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

Let $X\sim‬G(p_1)$, $Y\sim ‬G(p_2)$, $X$ 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.

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

  • $\begingroup$ Well, just follow the clue. Calculate P(min(X,Y) > k), for a general k. Did you try that? what did you get? $\endgroup$
    – Prateek
    Dec 12, 2011 at 13:05
  • 2
    $\begingroup$ note that $P(\min(X,Y)>k)$ is the same as $P(X>k,Y>k)$. $\endgroup$
    – deinst
    Dec 12, 2011 at 13:17
  • $\begingroup$ 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? $\endgroup$
    – yotamoo
    Dec 12, 2011 at 15:29
  • $\begingroup$ @yotamoo Is this the same as the probability that a $G(1-(1-p_1)(1-p_2))$ geometric variable is greater than $k$? $\endgroup$
    – deinst
    Dec 12, 2011 at 15:42
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    $\begingroup$ I found several later posts of exactly the same question. Judging by the content, it's hard to decide which one should be deemed the original and the others duplicate. In chronological order: 845706, 1040620, 1056296, 1169142, and 1207241. $\endgroup$ Mar 21, 2018 at 1:10

1 Answer 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 $\left[(1-p_1)(1-p_2)\right]^k$. 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 $1-\left[(1-p_1)(1-p_2)\right]^k$, 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 $1-(1-p_1)(1-p_2)$.

  • $\begingroup$ 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. $\endgroup$ Dec 13, 2011 at 16:46
  • $\begingroup$ A minor correction to my comment: in the first line, for "the probability of", please read "the event" $\endgroup$ Dec 13, 2011 at 18:56

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