Expectation of supremum of normal distributed random variables

Question

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!

Answer 1

1. 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 $$\begin{align*} \mathbb{E}(e^{\lambda W}) -1 &= \lambda \int_{\Omega} \left( \int_{(0,\infty)} e^{\lambda t} 1_{(0,t)}(W(\omega)) \, dt \right)\, d\mathbb{P}(\omega) \\ &= \lambda \int_{(0,\infty)} e^{\lambda t} \left( \int_{\Omega} 1_{(0,t)}(W(\omega)) \, d\mathbb{P}(\omega) \right) \,d t \\&= \lambda \int_0^{\infty} e^{\lambda t} \mathbb{P}(W>t) \, dt. \tag{1} \end{align*}$$
2. 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 $$\mathbb{P} \left( \sup_{k \geq 0} M_k \geq r \right) \leq \frac{1}{r}. \tag{2} $$
3. 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 $\lambda<\lambda_0$.