure space we need a set$X$, a$\sigma$-algebra$\mathcal{A}$on$X$and a function$\mu: \mathcal{A} \to [0, \infty]$which satisfies some properties. A set$A \subseteq X$is called$\mu$-measurable (or simply ... 4 The symmetrization is called Khintchine's inequality, see here: https://en.wikipedia.org/wiki/Khintchine_inequality For the contraction principle, I don't know any reference, but I can prove it. First note that if$Z\geq 0$and$p > 0, then \begin{align} \mathbb{E}[Z^p] = \mathbb{E}\Big[\int_0^{Z} p t^{p-1} dt\Big] = \mathbb{E}\Big[\int_0^{\infty} p t^{p-... 4 A set\mathfrak B$satisfies the definition in your book if and only if$\mathfrak B$is a$\sigma$-algebra (in the wikipedia sense) on$\cup\mathfrak B$: for the noteworthy algebraic detail, notice that$A\setminus B=A\triangle (A\cap B)$. This is also equivalent to the property of$\mathfrak B$being a$\sigma$-algebra in the wikipedia sense over some set.... 4 Let me discuss the case$n=1$only. Let me denote by$f_x$the weak derivative of$f$with respect to$x$. Let$\phi,\psi$be two smooth test functions. Then $$-\iint f_x(x,y) \phi(x)\psi(y)dx\ dy=\iint f(x,y) \phi_x(x)\psi(y)dx\ dy,$$ using Fubini on both sides gives $$\int \left( \int -f_x(x,y) \phi(x) + f(x,y) \phi_x(x)\ dx\right) \psi(y) \ dy = 0$$ ... 3 You prove that by giving an example of a set of subsets that is closed under intersections and differences but not closed under unions. E.g. on any set$X$with at least two points, the family$\{\{x\}: x \in X\} \cup \{\emptyset\}$is such. so being closed under union and differences (a ring) implies being closed under intersections, but being closed under ... 3 I have recently got interested in continued fractions, so I'll give it a try! CONTINUED FRACTIONS. It is a basic property of continued fractions that if$h_n/k_n$is the sequence of convergent to$x\in\mathbb R\setminus\mathbb Q$, then $$\frac 1{k_n(k_n+k_{n+1})}<\left|x-\frac{h_n}{k_n}\right|<\frac 1{k_nk_{n+1}}.$$ It follows that for all$h,k$... 3 The set$\mathcal{P}(X)$of all subsets of$X$is a sigma algebra because it satisfies the axioms for that kind of structure. That is independent of any discussion of measures. The definition of a measure space is a pair$(X,S)$where$S$is some subset of$\mathcal{P}(X)$that happens to be a sigma algebra. It may or may not be all of$\mathcal{P}(X)$. ... 3 Suppose that such measure exists. Then taking countable cover by open sets of diameter$\leq 1$, one of those sets, say$U_1$, must have measure$1$. Cover$U_1$by countable amount of open sets of diameter$\leq 1/2$and take the one having measure$1$. This process gives us a decreasing sequence of open sets$U_k$with diameter$\leq 1/k$and$\mu(U_k) = 1$... 3 This works. Your argument shows that if$f$is bounded on a set$E$of Lebesgue measure zero, then$\int_E f = 0$. In fact the stronger claim that$\int_E f = 0$without any hypothesis on$f$holds! Indeed the integral is, by definition, the supremum over$g$with$g$simple and$g \leq f$of$\int_E g$. But any such$g$has integral zero. 3 Welcome to MSE! Hint: It's often useful to split up a random variable into the places where it is "good" and "bad". Then we can control those regions separately to get whatever inequality we're interested in. For this, we might try looking at the decomposition $$X = X \cdot \mathbf{1}_{\{X \leq \lambda a\}} + X \cdot \mathbf{1}_{\{X >... 3 According to the p-Adics section of the reference manual there is nothing I could find to do with zeta functions nor integration. At any rate, here's a suggestion on algorithmically getting an arbitrary level of accuracy on your integral. In order to be called a p-adic measure we require it to be a bounded p-adic distribution, and this is enough to imply ... 3 By definition, \int f is the supremum of \int h for 0 \leq h \leq f bounded measurable of finite support. So at the very end, you just need to take the sup over such h to get \int f \leq \liminf \int f_n. He uses such h is order to apply the bounded dominated convergence theorem, which he proves earlier. This allows him to prove Fatou in terms of ... 3 You can also let g_n=inf_{i \geq n} \{f_n\} and note that \int g_n \leq \int f_n for each n. Moreover g_n increases monotonically to lim inf f_n . Hence by monotone convergence theorem \int g_n = \int lim inf f_n. Since \int g_n \leq \int f_n we have \int lim inf f_n \leq lim inf \int f_n. The final step is to note that pointwise convergence ... 3 If you are familiar with dominated convergence notice that f_n(X)\xrightarrow{n\rightarrow\infty} X pointwise, |f_n(X)|\leq |X|, X is assumed to be integrable (E[|X|]=\int |X|\,dP<\infty). Then by dominated convergence$$ \int f_n(X)\,dP\xrightarrow{n\rightarrow\infty}\int X\,dP=E[X]$$similarly for$Y$. There is still the issue of the ... 3 Let$f$be the limit of$\chi_{E_n}$in$L^p$. All you have to do is show that$f(x) = 0$or$f(x) = 1$almost everywhere. Then we see that$f = \chi_E$almost everywhere, where$E = \{x \in X | f(x) = 1\}$To do this, it is helpful to define$g(x) = \min(|x - 1|, |x|)$. We will show that$g \circ f = 0$almost everywhere. We first show that$\int g(f(x))^p \...

3

Let $f \geq 0$ and let $M=\sup \{f(x): x\in X\}$. Then $\|M-f\|\leq M$ since $f(x) \in [0,M]$ for all $x$. If $\phi(f)$ is real we can finish the proof as follows: $M-\phi (f)=\phi(M-f) \leq \|\phi\| \|M-f||\leq M$. Thus $\phi (f) \geq 0$. To show that $\phi (f)$ is real consider following: \$|\phi (f) \pm in|=|\phi(f \pm in)|\leq \|f\pm in\|\leq \sqrt {M^{...

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Calculating probability of 'at least one event occurring'

If I know the probability of A and the probability of B how can I calculate the probability of "at least one of them" occurring?

I was thinking that this is P(A or B) = P(A) + P(B) - P(AB). Is this correct?

If it is, then how can I solve the following problem taken from 'DeGroot - Probability and Statistics':

If 50 percent of families in a certain city subscribe to the morning newspaper, 65 percent of the families subscribe to the afternoon newspaper, and 85 percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?

Here the question is P(morning)=.5, P(afternoon)=.65 P(morning OR afternoon)=.85

P(morning OR afternoon) = .5 + .65 - .3 = .85

P(morning AND afternoon) = P(morning) + P(afternoon) - P(morning OR afternoon)

So the answer to the question is .3

Is my reasoning correct?

EDIT.

If this reasoning is correct:

How can I calculate the following:

If the probability that student A will fail a certain statistics examination is 0.5, the probability that student B will fail the examination is 0.2, and the probability that both student A and student B will fail the examination is 0.1,what is the probability that exactly one of the two students will fail the examination?

So this questions highlights the difference between !at least one of them! !or exactly one of them!

I understand that at least one of them = P(A or B)

but how can I work out the probability of exactly one of them?

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Somebody knows an algorith to calculate the OR for n probabilistic values? Thanks in advance. Regards, Daniel. –  Daniel Mejia Nov 8 '12 at 23:53
You need to specify the correlation between the $n$ probabilistic values - at least whether they are independent or not. –  dexter04 Nov 9 '12 at 0:25

You are correct.

To expand a little: if $A$ and $B$ are any two events then

$$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$

or, written in more set-theoretical language,

$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$

In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $P(A\cup B)$ and you need to work out $P(A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out.

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thanks for your answer! please have a look at my edited question, your help highlighted a further issue.. –  Dbr Oct 14 '11 at 11:10
"Exactly one of A and B" means "Either A or B, but not both" which you can calculate as P(A or B) - P(A and B). –  Chris Taylor Oct 14 '11 at 11:13
thank you, this clarifies everything. Just one last quick thing: Where can I learn the mathematical notation you use on this forum? For some questions I find it quite hard to understand them, is there a tutorial? –  Dbr Oct 14 '11 at 11:15
Are you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\LaTeX$ - you can find a tutorial by searching for "latex tutorial" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. –  Chris Taylor Oct 14 '11 at 11:21
So if I download LaTeX and paste your notation then it displays it in a more readable form? –  Dbr Oct 14 '11 at 11:24

For your second question, you know $\Pr(A)$, $\Pr(B)$, and $\Pr(A \text{ and } B)$, so you can work out $\Pr(A \text{ and not } B)$ and $\Pr(B \text{ and not } A)$ by taking the differences. Then add these two together.

Alternatively take $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$.

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For the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\cap B)$$

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probability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )

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# Calculating dice probabilities with multiple, overlapping success criteria

Alright, so I play a tabletop game where models get in fights and roll dice to see if they deal damage to the other models. I'm wondering how to calculate the probabilities of dealing 0, 1, and 2 damage in a 2v2 fight.

Here is the situation:
- We have Model A and Model B rolling to deal damage to Model C and Model D
- A needs to roll a 5 or 6 to damage C, and a 6 to damage D
- B needs to roll a 4, 5, or 6 to damage C, and a 5 or 6 to damage D
- C and D can only be damaged once each (i.e., they die when they get damaged)

So for starters, I know the probabilities for this scenario just by listing out the possibilities:

  A | 111111 222222 333333 444444 555555 666666
B | 123456 123456 123456 123456 123456 123456
----+------------------------------------------
Dmg | 000111 000111 000111 000111 111122 111222

So the probability of dealing 0 damage is $$\frac{12}{36}$$, (exactly) 1 damage is $$\frac{19}{36}$$, and 2 damage is $$\frac{5}{36}$$. One tricky outcome here is when A rolls a 5 and B rolls a 4. This only results in 1 damage since both rolls are only enough to damage C and C can only be damaged once.

I am hopelessly stumped on how to model this so that I can then apply it to other scenarios such as 3v3s or when some models use more than 1 die.

It is pretty easy to calculate the probabilities of doing damage in a 1v1 fight. I've used binomial distribution statistics to calculate those. Conveniently it works for multiple dice per model as well. However, when it comes to combining these probabilities ... I am at a loss. They don't seem to follow the rules for multiplication and addition.

For example, I naïvely expected the probability of doing 0 damage to be the probability of A doing 0 damage and the probability of B doing no damage. That sounds right intuitively, but no matter how I try to do it ... it never makes sense:

$$P(A\;miss\;C) = \binom{1}{0} \bullet 2^0 \bullet 4^1 = 4$$
$$P(A\;miss\;D) = \binom{1}{0} \bullet 1^0 \bullet 5^1 = 5$$
$$P(B\;miss\;C) = \binom{1}{0} \bullet 3^0 \bullet 3^1 = 3$$
$$P(B\;miss\;D) = \binom{1}{0} \bullet 2^0 \bullet 4^1 = 4$$

Therefore,

$$P(A\;miss\;C) \bigcap P(A\;miss\;D) \bigcap P(B\;miss\;C) \bigcap P(B\;miss\;D) = 4 \bullet 5 \bullet 3 \bullet 4 = 240$$

Wait what?!?! I expect the answer to be 12. Clearly there is something else going on here that I do not understand. :/

• You can't say $\Pr(X\cap Y)=\Pr(X)\Pr(Y)$ unless $X$ and $Y$ are independent events, and these are clearly not. When A doesn't damage C it's likely that he doesn't damage D, either, because he rolled a low number. Mar 16 '20 at 20:56
• Also, you're multiplying counts instead of probabilities. The product of the probabilities (even if you could just multiply them, i.e. if they were independent, which they aren't) would be $\frac46\cdot\frac56\cdot\frac36\cdot\frac46=\frac{240}{6^4}=\frac5{27}\approx0.185$, which is at least a lot closer to the $\frac{12}{6^2}\approx0.333$ that you were apparently expecting than $\frac{240}{6^2}\approx6.67$ would have been. Mar 16 '20 at 21:44

First I'd like to rephrase your question:

A and B each roll a die. The results of the die determine how much damage is dealt. The maximum{damage dealt by A, damage dealt by B} is the number we are interested in.

A Deals 1 damage by rolling a 5 or 2 damage by rolling a 6. B deals 1 damage by rolling a 4 or 2 damage by rolling a 5 or 6. Now take the maximum of the damage dealt by A or B.

I think it may be easier to frame it this way:

A misses and B Misses simultaneously:

P(0 Damage) = P(A Rolls 4 or less) * P(B rolls 3 or less) = 2/3*1/2= 1/3 = 3/9

A Hits both or B hits Both (don't double count):

P(2 Damage) = P(A Rolls 6) + P(B Rolls 5 or 6) - P(A Rolls 6 AND B rolls 5 or 6) = 1/6 + 1/3 - 1/6*1/3 = 4/9

P(1 Damage) = 1-P(0)-P(2) = 2/9

Alternatively,

A hits 1 (rolls 5) or B hits 1 (rolls 4)

P(1 Damage) = P(B Rolls 4)*P(A Rolls 5 or less) + P(B Rolls 3 or less)*P(A Rolls 5) = 1/6*5/6+ 3/6*1/6= 2/9

How I verified my answer: The table below shows how much damage occurs when (A rolls, B rolls) happens. Count the number of 0s and divide by 36 to determine how often 0 damage occurs. Repeat for 1s and 2s.

## | 6 | 2 | 2 | 2 | 2 | 2 | 2 |

Summary:

P( 0 damage ) = 3/9

P( 1 damage ) = 2/9

P( 2 damage ) = 4/9