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here's a question I got for homework (sorry if my translation is a bit unclear):

Let X‫~‬G(P1), Y~‬(P2), X and Y are independent. Prove that the minimum is also geometric, meaning: min(X,Y)~G(1-(1-P1)(1-p2)).

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

I have no idea where to start, even with the great clue that they've supplied. Any hints?

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Well, just follow the clue. Calculate P(min(X,Y) > k), for a general k. Did you try that? what did you get? –  Prateek Dec 12 '11 at 13:05
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note that $P(\min(X,Y)>k)$ is the same as $P(X>k,Y>k)$. –  deinst Dec 12 '11 at 13:17
    
Ok I did that and it was quite easy. P[min(X,Y)>k] = ((1-p1)^K)((1-p2)^K) - is that enough to prove what I need to prove? –  yotamoo Dec 12 '11 at 15:29
    
@yotamoo Is this the same as the probability that a $G(1-(1-p_1)(1-p_2))$ geometric variable is greater than $k$? –  deinst Dec 12 '11 at 15:42

1 Answer 1

up vote 1 down vote accepted

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 $1-(1-p1)(1-p2)$.

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 $\left[(1-p1)(1-p2)\right]^k$. 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 $1-\left[(1-p1)(1-p2)\right]^k$, and we can note that this is the cdf of a geometric random variable with probability parameter $1-(1-p_1)(1-p_2)$.

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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As an addendum and a slightly different way of calculating the results, the probability of a "success" on a compound trial is that either of the two separate sub-trials results in a success. This has probability $p_1 + p_2 - p_1p_2$ (using $$P(A\cup B) = P(A) + P(B) - P(A\cap B) = P(A) + P(B) - P(A)P(B)$$ by independence of individual sub-trials) and so we have a geometric random variable with parameter $$p_1 + p_2 - p_1p_2 = 1 - (1-p_1)(1-p_2)$$ just as you found. –  Dilip Sarwate Dec 13 '11 at 16:46
    
A minor correction to my comment: in the first line, for "the probability of", please read "the event" –  Dilip Sarwate Dec 13 '11 at 18:56

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