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# Proportion Standard Error

## Contents

repeatedly randomly drawn from a population, and the proportion of successes in each sample is recorded ($$\widehat{p}$$),the distribution of the sample

## Standard Error Of Proportion Formula

proportions (i.e., the sampling distirbution) can be approximated by a normal standard error of proportion definition distribution given that both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$. This is sample proportion formula known as theRule of Sample Proportions. Note that some textbooks use a minimum of 15 instead of 10.The mean of the distribution of sample proportions is equal

to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$## Sample Proportion Calculator SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("Standard Error")$$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Lesson 4: Probability - 2 Variables Lesson 5: Probability Distributions Lesson 6: Sampling Distributions6.1 - Simulation of a Sampling Distribution of a Proportion (Exact Method) 6.2 - Rule of Sample Proportions (Normal Approximation Method)6.2

0 otherwise. The standard deviation of any variable involves the https://onlinecourses.science.psu.edu/stat200/node/43 expression . Let's suppose there are m 1s (and n-m 0s) among the n subjects. Then, and is equal to (1-m/n) for m observations and 0-m/n http://www.jerrydallal.com/lhsp/psd.htm for (n-m) observations. When these results are combined, the final result is and the sample variance (square of the SD) of the 0/1 observations is The sample proportion is the mean of n of these observations, so the standard error of the proportion is calculated like the standard error of the mean, that is, the SD of one of them divided by the square root of the sample size or Copyright © 1998 Gerard E. Dallal

Tables Constants Calendars Theorems Standard Error of Sample Proportion Calculator https://www.easycalculation.com/statistics/standard-error-sample-proportion.php Calculator Formula Download Script Online statistic calculator allows you to estimate the accuracy of the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard standard error Deviation Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion= Code to add this calci to your website Just copy and paste the below code to your webpage where you standard error of want to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of successes in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. Related Calculators: Vector Cross Product Mean Median Mode Calculator Standard Deviation Calculator Geometric Mean Calculator Grouped Data Arithmetic Mean Calculators and Converters ↳ Calculators ↳ Statistics ↳ Data Analysis Top Calculators Standard Deviation Mortgage Logarithm FFMI Popular Calculators Derivative Calculator Inverse of Matrix Calculator Compound Interest Calculator Pregnancy Calculator Online Top Categories AlgebraAnalyticalDate DayFinanceHealthMortgageNumbersPhysicsStatistics More For anything contact support@easycalculation.com

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# Spectral Decomposition of A and B.

I was given the following question in my linear algebra course.

Let $A$ be a symmetric matrix, $c >0$, and $B=cA$, find the relationship between the spectral decompositions of $A$ and $B$.

From what I understand. If $A$ is a symmetric matrix, then $A=A^T$. A symmetric matrix has $n$ eigenvalues and there exist $n$ linearly independent eigenvectors (because of orthogonality) even if the eigenvalues are not distinct. Since $B=cA$ and $A=A^T$, then we can conclude that $B=cA^T$, which would imply that $B$ is also symmetric, meaning it also has a linearly independent eigenbasis.

Focusing on $A$, since it has a linearly independent eigenbasis, we have $A = PD_aP^{-1}$ by Spectral decomposition where $P$ is the eigenbasis and $D_a$ is the diagonal matrix of $A$ eigenvalues $\lambda_i$ \begin{array} d D_a & = & \begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} \end{array}

Now since $B=cA$, then we have $B=cPD_aP^{-1}$, which can be rewritten as $B = PD_bP^{-1}$, where

\begin{array} d D_b & = & cD_a & =c\begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} & = & \begin{bmatrix} c\lambda_1 & & \\ &\ddots&\\ & & c\lambda_i \end{bmatrix} \end{array}

From this I can conclude that $B$ and $A$ actually have the same linearly independent eigenbasis. Furthermore, the eigenvalues of $B$ are a scalar multiple of the eigenvalues of $A$ by a factor of $c$.

Have I fully describe the relationship between $A$ and $B$?

$\dfrac{\lambda}{c} I- A$ isn't invertible if and only if $\lambda I- cA$ isn't invertible.
Hence, $\lambda\in Sp(cA)$ if and only if $\dfrac{\lambda}{c}\in Sp(A)$
Yes, I would say that you have fully described the relationship between $A$ and $B$.


# Why Binomial Distribution formula includes the “not-happening” probability?

Suppose I have a dice with 6 sides, and I let a random variable $X$ be the number of times I get 3 points when I throw the dice.

So I throw the dice for $10$ times, I want to find the probability of getting 3 points from the dice for $4$ times, ie: $P(X=4)$.

Since the order doesn't matter, there are $\binom{10}{4}=210$ ways and the chance of getting a 3 point is $\frac { 1 }{ 6 }$. Also, because I want to have $4$ of such occurrence, it would be $\frac{1}{6}^4$. So, I could just calculate $P(X=4)=\binom{10}{4}\frac { 1 }{ 6 }^4 =0.027006173\approx 2.7\%$.

But, suppose if I use the Binomial Distribution formula, it would be a little different because it needs to multiply the "not-happening" probability to it. The Binomial Distribution looks like this: $$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

So if I plug in my values, it would be: $$P(X=4)=\binom{10}{4}(\frac{1}{6})^4(\frac{5}{6})^{6}=0.054=5.4\%$$

Here, $2.1\%$ is lesser than $5.4\%$. What's the difference between the two values? Which is the correct value?

Intuitively, I find the Binomial Distribution may be more accurate since it dictates the situation to consider both the happening and not-happening outcomes. But usually, I thought we just multiply the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay too. Eg: What's the probability to get 2 heads out of 5 flips of a fair coin, I just use $\frac { 1}{ 2} \times \frac { 1}{ 2}$. The not-happening probabilities are not cared of.

-
To get exactly 4 "3 points", the other 6 throws have to not be "3 points". The other 6 throws not being 3 points is also part of what you want to happen if exactly 4 of the 10 throws are 3 points. So, you need to multiply by $(5/6)^6$. By the way, in your second displayed equation, you should have ${10\choose 4}(1/6)^4(5/6)^6$ – David Mitra Feb 18 '12 at 16:33
oh yea, thanks. It should ${10\choose 4}(1/6)^4(5/6)^6$ in the second line. Updated the equation. If I need to multiply the "not-happening" probability, then suppose if I want to find the probability of getting 2 heads out of 5 flip of a fair coin, it wouldn't it be $\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }=\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$? But shouldn't it be just $\frac { 1}{ 2} \times \frac { 1}{ 2}$? – xenon Feb 18 '12 at 16:51
No, not just $(1/2)(1/2)$ for exactly two heads. You $need$ the other flips to not be heads; which means you multiply by $(1/2)^3$. – David Mitra Feb 18 '12 at 17:10
@xEnOn: The calculation that gave about $5.4$% is based on the correct analysis, and the calculator work is right. The first calculation, the one that gave about $2.7$%, is based on an incorrect analysis (formula). In addition, there was a calculator mistake. The wrong formula gives an answer of about $16.2$%, not $2.7$%. This is clear if you compare the two expressions. The second multiplies the first by $(5/6)^6$, which is less than $1$. – André Nicolas Feb 18 '12 at 17:18
@xEnOn : Please don't write $\frac 16 ^4$ if you mean $\left(\frac16\right)^4$. The former expression should be used only if you mean $(1^6)/4$. – Michael Hardy Feb 18 '12 at 17:53

Your first argument is a bit off. $X=4$ means exactly four of the tosses resulted in $3$-points To get exactly four $3$-points, the other six throws have to not be $3$- points. The other six throws not being $3$-points is also part of what you want to happen if exactly four of the 10 throws are $3$-points.

Look at a particular case: the first four throws are 3-points and there are exactly four 3-points. Then the throw sequence was $$3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \$$ The probability of this occurring is $(1/6)^4(5/6)^6$.

So, you need to multiply your proposed answer by $(5/6)^6$. So, the correct answer is what is given by the Binomial distribution: ${10\choose4}(1/6)4(5/6)6$.

In your second argument (of the original post), you are correct assuming that you only drew two cards. If you drew three cards, the probability that exactly two were hearts would be $$\underbrace{{13\over52}{12\over 51}{39\over 50}}_{ \text{two hearts, then non heart}}+ \underbrace{{13\over52}{39\over 51}{12\over 50}}_{ \text{ heart, non heart, heart}}+ \underbrace{{39\over52}{13\over 51}{12\over 50}}_{ \text{ non heart, then two hearts}}$$

To be brief you do care about the "not happening" probabilities when you consider the probability that an event happens exactly $n$ times. Knowing that something does not happen is concrete information. In the die example, "not 3-pt" means that six of the flips were 1,2,4,5, or 6 points.

In your coin example, if you want the probability of exactly two heads in five flips, then the three non-head flips must be tails. What is the probability of flipping $H\ H\ T\ T\ T$? It's not $1/4$... It is $1/32$, as you should be able to see from the multiplication rule for a sequence of independent events.

The number of ways in which you can have exactly two heads in five flips is ${5\choose2}=10$ (it's the number of ways to choose two slots from five in which to put the two heads in; note the other three slots must be tails). The probability of exactly two heads in five flips is ${5\choose2}\cdot(1/2)^2(1/2)^3={10\over 32}={5\over16}$.

To see more concretely why your argument fails, consider this simple case. Toss a fair coin three times. What is the probability of having exactly one head? By your reasoning it is $3\cdot(1/2)^1>1$; which is nonsense. Perhaps it's just $1/2$? This is incorrect also:

The equally likely outcomes here are: $$HHH \ \ \ HHT \ \ \ HTT\ \ \ HTH$$ $$THH \ \ \ THT \ \ \ TTT\ \ \ TTH$$ And we see the probability of exactly one head is ${3\over8}$, which is exactly what the Binomial formuls gives: ${3\over8}={3\choose2}(1/2)^1(1/2)^2$.

Incidentally, you're proposed method would not even give the probability of having at least one head. Here, that probability is $7/8$.

-


# Homework Help: Give a big-O estimate of the product of the first n odd positive integers

1. Jul 17, 2011

### pc2-brazil

1. The problem statement, all variables and given/known data
Give a big-O estimate of the product of the first n odd
positive integers.

2. Relevant equations
Big-O notation:
f(x) is O(g(x)) if there are constants C and k such that
|f(x)| ≤ C|g(x)| whenever x > k.

3. The attempt at a solution
The product of the first n odd integers can be given by:
$$P(n)=1\times 3\times 5\times 7\times...\times (2n-1)$$
For n > 0, no element in the above sequence will be greater than (2n-1). Thus:
$$1\times 3\times 5\times 7\times...\times (2n-1)\leq (2n-1)\times (2n-1)...\times (2n-1)=(2n-1)^n$$
So:
P(n) ≤ (2n-1)n whenever n > 0
I could stop here and say that
P(n) is O((2n-1)n)
But to simplify I think I could consider that:
P(n) ≤ (2n-1)n ≤ (2n)n
Thus,
P(n) is O((2n)n)

Is this reasoning correct?

Last edited: Jul 17, 2011
2. Jul 17, 2011

### tiny-tim

hi pc2-brazil!

it's correct, but it's not very accurate, is it?

do you know a big-O estimate for n! ?

3. Jul 17, 2011

### pc2-brazil

A big-O estimate for n! would be O(nn).
I could say that, for n > 0,
$$1\times 3\times 5\times 7...\times (2n-1)\leq 1\times 2\times 3\times 4\times...\times (2n-1)=(2n-1)!\leq (2n)!=2^n n!$$
Thus, P(n) is O(2nn!). Since n! is O(nn), this estimate seems more accurate than the previous one (O(2nnn)).

4. Jul 17, 2011

### tiny-tim

Last edited by a moderator: Apr 26, 2017
5. Jul 17, 2011

### Ray Vickson

If En = product of the even numbers from 2 to 2n - 2, your product is (2n-1)!/En, and En = 2^(n-1) * (n-1)! Now apply Stirling's formula to both factorials. Note: if you want a true upper bound, rather than just an *estimate* you can use the fact that if St(n) is defined as
sqrt(2pi)*n^(n + 1/2)*exp(-n), then we have St(n) <n! < St(n)*exp(1/(12n)), even if n is not large. You can use the upper bound on (2n-1)! in the numerator and the lower bound on (n-1)! in the denominator.

RGV

sqrt(2pi)*n


rational numbers symbol

# rational numbers symbol

The real line consists of the union of the rational and irrational numbers. We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. Formally, rational numbers are the set of all real numbers that can be written as a ratio of integers with nonzero denominator. The OP asked in his title only, if he can use the symbol for rational numbers. Hence, we can say that ‘0’ is also a rational number, as we can represent it in many forms such as 0/1, 0/2, 0/3, etc. In other words, most numbers are rational numbers. Answer: yes. Some examples of irrational numbers are $$\sqrt{2},\pi,\sqrt[3]{5},$$ and for example $$\pi=3,1415926535\ldots$$ comes from the relationship between the length of a circle and its diameter. Numbers that are not rational are called irrational numbers. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. When we put together the rational numbers and the irrational numbers, we get the set of real numbers. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. By using this website, you agree to our Cookie Policy. A rational number is a number that can be expressed as a fraction p/q where p and q are integers and q!=0. Rational numbers are indicated by the symbol . For instance, the decimal version of the therefore symbol (∴) would be ∴ The hexadecimal version of the therefore symbol (∴) would be ∴ Note that the hexadecimal numbers include x as part of the code. In Maths, rational numbers are represented in p/q form where q is not equal to zero. RapidTables. 1 $\begingroup$ I think your answer is fine. Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number set. $\endgroup$ – Dietrich Burde Aug 15 '19 at 18:42. Rational Numbers All positive and negative fractions, including integers and so-called improper fractions. Irrational numbers are a separate category of their own. Many people are surprised to know that a repeating decimal is a rational number. I just disagree that the notations should not be used for fields. Note that the set of irrational numbers is the complementary of the set of rational numbers. Expressed as an equation, a rational number is a number. where a and b are both integers. This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. The denominator in a rational number cannot be zero. ... rational numbers set = … The ancient greek mathematician Pythagoras believed that all numbers were rational, but one of his students Hippasus proved (using geometry, it is thought) that you could not write the square root of 2 as a fraction, and so it was irrational. a/b, b≠0. It is also a type of real number. Top of Page The letters R, Q, N, and Z refers to a set of numbers such that: R = real numbers includes all real number [-inf, inf] Q= rational numbers ( numbers written as ratio) The Unicode numeric entity codes can be expressed as either decimal numbers or hexadecimal numbers. A rational number p/q is said to have numerator p and denominator q. Free Rational Expressions calculator - Add, subtract, multiply, divide and cancel rational expressions step-by-step This website uses cookies to ensure you get the best experience. Any fraction with non-zero denominators is a rational number.


# Conditional probability exercise - apples and oranges

We have two crates, crate 1 and crate 2. Crate 1 has 2 oranges and 4 apples, and crate 2 has 1 orange and 1 apple. We take 1 fruit from crate 1 and put it in crate 2, and then we take a fruit from crate 2.

The first point of this exercise asks me to calculate the probability that the fruit taken from crate 2 is an orange. I did this by calculating the probability that the fruit we took from crate 1 was an orange(which is $\frac{2}{6}$) and then saying that I have 3 fruits in crate 2, $1+\frac{2}{6}$ oranges and the rest apples, which lead me to a $44.44\%$ probability that the fruit we take from crate 2 was an orange. The probability I got seems reasonable, but I don't know for sure if what I did was correct.

Anyway, point 2 of this problem is a little bit harder and I'm stuck. It tells me to calculate the probability that the fruit we took from crate 1 was an orange, if we know that the fruit we took out from crate 2 was also an orange. So if I consider A: Fruit taken from crate 1 was an orange, and B: Fruit taken from crate 2 was an orange, I think I have to calculate $\:P\left(A|B\right)$ I think, which means "Probability that A happens if we know B happened", but I'm not so sure about this.

Could anyone give me a hint on how to go about solving this problem?

• Your answer to the first question is correct. For the second question, $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ where $A$ is the event that an orange was selected from the first crate and $B$ is the event that an orange was selected from the second crate. – N. F. Taussig Jan 29 '17 at 12:18
• But what is $P\left(A\cap B\right)$ ? I mean, I know it is the probability of the joint events, but how do I determine that? – MikhaelM Jan 29 '17 at 12:32
• It is the probability that you put an orange from the first crate into the second, and having done that then took an orange from the second . $$\mathsf P(A\cap B)= \mathsf P(A)~\mathsf P(B\mid A)$$ – Graham Kemp Jan 29 '17 at 12:34
• Doing that I get $\:P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(A\right)\cdot P\left(B|A\right)}{P\left(B\right)}=\frac{\frac{2}{6}\cdot \frac{2}{3}}{\frac{4}{9}}=\frac{1}{2}$ so $50\%$. Is this correct or am I making a mistake somewhere? – MikhaelM Jan 29 '17 at 12:45
• The answer you obtained in your comment is correct. – N. F. Taussig Jan 29 '17 at 13:15

Let $A,B$ be the events of removing an orange from the first and second crates, respectively.

You have calculated $\mathsf P(B) = 4/9$ correctly. Another way to look at it is through the law of total probability. \begin{align}\mathsf P(B) ~&=~\mathsf P(A)~\mathsf P(B\mid A)+\mathsf P(A^\complement)~\mathsf P(B\mid A^\complement) \\ &=~ \tfrac 26\cdot\tfrac 23+\tfrac 46\cdot\tfrac 13 & =&~ \frac{\tfrac 2 6+1}3 \\ &=~ \tfrac 49 \end{align}

Where $\mathsf P(A)$ is the probability of taking an orange from cart 1, $\mathsf P(A^\complement)$ is that of taking an apple from cart 1, $\mathsf P(B\mid A)$ is the probability of taking a orange from cart 2 when given that you have added an orange to that cart, and $\mathsf P(B\mid A^\complement)$ is the probability of taking a orange from cart 2 when given that you have added an apple to that cart.

Now you just need to calculate $\mathsf P(A\mid B)$ the probability of having taken an orange from cart 1 when given that you took an orange from cart 2.

Use Bayes' Rule.

Yes for second part solve P(A|B).

Probability of A happens if we know B happened.


# general solution for a recurrence relation

I have the following recurrence relation: $$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\hspace{1cm}(*)$$

If we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that $r=\frac{a+\sqrt{a^2-4}}{2}$ or $r=\frac{a-\sqrt{a^2-4}}{2}$.

By using the initial values $x_1=1, x_2=a$, I found that $$x_n=\frac{1}{\sqrt{a^2-4}}\left(\frac{a+\sqrt{a^2-4}}{2}\right)^n-\frac{1}{\sqrt{a^2-4}}\left(\frac{a-\sqrt{a^2-4}}{2}\right)^n$$ is a solution for the recurrence relation (*).

Question: How do we know whether this is the only solution for the recurrence relation $(*)$? Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions will have this form.

• Use method of difference on the recursive definition $x_{n+2}-x_{n+1}=(a-2)x_{n+1}+x_{n+1}-x_{n}$ and add these equations for all values of n. – Pi_die_die Jul 8 '18 at 20:57
• In other words, prove that it satisfies condition $(*)$. – steven gregory Jul 8 '18 at 20:59
• Yes you would get the sum which would be the type of sum you would expect from a gp – Pi_die_die Jul 8 '18 at 21:01

You have specific initial values $x_0$ and $x_1$. The recurrence relation fully determines all other values. Thus, there's exactly one solution. If you've found it, that's it, there can't be any others, no matter what approach you took in order to find it.

If you want to show that all solutions of $x_{n+2}=ax_{n+1}-x_n$ are combinations of the geometric series you found, you can argue as follows:

This is a linear second-order recurrence. Its solutions form a two-dimensional vector space. (A vector space because of the linearity of the recurrence, and a two-dimensional one because the solution space is spanned by the two solutions for the initial conditions $x_1=1$, $x_2=0$ and $x_1=0$, $x_2=1$.) You've found two linearly independent solutions; hence they span the entire solution space.

• Hi @joriki I would appreciate some clarifications before I accept your answer. I understand that the set of solutions is a subspace of the vector space of sequences $\{f:\mathbb{N}\to\mathbb{C}: f\mbox{ is a function}\}$. I also understand that I found two solutions that are linearly independent. I do not know why the dimension of the solution space is 2 and not greater than 2. Also, what do you mean by $a_0$ and $a_1$? – Chilote Jul 8 '18 at 21:23
• @Chilote: I'm sorry, I meant $x_1$ and $x_2$; I've fixed that. The solution for given initial conditions $x_1=\hat x_1$ and $x_2=\hat x_2$ is $\hat x_1s_1+\hat x_2 s_2$, where $s_1$ is the solution for $x_1=1$ and $x_2=0$ and $s_2$ is the solution for $x_1=0$ and $x_2=1$; so $s_1$ and $s_2$ span the solution space; hence it's at most two-dimensional (in fact exactly two-dimensional, since $s_1$ and $s_2$ are linearly independent). – joriki Jul 8 '18 at 21:31
• Got it, the solution space of $x_{n+2}=ax_{n+1}-x_n$ is $S=\{(x_n)_n:x_{n+2}=ax_{n+1}-x_n\}$ which is a vector space. Any vector in this space is of the form $x=(x_1,x_2,ax_2-x_1,a(ax_2-x_1)-x_2,\dots)=x_1 s_1+x_2 s_2$ where $s_1$ and $s_2$ are two particular solutions that are linearly independent. – Chilote Jul 8 '18 at 21:49
• You've found two linearly independent solutions Just to nitpick, that's iff $a \ne \pm 2$. – dxiv Jul 8 '18 at 23:29
• @Chilote: This was actually overkill, since you only asked about the uniqueness of one particular solution, not about all solutions being combinations of the geometric series. I edited the answer accordingly. – joriki Jul 9 '18 at 5:22

A bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy $$x_n^2 - a \, x_n \, x_{n+1} + x_{n+1}^2 = 1$$

Try consecutive values in $$1, \; \; a, \; \; a^2 - 1, \; \; a^3 - 2a, \; \ldots$$

This comes from the matrix $$\left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right)$$ which has determinant $1$ and trace $a.$ It also gives an automorphism of the quadratic form $x^2 - a xy+y^2.$ An automorphism matrix $P$ for a quadratic form that has Hessian matrix $H$ satisfies $P^T HP = H$

In this case $$\left( \begin{array}{cc} 0 & -1 \\ 1 & a \end{array} \right) \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) = \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right)$$

The explicit relation with the sequence is

$$\left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) \left( \begin{array}{c} x_n \\ x_{n+1} \end{array} \right) = \left( \begin{array}{c} x_{n+1} \\ x_{n+2} \end{array} \right)$$

• How can I get the first equality? – Chilote Jul 8 '18 at 21:52
• @Chilote the automorphism group for an (indefinite) binary quadratic form, along with the Pell type equation $\tau^2 - \Delta \sigma^2 = 4,$ is discussed in many number theory and quadratic forms books. – Will Jagy Jul 8 '18 at 23:24
• I think the quadratic form corresponding to the matrix you wrote above is $ay^2$ – Chilote Jul 8 '18 at 23:59
• @Chilote I put in a little about the automorphism. The matrix that gives the quadratic form is the symmetric one; it is just the Hessian matrix of second partial derivatives – Will Jagy Jul 9 '18 at 0:23


Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share â¦ In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. If P A Ais nonsingular then the eigenvectors of P 1 U Acorresponding to are of the form [0 T;vT] where v is any eigenvector of P 1 S Cthat corresponds to its unit eigenvalue. T is diagonal iff A is symmetric. Every square real matrix A is orthogonally similar to an upper block triangular matrix T with A=Q T TQ where each block of T is either a 1#1 matrix or a 2#2 matrix having complex conjugate eigenvalues. Moreover, the eigenvectors of P 1 U Acorresponding to are of the form [uT;((P S+ C) 1Bu) T] . Block lower triangular matrices and block upper triangular matrices are popular preconditioners for $2\times 2$ block matrices. Moreover, the eigenvectors of P 1 Based on the lemma, we can derive the following main results about the SBTS iteration method. Yes. First of all: what is the determinant of a triangular matrix? However, a 2 by 2 symmetric matrix cannot have imaginary eigenvalues, so R must be diagonal. The second consequence of Schurâs theorem says that every matrix is similar to a block-diagonal matrix where each block is upper triangular and has a constant diagonal. The determinant of a block-diagonal matrix is the product of the determinants of the blocks, so, by considering the definition of the characteristic polynomial, it should be clear that the eigenvalues of a block-diagonal matrix are the eigenvalues of the blocks. Assume that Î± is a positive constant and S = W â 1 T. These eigenvectors form an orthonormal set. Theorem 3.2. Then the eigenvalues of the matrix S = W â 1 T are all real, and S is similar to a diagonal matrix. 2 AQ = QÎ A(Qe i)=(Qe i)Î» i Qe i is an eigenvector, and Î» i is eigenvalue. TRIANGULAR PRECONDITIONED BLOCK MATRICES 3 P 1 A Athat corresponds to its unit eigenvalue. Theorem 6. This is an important step in a possible proof of Jordan canonical form. Developing along the first column you get $a_{11} \det(A_{11}'),$ where $A_{11}'$ is the minor you get by crossing out the first row and column of [math]A. This method can be impractical, however, due to the contamination of smaller eigenvalues by This decouples the problem of computing the eigenvalues of Ainto the (solved) problem of computing 1, and then computing the remaining eigenvalues by focusing on the lower right (n 1) (n 1) submatrix. Let W, T â R n × n be symmetric positive definite and symmetric, respectively. Hence R is symmetric block diagonal with blocks that either are 1 by 1 or are symmetric and 2 by 2 with imaginary eigenvalues. upper-triangular, then the eigenvalues of Aare equal to the union of the eigenvalues of the diagonal blocks. If each diagonal block is 1 1, then it follows that the eigenvalues of any upper-triangular matrix are the diagonal elements. 1 is a matrix with block upper-triangular structure. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. In this note we show that a block lower triangular preconditioner gives the same spectrum as a block upper triangular preconditioner and that the eigenvectors of the two preconditioned matrices are related.
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# A Seemingly Impossible Problem

Given that $w, x, y, z$ take on values $0$ and $1$ with equal probability, what is the probability that $w+x+y+z$ is odd?

Which of the following arguments is correct?

Furthermore, can you generalize this result?

Argument 1:
If all 4 numbers are even, the sum is even.
If 3 numbers are even, the sum is odd.
If 2 numbers are even, the sum is even.
If 1 number is even, the sum is odd.
If 0 numbers are even, the sum is even.
In 2 of the 5 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\frac{ 2}{5}$.

Argument 2:
$w + x + y + z$ is either odd or even.
In 1 of the 2 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\frac{ 1}{2}$.

Argument 3:
By listing out all the $2^4$ cases, we find that there are ${ 4 \choose 0} + { 4 \choose 2} + { 4 \choose 4 } = 8$ cases with an even sum, and ${ 4 \choose 1} + {4 \choose 3} = 8$ cases with an odd sum.
Hence, $w + x +y + z$ is odd with probability $\frac{ 1}{2}$.

Context: In a recent problem, Argument 1 was given, and many people agreed with it.

Note by Calvin Lin
5 years, 5 months ago

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The problem with the first argument is that the probabilities of each of the cases is different. In reality, the first case has one way, the second has 4 ways, the third has 6 ways, the fourth has 4 ways, and the last one has only 1 way. Thus the actual probability is $\dfrac{4+4}{1+4+6+4+1}=\dfrac{1}{2}$.

- 5 years, 5 months ago

I disagree with Argument 1:

In case 1 there is only 1 outcome to make an even sum.

In case 2 there are 4 outcomes that make an odd sum

In case 3 there are 6 outcomes that make an even sum

In case 4 there are 4 outcomes that make an odd sum

In case 5 there is only 1 outcome to make an even sum.

Thus out of the 16 equally likely outcomes 8 will yield an odd sum and 8 will yield an even sum.

Therefore the probability that w + x + y + z is odd is 1/2.

- 5 years, 5 months ago

Reminds me of Bertrand paradox, where three ways of choosing a seemingly the same uniformly random chord on a circle give different probability distributions (which means they are different).

- 5 years, 5 months ago

I am not very sure if these two are similar indeed. The problem stated above seems to have objective solution and other two are simply counting errors.

- 5 years, 3 months ago

They are not similar, yes. It's just the fact that there are three solutions arriving at different answers immediately reminded me to that.

- 5 years, 3 months ago

Well, this is how i solve this problem.

Assume that we choose w, x, y first.

w + x + y = A.

Doesn't matter whether A is odd or even. z will be the one that decide whether the sum of w, x, y, z (assume w +x+y+z= B) is odd or even

=> With that the possibility for B to be odd = 50%

=> Argument 1: Wrong

Argument 2: Correct (but lack at explaining)

Argument 3: Correct

- 5 years, 5 months ago

While argument 2 produces the correct numerical answer, the logic presented is incorrect.

See @SAMARTH M.O. Comment.

Staff - 5 years, 5 months ago

Argument 1 is wrong because it assumes they all happen with equal probability, which they don't.

- 5 years, 5 months ago

Argument 3 is arguably the right one. Arguments 1 and 2 assume that their mentioned respective events are equally likely.

- 5 years, 5 months ago

Was this problem perhaps my problem All That Glitters Is Gold?

- 5 years, 5 months ago

No. I like your problem, which highlights a common mistake made by those who are first introduced to probability.

The problem that I was referencing had a similar title to my note. It has since been deleted.

Staff - 5 years, 5 months ago

Dude Calvin, I just found this. You didn't even tag me in it! But thanks, it's still awesome! :D

- 5 years, 5 months ago

The last argument is correct. In general for n numbers the probability is $\frac {{2}^{n-1}}{{2}^n} = \frac {1}{2}$

- 5 years, 5 months ago

argument 3 feels more logical

- 5 years, 5 months ago

.5

- 5 years, 5 months ago

Argument 3 is perfect.

- 5 years, 5 months ago

1\2

- 5 years, 5 months ago

2

- 5 years, 5 months ago

Obviously universal space= 2^4 Sample space= 4C0+ 4C2+ 4C4= 8 So probability= 8/16 =1/2

- 5 years, 5 months ago

let a(n) = P(Bin(n,0.5) is even), then a(n+1) = 0.5a(n) + (1-0.5)(1-a(n)) = 0.5

- 5 years, 5 months ago

We have 2^{4} case. Odd sum is 2.\frac{4!}{3!}=2^{3} case. There for sum is odd with probability \frac{1}{2}

- 5 years, 4 months ago

An even easier argument which is easily adapter to be generalisation for the number of: picking the first 3 numbers randomly, you'll find either an even or an uneven number. The last term will change it to either even or uneven, with an equal chance of either.

This will generalise to any number of variables easily, and also adapt to the odds of the result being k (mod n) if your variables can have values 1, 2, ... , n.

One problem it can't help with is: if the variables are binary, what are the odds of the result being a multiple of k (for a k greater than 2).

- 5 years, 4 months ago

1

- 5 years, 4 months ago

The first arguement is wrong as each case doesnot have equal probability .. The second solution (though gives right answer ) is an insufficient one. ARGUEMENT 3is descent

- 4 years, 7 months ago

The second solution doesn't give a correct argument. The point isn't to "get to the correct answer". The point is to "substantiate your argument".

Staff - 4 years, 7 months ago

Total number of cases = 222*2 Even cases are If number of ONEs is 2 or 4, So Favourable number of cases : 4P2 + 4P4 = 6 + 1 = 7 So Probability for odd is (16 - 7)/16 = 9/16

- 4 years, 6 months ago

The answer is 1/2.

When dealing with even cases, you forgot that 0 is an even number.

Staff - 4 years, 6 months ago

Argument 1 is incorrect because each case is not equally likely. Argument 2 gives the correct answer but for the wrong reason. Consider the probability of drawing the Ace of spades from a deck of cards. The card drawn is either the ace or it is not. One out of two cases is favourable hence 1/2. (?) Argument 3 would be correct if it was asserted that each case was equally probable. Which of course it is. J

- 4 years, 6 months ago

Great :)

Staff - 4 years, 6 months ago

The issue with the first argument is that of frequency distribution

- 5 years, 5 months ago

now since every single variable counts and is equally important we should be going with the the third one .. ARGUMENT 3 is correct .... nearly similar way to solve is .. there are two possibilities:-

1.) - three variables are 1

2.)- only one variable is 1 .. both ways we get odd

hence- 4C3[ any three numbers] (1/2)^4[the probability of getting 1 is 1/2 and zero is also the same AND we need to get 1 in three cases AS WELL AS 0 in the other case] => 4C3 * (1/2)^4 +4C1[only one variable](1/2)^4 AND HENCE

4C3 * (1/2)^4 + 4C1 * (1/2)^4 = 1/2 ..

- 5 years, 5 months ago

1

- 5 years, 5 months ago

10

- 5 years, 5 months ago


What is the remainder of $(14^{2010}+1) \div 6$?

What is the remainder of $(14^{2010}+1) \div 6$?

Someone showed me a way to do this by finding a pattern, i.e.:

$14^1\div6$ has remainder 2
$14^2\div6$ has remainder 4
$14^3\div6$ has remainder 2
$14^4\div6$ has remainder 4

And it seems that when the power is odd, the answer is 2, and when it's even, the answer is 4.

2010 is even, so the remainder is 4, but we have that +1, so the final remainder is 5. Which is correct.

But this method doesn't seem very concrete to me, and I have a feeling the pattern may not be easy to find (or exist?) for every question. What theorem or algorithm can I use to solve this instead?

-

$14\equiv 2\pmod 6$, so $14^{2010}+1\equiv 2^{2010}+1\pmod 6$. Now $2\cdot 2^2=2^3=8\equiv 2\pmod 6$, so $2\cdot (2^2)^k\equiv 2\pmod 6$ for any non-negative integer $k$. This shows that the pattern that you observed is real: $2^{2k+1} \equiv 2\pmod 6$ for any non-negative integer $k$. In particular, $$2^{2010}+1\equiv 2\cdot 2^{2009}+1 \equiv 2\cdot 2+1 \equiv 5\pmod 6\;.$$

The same basic idea can be used in similar problems, though the cycle of the pattern may not be nearly so short. To go much deeper than this kind of analysis, you want to look into the Chinese Remainder Theorem.

-

$14=0\pmod{2}$ and $14=2\pmod{3}$. Thus, because $2^2=1\pmod{3}$, for any $k>0$, we have $$14^k=0\pmod{2}$$ and $$14^k=\left\{\begin{array}{}1\pmod{3}&\text{when }k=0\pmod{2}\\2\pmod{3}&\text{when }k=1\pmod{2}\end{array}\right.$$ Therefore, for any $k>0$, we have by the Chinese Remainder Theorem, $$14^k=\left\{\begin{array}{}4\pmod{6}&\text{when }k=0\pmod{2}\\2\pmod{6}&\text{when }k=1\pmod{2}\end{array}\right.$$ So, as you surmised, $14^{2010}+1=5\pmod{6}$.

-

To solve this problem, in general, I'd use use two facts:

1. If $a$ and $n$ are relatively prime, $a^{\phi(n)} \bmod n = 1$. Here $\phi$ is the Euler phi function; when $n$ is prime, $\phi(n) = n-1$. This lets you reduce exponents to ones with power less than $\phi(n)$; for instance $$2^{2010} \bmod 5 = (2^4)^{502} 2^2 \bmod 5 = 1^{502} 4 \bmod 5 = 4.$$ (Here $2^2$ happened to be trivial to compute by hand; if the exponent were larger I would calculate it using the technique of repeated squaring.)

2. Unfortunately 2 and 6 are not relatively prime; in this case I would factor $6$ into primes and use the Chinese remainder theorem: $$2^{2010} \bmod 3 = (2^2)^{1005} \bmod 3 = 1$$ $$2^{2010} \bmod 2 = 0$$ so applying the Chinese remainder theorem, $2^{2010}$ must be congruent to 4 mod 6.

-

We may write following equalities :

$14^{2010}=6\cdot k_1+4$

$14^{2010}+1=6\cdot k_2+r$

$6\cdot k_1+4+1=6\cdot k_2+r\Rightarrow 6k_1+5=6k_2+r$

The last equality is true only if $k_1=k_2$ and $r=5$

-
Please excuse my ignorance, but it looks to me like you assumed that $14^{2010} = 4 \mod 6$, and then use this to 'prove' that $14^{2010} + 1 = 5 \mod 6$. How do you justify the first equality given $k_1$ is an integer? –  tom Nov 6 '11 at 9:01
@tom,read carefully text of the question... –  pedja Nov 6 '11 at 10:03

HINT $\rm\quad (m,n) = 1,\ m\:|\:a,\ n\:|\:a+1\ \ \Rightarrow\ \ a^{2\:k}\ \equiv\ m\:(m^{-1}\ mod\ n)\ \pmod{mn}\$ by CRT.

Therefore for $\rm\:m,n,a\ =\ 2,3,14\:$ we infer $\rm\: 14^{\:2\:k} \equiv\: 2\ (2^{-1}\ mod\ 3)\equiv -2\pmod 6$

-


# What's the probability of “at least” and “exactly” one event occurring?

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

I was thinking that this is $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and }B)$.

Is this correct?

If it is, then how can I solve the following problem taken from DeGroot's 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?

In a more mathematical language, we are given $P(\text{morning})=.5$, $P(\text{afternoon})=.65$, $P(\text{morning or afternoon}) = .5 + .65 - P(\text{morning and afternoon}) = .85$, which implies that $P(\text{morning and afternoon}) = .3$, which should be the answer to the question.

Is my reasoning correct?

If it 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?

These problems and questions highlight the difference between "at least one of them" and "exactly one of them". Provided that "at least one of them" is equivalent to $P(A \text{ or } B)$, but how can I work out the probability of "exactly one of them"?

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.

• "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
• 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? – upabove 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 $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$.

• Pr(A and not B) + Pr(B and not A) is not the same as Pr(A) + Pr(B) - Pr(A and B) – Petr Peller Apr 4 '15 at 17:44
• Pr(A and not B) + Pr(B and not A) + Pr(A and B) is what you are looking for, but calculating Pr(A) + Pr(B) - Pr(A and B) seems to be much easier. – Petr Peller Apr 4 '15 at 18:06
• $\Pr(A \text{ and not } B)+\Pr(B \text{ and not } A)$ is the answer to "but how can I work out the probability of exactly one of them?" which is what I meant by "your second question" – Henry Apr 4 '15 at 18:08

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 $$P(A)+P(B)-2P(A\cap B)$$

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 )