Suppose $X_i$ $i=1,2,3..$ are indepednent random variables with common distribution F(x). Let $N$ be a geometric with parameter $\alpha$ and suppose $X_i$ is independent of $N$ for all $i$. Let $M = \max(X_1,…,X_N)$. Find $P(X \leq x ) $ by conditioning on $N$ and find it in different way (Hint: for second part compute $P(M \leq x \mid N = 1)$ and $P(M \leq x \mid N > 1 $). )
Attempt:
By conditioning, we have
$$ P(M \leq x) = \sum_{i=1}^{\infty} P(M \leq x \mid N=i ) P(N=i) $$
$$ = \sum \frac{ P(M \leq x \cap N = i ) \cdot P(X=i) }{P(X=i)} = \sum P( X_1 \leq x, X_2 \leq x,…,X_N \leq x, N=i)$$
since they are all indepdenent, we have
$$ \sum F(x)^N P(N=i) = F^N(x) \sum P(N=i) = F^N(x) $$
Now, for the second part, Im not quite sure what they mean. Notice by independence, we have
$$ P(M \leq x \mid N = 1) = P(M \leq x ) $$
$$ P(M \leq x \mid N > 1) = P(M \leq x )$$
am I misunderstanding the hint?
Best Answer
\begin{align} P(M \le x) &= \sum_{n=1}^\infty P(M \le x|N=n)P(N=n)\\ &=\sum_{n=1}^\infty P(\max(X_1, \ldots, X_n )\le x )(1-\alpha)^{n-1}\alpha\\ &= \frac{\alpha}{1-\alpha}\sum_{n=1}^\infty [F(x)(1-\alpha)]^n \\ &= \frac{\alpha}{1-\alpha}\frac{F(x)(1-\alpha)}{1-F(x)(1-\alpha)}\\ &= \frac{F(x)\alpha}{1-F(x)(1-\alpha)} \end{align}
Notice that $N$ follows a geometric distribution, it has memoryless property, that is
$$P(N-1 > n|N>1)=\frac{P(N>1+n)}{P(N>1)}=\frac{(1-\alpha)^{n+1}}{(1-\alpha)}=P(N > n)$$
Hence, \begin{align} P(M \le x) &= P(M \le x|N=1)P(N=1)+P(M \le x|N\ge1)P(N>1)\\ &= F(x)\alpha + (1-\alpha) P(M \le x|N > 1)\\ &= F(x) \alpha +(1-\alpha)P(X_1 \le x)P(\max(X_2, \ldots, X_N) \le x|N > 1)\\ &= F(x)\alpha + (1-\alpha)F(x)P(\max(X_1, \ldots, X_{N-1} )\le x|N >1)\\ &= F(x) \alpha + (1-\alpha) F(x) P(M \le x) \end{align}
Hence $$(1-(1-\alpha)F(x))P(M \le x) = F(x) \alpha$$
$$P( M \le x) = \frac{F(x) \alpha}{1-(1-\alpha)F(x)}$$