Prove that if $X\sim $ Geometric (p) then, $E(X)=\frac{q}{p}\phantom{\rule{1em}{0ex}}\mathrm{Var}(X)=\frac{q}{{p}^{2}}\phantom{\rule{1em}{0ex}}{m}_{X}(t)=p(1-q{e}^{t})$

Jensen Mclean
2022-09-03
Answered

Prove that if $X\sim $ Geometric (p) then, $E(X)=\frac{q}{p}\phantom{\rule{1em}{0ex}}\mathrm{Var}(X)=\frac{q}{{p}^{2}}\phantom{\rule{1em}{0ex}}{m}_{X}(t)=p(1-q{e}^{t})$

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Belen Solomon

Answered 2022-09-04
Author has **5** answers

Step 1

For the variance, it suffices to compute $E[{X}^{2}]$. You can adapt the approach you used for computing E[X]; it may help to manipulate the series for $\frac{1}{(1-x{)}^{3}}$.

Here is a direct derivation:

$\begin{array}{rl}E[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}\\ (1-q)E[{X}^{2}]=E[{X}^{2}]-qE[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{m\ge 0}{m}^{2}p{q}^{m+1}\\ & =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{k\ge 1}(k-1{)}^{2}p{q}^{k}\\ & =\sum _{k\ge 1}(2k-1)p{q}^{k}\\ \\ & =2\sum _{k\ge 1}kp{q}^{k}-\sum _{k\ge 1}p{q}^{k}\\ & =2E[X]-(1-p)=2\frac{q}{p}-q=\frac{q(2-p)}{p}.\end{array}$

Given the above expression for $(1-q)E[{X}^{2}]$, you can then do a few more steps compute the variance of X.

Step 2

For the MGF, ${m}_{X}(t)=E[{e}^{tX}]=\sum _{k\ge 0}{e}^{tk}p{q}^{k}=p\sum _{k\ge 0}({e}^{t}q{)}^{k}$ which is a geometric series that you should already know how to compute. By the way, there is a typo in your expression for the MGF.

For the variance, it suffices to compute $E[{X}^{2}]$. You can adapt the approach you used for computing E[X]; it may help to manipulate the series for $\frac{1}{(1-x{)}^{3}}$.

Here is a direct derivation:

$\begin{array}{rl}E[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}\\ (1-q)E[{X}^{2}]=E[{X}^{2}]-qE[{X}^{2}]& =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{m\ge 0}{m}^{2}p{q}^{m+1}\\ & =\sum _{k\ge 0}{k}^{2}p{q}^{k}-\sum _{k\ge 1}(k-1{)}^{2}p{q}^{k}\\ & =\sum _{k\ge 1}(2k-1)p{q}^{k}\\ \\ & =2\sum _{k\ge 1}kp{q}^{k}-\sum _{k\ge 1}p{q}^{k}\\ & =2E[X]-(1-p)=2\frac{q}{p}-q=\frac{q(2-p)}{p}.\end{array}$

Given the above expression for $(1-q)E[{X}^{2}]$, you can then do a few more steps compute the variance of X.

Step 2

For the MGF, ${m}_{X}(t)=E[{e}^{tX}]=\sum _{k\ge 0}{e}^{tk}p{q}^{k}=p\sum _{k\ge 0}({e}^{t}q{)}^{k}$ which is a geometric series that you should already know how to compute. By the way, there is a typo in your expression for the MGF.

Austin Rangel

Answered 2022-09-05
Author has **2** answers

Step 1

I would do the MGF calculation first, then use the result to compute the expectation and variance. That's probably not the intent of the problem, but it is mathematically valid.

${M}_{X}(t)=\mathrm{E}[{e}^{tX}]=\sum _{x=0}^{\mathrm{\infty}}{e}^{tx}(1-p{)}^{x}p.$

Step 2

Then, $\mathrm{E}[X]={M}_{X}^{\prime}(0),\phantom{\rule{1em}{0ex}}\mathrm{E}[{X}^{2}]={M}_{X}^{\u2033}(0),$, and $\mathrm{Var}[X]=\mathrm{E}[{X}^{2}]-\mathrm{E}[X{]}^{2}.$

I would do the MGF calculation first, then use the result to compute the expectation and variance. That's probably not the intent of the problem, but it is mathematically valid.

${M}_{X}(t)=\mathrm{E}[{e}^{tX}]=\sum _{x=0}^{\mathrm{\infty}}{e}^{tx}(1-p{)}^{x}p.$

Step 2

Then, $\mathrm{E}[X]={M}_{X}^{\prime}(0),\phantom{\rule{1em}{0ex}}\mathrm{E}[{X}^{2}]={M}_{X}^{\u2033}(0),$, and $\mathrm{Var}[X]=\mathrm{E}[{X}^{2}]-\mathrm{E}[X{]}^{2}.$

asked 2022-07-19

Geometric Sequence with Normal Distribution Problem

The running time (in seconds) of an algorithm on a data set is approximately normally distributed with mean 3 and variance 0.25.

a. What is the probability that the running time of a run selected at random will exceed 2.6 seconds?

Answer for (a): I've computed this and found it to be $p=\mathrm{0.7881.}$.

b. What is the probability that the running time of exactly one of four randomly selected runs will exceed 2.6 seconds?

Answer for (b): Not sure, I know it's a geometric sequence and I believe the formula that I need to use to be $p(1-p{)}^{3}$, thus giving me $0.7881(1-0.7881{)}^{3}$ however this was marked incorrect. I'm trying to figure out why, can someone explain my error and how to arrive at the correct solution?

The running time (in seconds) of an algorithm on a data set is approximately normally distributed with mean 3 and variance 0.25.

a. What is the probability that the running time of a run selected at random will exceed 2.6 seconds?

Answer for (a): I've computed this and found it to be $p=\mathrm{0.7881.}$.

b. What is the probability that the running time of exactly one of four randomly selected runs will exceed 2.6 seconds?

Answer for (b): Not sure, I know it's a geometric sequence and I believe the formula that I need to use to be $p(1-p{)}^{3}$, thus giving me $0.7881(1-0.7881{)}^{3}$ however this was marked incorrect. I'm trying to figure out why, can someone explain my error and how to arrive at the correct solution?

asked 2022-09-03

How can i solve this question with Geometric distribution or random variables?

I tried have tried using a 'Geometric distribution', but it hasn't worked.

John and Ron play basketball 10 times. The probability that John wins in single round $=0.4$. The probability that Ron wins in single round $=0.3$. The probability that there is equality between them in a single round $=0.3$. The "Winner" is defined to be the first to win a single round.

The rotations are different and independent.

What is the probability that John is the Winner?

I tried have tried using a 'Geometric distribution', but it hasn't worked.

John and Ron play basketball 10 times. The probability that John wins in single round $=0.4$. The probability that Ron wins in single round $=0.3$. The probability that there is equality between them in a single round $=0.3$. The "Winner" is defined to be the first to win a single round.

The rotations are different and independent.

What is the probability that John is the Winner?

asked 2022-10-23

Find probability using geometric distribution

Two boys play basketball in the following way. They take turns shooting and stop when a basket is made. Player A goes first and has probability ${p}_{1}$ of making a basket on any throw. Player B, who shoots second, has probability ${p}_{2}$ of making a basket. The outcomes of the successive trials are assumed to be independent.

a. Find the frequency function for the total number of attempts.

b. What is the probability that player A wins?

My solution: a. suppose that they make k attempts. Then the answer depends on parity of k. If $k=2n+1$ it means that first player made $n+1$ attempts and n first attempts were unsuccessful, while the last one was successful (he made a basket). As for second player, he made n unsuccessful attempts. Bot these probabilities are easily found by multiplication of respective probabilities for the first and second player.

The case when k is even is solved in a similar way.

b. According to part a., we know how to find the probability that the game is over at $2n+1$ attempt. Therefore we sum these probabilities over $n\in \mathbb{N}$.

Is it correct?

Two boys play basketball in the following way. They take turns shooting and stop when a basket is made. Player A goes first and has probability ${p}_{1}$ of making a basket on any throw. Player B, who shoots second, has probability ${p}_{2}$ of making a basket. The outcomes of the successive trials are assumed to be independent.

a. Find the frequency function for the total number of attempts.

b. What is the probability that player A wins?

My solution: a. suppose that they make k attempts. Then the answer depends on parity of k. If $k=2n+1$ it means that first player made $n+1$ attempts and n first attempts were unsuccessful, while the last one was successful (he made a basket). As for second player, he made n unsuccessful attempts. Bot these probabilities are easily found by multiplication of respective probabilities for the first and second player.

The case when k is even is solved in a similar way.

b. According to part a., we know how to find the probability that the game is over at $2n+1$ attempt. Therefore we sum these probabilities over $n\in \mathbb{N}$.

Is it correct?

asked 2022-07-17

What is the expected volume of the simplex formed by $n+1$ points independently uniformly distributed on ${\mathbb{S}}^{n-1}$?

asked 2022-11-08

An upper bound on the expected value of the square of random variable dominated by a geometric random variable

Let X and Y be two random variables such that:

1. $0\le X\le Y$

2. Y is a geometric random variable with the success probability p (the expected value of Y is 1/p).

I would be grateful for any help of how one could upperbound $\mathbb{E}({X}^{2})$ in terms of p.

Let X and Y be two random variables such that:

1. $0\le X\le Y$

2. Y is a geometric random variable with the success probability p (the expected value of Y is 1/p).

I would be grateful for any help of how one could upperbound $\mathbb{E}({X}^{2})$ in terms of p.

asked 2022-09-01

Suppose a dart lands at random on the dartboard.

Find the probability that the dart will land in the purpleregion.

Four rings to dart board inner ring is green, then the next out ispurple, then the next is green, then the outer/last one ispurple.

The width of each ring = 2 in.

r = 2 in (inner most circle)

The answer is 62.5% WHY?

Find the probability that the dart will land in the purpleregion.

Four rings to dart board inner ring is green, then the next out ispurple, then the next is green, then the outer/last one ispurple.

The width of each ring = 2 in.

r = 2 in (inner most circle)

The answer is 62.5% WHY?

asked 2022-07-17

Let ${X}_{n}$ be a geometric random variable with parameter $p=\lambda /n$. Compute $P({X}_{n}/n>x)$

$x>0$ and show that as n approaches infinity this probability converges to $P(Y>x)$, where Y is an exponential random variable with parameter $\lambda $. This shows that ${X}_{n}/n$ is approximately an exponential random variable.

$x>0$ and show that as n approaches infinity this probability converges to $P(Y>x)$, where Y is an exponential random variable with parameter $\lambda $. This shows that ${X}_{n}/n$ is approximately an exponential random variable.