# Conditional probability

• Apr 29th 2010, 08:12 AM
ChaosticMoon
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!
• Apr 29th 2010, 11:43 AM
Laurent
Quote:

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.

Quote:

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 $N$).