Show that $P(\max_{i=N(s)+1,..., N(t)} (X_i)\leq x)=e^{-(t-s)e^{-x}}$

Question

Let $N$ be a standard homogenous Poisson process, independent of the iid standard exponential claim sizes $\left(X_i\right)$ Define:

$$M_{s,t}=\displaystyle\max_{i=N(s)+1,...,N(t)} \left(X_i\right), \text{ } 0\leq s<t$$

(Keep in mind that if $N(s,t]=0$ we write $M_{s,t}=0$)

Show that $P(M_{s,t}\leq x)=P\left(\displaystyle\max_{i=N(s)+1,...,N(t)} (X_i) \leq x\right)=e^{-(t-s)e^{-x}}$

Attempt:

$$P(M_{s,t}\leq x)=P(\displaystyle\max_{i=N(s)+1,...,N(t)}(X_i)\leq x)=P(X_{N(s)+1}\leq x,X_{N(s)+2}\leq x, ... , X_{N(t)}\leq x)\stackrel{iid}{=}P(X_{N(s)+1}\leq x)\cdots P(X_{N(t)}\leq x)$$ But this doesnt help me at all...

Any hints?

Notation: $$N(s,t]=N(t)-N(s)$$

Let $N$ be a standard homogenous Poisson process $\Rightarrow$ the mean value function $\mu(t)=t$

Answer 1

\begin{align} \mathbb P\left[M_{s,t}\leq x\right] &=\mathbb P\left[\displaystyle\max_{i=N(s)+1,\ldots,N(t)} X_i \leq x\right]\\ &= \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \mathbb P\left[\left\{\displaystyle\max_{i=N(s)+1,\ldots,N(t)} X_i \leq x\right\} \cap \left\{N(s) = n\right\}\cap \left\{N(t) - N(s) = m\right\}\right]\\ &= \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \mathbb P\left[\left.\left\{\displaystyle\max_{i=N(s)+1,\ldots,N(t)} X_i \leq x\right\}\right| \left\{N(s) = n\right\}\cap \left\{N(t) - N(s) = m\right\} \right]\mathbb P\left[\left\{N(s) = n\right\}\cap \left\{N(t) - N(s) = m\right\}\right]\\ &= \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \mathbb P\left[\left.\left\{\displaystyle\max_{i=n+1,\ldots,n+m} X_i \leq x\right\}\right. \right]\mathbb P\left[\left\{N(s) = n\right\}\right]\mathbb P\left[ \left\{N(t) - N(s) = m\right\}\right]\\ &= \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \mathbb P\left[\left.\left\{\displaystyle\max_{i=n+1,\ldots,n+m} X_i \leq x\right\}\right. \right]\mathbb P\left[\left\{N(s) = n\right\}\right]\mathbb P\left[ \left\{N(t) - N(s) = m\right\}\right]\\ &= \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \left(\prod_{i=n+1}^{n+m} \mathbb P\left[\left.\left\{X_i \leq x\right\}\right. \right]\right)\frac{s^n}{n!} e^{-s} \frac{(t-s)^m}{m!}e^{-(t-s)}\\ &= \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \left(1 - e^{-x}\right)^m\frac{s^n}{n!} e^{-s} \frac{(t-s)^m}{m!}e^{-(t-s)}\\ &= \sum_{m=0}^{\infty} \left(1 - e^{-x}\right)^m\frac{(t-s)^m}{m!}e^{-(t-s)}\\ &=e^{\left(1 - e^{-x}\right)(t-s) - (t-s)} = e^{-e^{-x}(t-s)} \end{align}