1. ## Conditional probability

Let X1,..be independent random variables with the common distribution function F, and suppose they are independent of N, a geometric random variable with parameter p. Let M=max(X1,...Xn).

Find $P(M\leq x)$ by conditioning on N

At the first glance I didn't quite understand the information given in this problem. But after a while I wrote down what I believe is asked to solve in the problem

$P(M\leq x)=\sum_{k=0}^{\infty} P(M\leq x|N=k)P(k)$

But this is as far as I can go, I know P(k) = p(1-p)^(k-1), and I'm clueless on what's $P(M\leq x|N=k)$
I want to say that $M\leq x$ is independ of $N=k$, but I'm not sure whether it's true.

Helps would be appreciated, thanks!

2. Originally Posted by ChaosticMoon
Let X1,..be independent random variables with the common distribution function F, and suppose they are independent of N, a geometric random variable with parameter p. Let M=max(X1,...Xn).
It seems like it should be $M=\max(X_1,\ldots,X_N)$ where $N$ is the given random variable.

Find $P(M\leq x)$ by conditioning on N

At the first glance I didn't quite understand the information given in this problem. But after a while I wrote down what I believe is asked to solve in the problem

$P(M\leq x)=\sum_{k=0}^{\infty} P(M\leq x|N=k)P(k)$

But this is as far as I can go, I know P(k) = p(1-p)^(k-1), and I'm clueless on what's $P(M\leq x|N=k)$
I want to say that $M\leq x$ is independ of $N=k$, but I'm not sure whether it's true.
Since $X_1,X_2,\ldots$ are independent of $N$, conditioning by $N=k$ just gives $P(\max(X_1,\ldots,X_N)\leq x|N=k)=P(\max(X_1,\ldots,X_k)\leq x)$ (the conditioning doesn't affect the $X_i$'s by independence; it just fixes the value of $N$).