Show that the stopping moment $\tau$ is finite, find the distribution of $X_{\tau}$, calculate $\mathbb{E}e^{\tau}$. Let $X = (X_{t})_{t \geqslant 0}$ be a stochastic process with independent increments. $X$ is continuous and $\mathbb{E}X_{t} = 1$, $Var X_t = e^t -1$, $t \geqslant 0$. Let $\tau = \inf\{t \geqslant 0 : X_t = 3$ or $X_t = 0\}$.
a) Find a function $f: \mathbb{R_{+}} \rightarrow \mathbb{R}$ such that the process $((X_t -1)^2 - f(t))_{t \geqslant 0}$ is a martingale.
b) Show that $\tau < \infty$ and find the distribution of $X_\tau$
c) Calculate $\mathbb{E}e^\tau$.
I managed to do a) ! :) It's just $f(x) = e^{x}$ and I found it by checking the condition $\mathbb{E}((X_t -1)^2 - f(t)|\mathcal{F}_s) = (X_s - 1)^2 - f(s)$.
Although I don't have any ideas on b) and c).
 A: Hints:


*

*Conclude from $\mathbb{E}(X_t)=1$ and $\text{var}(X_t)=1$ that $X_0=1$ almost surely.

*Use the optional stopping theorem to show that $$\mathbb{E}((X_{t \wedge \tau}-1)^2) = \mathbb{E}(e^{t \wedge \tau})-1. \tag{1}$$

*Using the continuity of the sample paths, prove that $|X_{t \wedge \tau}| \leq 3$. Conclude from $(1)$ and the monotone convergence theorem that $$\mathbb{E}(e^{\tau})< \infty.$$ In particular, $\tau<\infty$ almost surely.

*Show that $(X_t)_{t \geq 0}$ is a martingale. Apply once more the optional stopping theorem and the dominated convergence theorem to deduce that $$\mathbb{E}(X_{\tau})=\mathbb{E}(X_0)=1. \tag{2}$$

*Since $\tau<\infty$ almost surely, it follows from the continuity of the sample paths that $X_{\tau} \in \{0,3\}$. Hence, $(2)$ can be written as $$3 \mathbb{P}(X_{\tau}=3) + 0 \cdot \mathbb{P}(X_{\tau}=0)=1.$$ Using $\mathbb{P}(X_{\tau}=3)+\mathbb{P}(X_{\tau}=0)=1$ calculate both probabilities.

*Letting $t \to \infty$ in $(1)$ gives $$\mathbb{E}((X_{\tau}-1)^2)=\mathbb{E}(e^{\tau})-1.$$ Use Step 5 to calculate the left-hand side explicitly.

