Given two iid geometric random variables with $p=\frac{1}{5}$, what is the probability that variates from them equate?

benatudq
2022-10-16
Answered

Given two iid geometric random variables with $p=\frac{1}{5}$, what is the probability that variates from them equate?

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Plutbantonavv

Answered 2022-10-17
Author has **15** answers

Step 1

No need for the joint mass function. If the variables count the number of failures before the first success:

- The probability that both variates are 0 is ${\left(\frac{1}{5}\right)}^{2}$

- The probability that both variates are 1 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2}$

- The probability that both variates are 2 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{4}$, etc.

Step 2

Thus the probability that both variates are equal is

$\sum _{n=0}^{\mathrm{\infty}}{\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2n}$

$=\frac{1}{25}\sum _{n=0}^{\mathrm{\infty}}{\left(\frac{16}{25}\right)}^{n}$

$=\frac{1}{25}\cdot \frac{1}{1-\frac{16}{25}}=\frac{1}{9}$

No need for the joint mass function. If the variables count the number of failures before the first success:

- The probability that both variates are 0 is ${\left(\frac{1}{5}\right)}^{2}$

- The probability that both variates are 1 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2}$

- The probability that both variates are 2 is ${\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{4}$, etc.

Step 2

Thus the probability that both variates are equal is

$\sum _{n=0}^{\mathrm{\infty}}{\left(\frac{1}{5}\right)}^{2}{\left(\frac{4}{5}\right)}^{2n}$

$=\frac{1}{25}\sum _{n=0}^{\mathrm{\infty}}{\left(\frac{16}{25}\right)}^{n}$

$=\frac{1}{25}\cdot \frac{1}{1-\frac{16}{25}}=\frac{1}{9}$

Vincent Norman

Answered 2022-10-18
Author has **1** answers

Step 1

Think of it this way: You want to find

$P(X=Y)=P(X=Y=1\text{or}X=Y=2\text{or}X=Y=3\text{or}...)$

These are clearly disjoint cases, so we break apart into a sum of probabilities while also realizing that each such statement $X=Y=n$ is equivalent to saying $X=n$ AND $Y=n$. Thus we have $P(X=Y)=P(X=1,Y=1)+P(X=2,Y=2)+P(X=3,Y=3)+...$

Step 2

Now, we use the fact that the two variates were considered IID. As such, each term $P(X=n,Y=n)={P}^{2}(X=n)$ and we ultimately end up with the infinite sum $P(X=Y)={P}^{2}(X=1)+{P}^{2}(X=2)+{P}^{2}(X=3)+...$

Think of it this way: You want to find

$P(X=Y)=P(X=Y=1\text{or}X=Y=2\text{or}X=Y=3\text{or}...)$

These are clearly disjoint cases, so we break apart into a sum of probabilities while also realizing that each such statement $X=Y=n$ is equivalent to saying $X=n$ AND $Y=n$. Thus we have $P(X=Y)=P(X=1,Y=1)+P(X=2,Y=2)+P(X=3,Y=3)+...$

Step 2

Now, we use the fact that the two variates were considered IID. As such, each term $P(X=n,Y=n)={P}^{2}(X=n)$ and we ultimately end up with the infinite sum $P(X=Y)={P}^{2}(X=1)+{P}^{2}(X=2)+{P}^{2}(X=3)+...$

asked 2022-09-27

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$|AB|=8,|CD|=6$. We choose the point P by chance on AB and again a point Q by chance on CD find the probability so the the triangle whose height is AP, and base is CQ is larger than 12. The answer is in the picture in it's condesed form, can anyone provide an explanation?

$y=\frac{24}{x},p={\int}_{4}^{8}\frac{(6-\frac{24}{x})dx}{48}$

$|AB|=8,|CD|=6$. We choose the point P by chance on AB and again a point Q by chance on CD find the probability so the the triangle whose height is AP, and base is CQ is larger than 12. The answer is in the picture in it's condesed form, can anyone provide an explanation?

$y=\frac{24}{x},p={\int}_{4}^{8}\frac{(6-\frac{24}{x})dx}{48}$

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Clearly when $n=1$ the expected distance is 1. When $n=2$ we can find the expected distance by integrating

$\frac{1}{2\pi}{\int}_{0}^{2\pi}2\mathrm{sin}\left(\frac{x}{2}\right)dx=\frac{4}{\pi}$

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Can one find a general formula for any n?

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$\frac{1}{2\pi}{\int}_{0}^{2\pi}2\mathrm{sin}\left(\frac{x}{2}\right)dx=\frac{4}{\pi}$

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Can one find a general formula for any n?

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However, the next question asks what the probability of choosing a nonnegative real number is, and I do understand it is also 1/2

After finishing my homework, I'm wondering: Is it possible to choose 0, and what would the probability of choosing 0 be?

Find the probability that the number is positive, and using geometric probability, I get 1/2. Simple, right?

However, the next question asks what the probability of choosing a nonnegative real number is, and I do understand it is also 1/2

After finishing my homework, I'm wondering: Is it possible to choose 0, and what would the probability of choosing 0 be?

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1. What is the probability that a drought lasts exactly 3 intervals? (0.0844)

2. What is the probability that a drought lasts at most 3 intervals? (0.878)

Drought length is referred to as the number of consecutive time intervals in which the water supply remains below a critical value. Consider the drought length as a random variable, denoted as Y, which is assumed to have a geometric distribution with $p=0.409$

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2. What is the probability that a drought lasts at most 3 intervals? (0.878)

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