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?
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 $1-(1-p_1)(1-p_2)$.
