Let X1, X2, . . . be i.i.d. random variables with the common CDF F , and suppose they are independent of N, a geometric random variable with parameter p. Let $M = max\{X_1, …, X_N \}$.
(a) Find Pr{M ≤ x} by conditioning on N.
(b) Find Pr{M ≤ x|N = 1}.
(c) Find Pr{M ≤ x|N > 1}.
Could anyone explain why in (c) $Pr\{M ≤ x|N > 1\}=F(x)Pr\{M\leq x\}$
Best Answer
If you have solved (a) and (b) then you can find the answer of (c) on base of the equality:$$P\left(M\leq x\right)=P\left(M\leq x\mid N=1\right)P\left(N=1\right)+P\left(M\leq x\mid N>1\right)P\left(N>1\right)$$
edit:
I found $$P\left(M\leq x\mid N>1\right)=\frac{pF\left(x\right)^{2}}{1-\left(1-p\right)F\left(x\right)}$$ and: $$P\left(M\leq x\right)=\frac{pF\left(x\right)}{1-\left(1-p\right)F\left(x\right)}$$
showing that indeed:$$P\left(M\leq x\mid N>1\right)=F\left(x\right)P\left(M\leq x\right)$$