Jtdylktuy

Let $X_1,...$ be indepedent rv with common distribution $F(x)$ and $N$ geometric random variable with parameter $p$. $N$ is indepedent from all the $X_i$ . let $M = max(X_1,X_2,...,X_N)$. Find $P(M leq x mid N > 1 ) $

solution sketch

We have

$$ P(M leq x mid N > 1 ) = frac{P( X_1 leq x, X_2 leq x, ..., X_N leq x, N > 1 ) }{P(N > 1) }$$

$$ = frac{P(X_1 leq x )P( X_2 leq x, X_3 leq x, ..., X_N leq x, N > 1 ) }{P(N > 1) } $$
$$ = P(X_1 leq x ) P( max( X_2,X_3,...,X_N leq x mid N > 1 ) $$

$$ = P(X_1 leq x ) P( max( X_1,X_2,...,X_{N-1} leq x mid N > 1 ) $$

$$ = F(x) P( M leq x ) $$

My question is: Why can we pull out $P(X_1 leq x)$?? I mean by this reasoning, we can also do the following

$$ P(M leq x mid N > 1 ) = frac{ P(X_1 leq x) P(X_2 leq x) ... P(X_N leq x ) P(N > 1) }{P(N>1) } = F^n(x) $$

Also, in the equality before last one, why is it that

$$ P( max(X_2,...,X_N) leq x mid N > 1) = P( max(X_1,...,X_{N-1} ) leq x mid N > 1 )$$ Why can we do this ?

This can be done directly using the law of total probability. Note that

$$ begin{align} P(Mleq x) & = P(Mleq x|N = 1)P(N=1) + P(Mleq x | N > 1)P(N>1) \ & = F(x)p + (1-p)P(Mleq x| N > 1)end{align}$$

We can calculate $P(Mleq x)$ as follows using a geometric sum:

$$ begin{align} P(Mleq x) & = sum_{ngeq 1} P(Mleq x|N=n)P(N=n)\ &= sum_{ngeq 1} F(x)^n (1-p)^{n-1} p \ & = frac{F(x)p}{1 - F(x)(1-p)} end{align} $$

where $P(Mleq x | N=n) = F(x)^n$ since the $X_i$ are independent. Some quick rearranging gives $$ P(Mleq x|N>1) = frac{F(x)^2p}{1-F(x)(1-p)} = F(x)P(Mleq x)$$

For your method : we can pull out the $P(X_1 leq x)$ since $X_1$ is independent from $X_i$ for $i > 1$ and from $N$. We can't pull out all of the $X_i$ because how many there are depends on $N$. In the equality before last, we are just relabelling $X_i$ to $X_{i-1}$ which we are allowed to do as the $X_i$ are i.i.d. However, I'm not a hundred percent sure how to justify the last equality.