Haiden Meyer

2022-10-03

The minimum of two independent geometric random variables

Let $X\sim \u202cG({p}_{1})$, $Y\sim \u202cG({p}_{2})$, X and Y are independent. Prove that the minimum is also geometric, meaning: $min(X,Y)\sim G(1-(1-{p}_{1})(1-{p}_{2}))$.

Instructions: first calculate the probability $P(min(X,Y)>k)$ and compare it to the parallel probability in (of?) a geometric random variable.

Let $X\sim \u202cG({p}_{1})$, $Y\sim \u202cG({p}_{2})$, X and Y are independent. Prove that the minimum is also geometric, meaning: $min(X,Y)\sim G(1-(1-{p}_{1})(1-{p}_{2}))$.

Instructions: first calculate the probability $P(min(X,Y)>k)$ and compare it to the parallel probability in (of?) a geometric random variable.

antidootnw

Beginner2022-10-04Added 10 answers

Step 1

Let X and Y be independent random variables having geometric distributions with probability parameters ${p}_{1}$ and ${p}_{2}$ respectively. Then if Z is the random variable min(X,Y) then Z has a geometric distribution with probability parameter $1-(1-{p}_{1})(1-{p}_{2})$.

There are essentially two ways to see this:

First, the method outlined by the hint in your homework - Note that the cdf of X is $1-(1-{p}_{1}{)}^{k}$ and the cdf of Y is $1-(1-{p}_{2}{)}^{k}$, so the probability that $X>k$ is $(1-{p}_{1}{)}^{k}$ and the probability that $Y>k$ is $(1-{p}_{2}{)}^{k}$ and so the probability that both are greater than k is ${[(1-{p}_{1})(1-{p}_{2})]}^{k}$. But the probability that both are greater than k is the same as the probability that the minimum of the two is greater than k. From this we can get the cdf of Z as $1-{[(1-{p}_{1})(1-{p}_{2})]}^{k}$, and we can note that this is the cdf of a geometric random variable with probability parameter $1-(1-{p}_{1})(1-{p}_{2})$.

Step 2

Second, and more intuitively to me, we can go back to the definition of a geometric random variable with probability parameter p : the number of Bernoulli trials with probability p needed to get one success. So min(X,Y) is the number of trials of simultaneously running a Bernoulli experiment with probability ${p}_{1}$ and one with probability ${p}_{2}$ before one or the other experiments succeeds. The probability of one of the two experiments succeeding at any step is just $1-(1-{p}_{1})(1-{p}_{2})$, so Z is a geometric random variable with probability parameter $1-(1-{p}_{1})(1-{p}_{2})$

Let X and Y be independent random variables having geometric distributions with probability parameters ${p}_{1}$ and ${p}_{2}$ respectively. Then if Z is the random variable min(X,Y) then Z has a geometric distribution with probability parameter $1-(1-{p}_{1})(1-{p}_{2})$.

There are essentially two ways to see this:

First, the method outlined by the hint in your homework - Note that the cdf of X is $1-(1-{p}_{1}{)}^{k}$ and the cdf of Y is $1-(1-{p}_{2}{)}^{k}$, so the probability that $X>k$ is $(1-{p}_{1}{)}^{k}$ and the probability that $Y>k$ is $(1-{p}_{2}{)}^{k}$ and so the probability that both are greater than k is ${[(1-{p}_{1})(1-{p}_{2})]}^{k}$. But the probability that both are greater than k is the same as the probability that the minimum of the two is greater than k. From this we can get the cdf of Z as $1-{[(1-{p}_{1})(1-{p}_{2})]}^{k}$, and we can note that this is the cdf of a geometric random variable with probability parameter $1-(1-{p}_{1})(1-{p}_{2})$.

Step 2

Second, and more intuitively to me, we can go back to the definition of a geometric random variable with probability parameter p : the number of Bernoulli trials with probability p needed to get one success. So min(X,Y) is the number of trials of simultaneously running a Bernoulli experiment with probability ${p}_{1}$ and one with probability ${p}_{2}$ before one or the other experiments succeeds. The probability of one of the two experiments succeeding at any step is just $1-(1-{p}_{1})(1-{p}_{2})$, so Z is a geometric random variable with probability parameter $1-(1-{p}_{1})(1-{p}_{2})$

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