Let $X_1,X_2,...$ be independent random variables with common distribution function $F$ and suppose that they are independent of $N$, a geometric random variable with parameter $p$ (the probability mass function of $N$ is is $P(N = n) = (1-p)^{n-1}p$, for $n = 1,2,...$). Let $M = min(X_1,...,X_N)$.
(a) Find $P(M \geq x | N = 1)$.
(b) Find $P(M \geq x | N > 1)$.
Here are my thoughts so far:
For part (a), I tried to attack it using the basic definition of conditional probability. That is, $P(M \geq x | N = 1) = \frac{P(M \geq x, N = 1)}{P(N = 1)} = \frac{P(X_1 \geq x)}{P(N = 1)} = \frac{1-P(X_1 < x)}{P(N=1)} = \frac{1-F(x)}{p}$.
Is this the correct approach here ? If so, can my answer be more simplified ?
I'm not sure how to approach part (b). Am I supposed to find a conditional probability mass function here ? Or, a conditional moment generating function ? Or, can I get away with the approach in part (a) ? What makes it difficult is that I do not know the distribution of $X_i$ -- which makes me weary of creating a moment generating function or conditional probability mass function here for part (b).
Thanks for all your help. (=
$P(M \geq x|N>1)=\frac {P(M \geq x, N>1)} {P(N>1}=\frac {P(M \geq x)-P(M \geq x, N=1)} {1-P(N=1)}$. Use the fat that $P(M \geq x)=\sum_n P(M \geq x ,N=n) =\sum _n P(X_1 \geq x)^{n} (1-p)^{n-1}p$.