Expectation of supremum of normal distributed random variables I'm working on a exercise and at the moment I'm stuck on this part:

Let $m<0$ and $(X_n)$ a sequence of independent normal distributed random variables with parameters $m$ and $\sigma²$. Define $S_n=X_1+\dots+X_n,  F_n=\sigma(S_0,\dots,S_n) \text{ and } W=sup_{n\geq 0} S_n$. Show that $$\mathbb{E}[e^{\lambda W}]=1 + \lambda \int^\infty_0e^{\lambda t}P(W>t)dt.$$ and conclude that  $\mathbb{E}[e^{\lambda W}]<\infty$ for every $\lambda<\lambda_0$.

The two things I have shown so far are that


*

*there exists a $\lambda_0 >0$ such that $(e^{\lambda_0 S_n})$ is a martingale, specifically $\lambda_0=-\frac{2m}{\sigma²}$

*$P(e^{\lambda_0 W} > a)\leq \frac{1}{a}$ for every $a>1$ so that $P(W>t)\leq e^{-\lambda_0t}$ for $t>0$.


But now I am not sure how to continue with this information to find the right approach for solving this last step described at the beginning. Any help for is appreciated, thanks in advance!
 A: *

*Writing $$e^{\lambda W(\omega)} -1 =\lambda \int_0^{W(\omega)} e^{\lambda t} \, dt = \lambda \int_{(0,\infty)} e^{\lambda t} \underbrace{1_{(0,W(\omega))}(t)}_{=1_{(0,t)}(W(\omega))} \, dt$$ we find from Fubini's theorem that $$\begin{align*} \mathbb{E}(e^{\lambda W}) -1 &= \lambda \int_{\Omega} \left( \int_{(0,\infty)} e^{\lambda t} 1_{(0,t)}(W(\omega)) \, dt \right)\, d\mathbb{P}(\omega)  \\ &= \lambda \int_{(0,\infty)} e^{\lambda t} \left( \int_{\Omega} 1_{(0,t)}(W(\omega)) \, d\mathbb{P}(\omega) \right) \,d t \\&= \lambda \int_0^{\infty} e^{\lambda t} \mathbb{P}(W>t) \, dt. \tag{1} \end{align*}$$

*For $\lambda_0 := -2m/\sigma^2>0$ the process $M_n := e^{\lambda_0 S_n}$ is a non-negative martingale. Applying the maximal inequality we find that $$\mathbb{P} \left( \sup_{k \leq n} M_k \geq r \right) \leq \frac{1}{r} \underbrace{\mathbb{E}(M_n)}_{=\mathbb{E}(M_0)=1}$$ for any $r>0$, and so $$\mathbb{P} \left( \sup_{k \geq 0} M_k \geq r \right) \leq \frac{1}{r}. \tag{2} $$

*By the monotonicity of $x \mapsto e^{\lambda_0 x}$ we have $$\mathbb{P}(W>t) = \mathbb{P} \left( \sup_{k \geq 0} M_k > e^{\lambda_0 t} \right) \stackrel{\text{(2)}}{\leq} e^{-\lambda_0 t}.$$ Plug this into $(1)$ to conclude that $\mathbb{E}e^{\lambda W}< \infty$ for any $\lambda<\lambda_0$.

