Solving equation with Wiener process

Question

I want to show that if $E(f(X_{t}))=E(f(W_{t})e^{\lambda W_{t}-0.5*\lambda^2*t})$, where $W_{t}$ is a Wiener Process, then $X_{t}\sim N(\lambda t,t)$. Does anyone have a clue how to solve this problem?

Answer 1

Let $f:\mathbb R \to \mathbb R^{+}$ be $x \mapsto e^{\theta x}$. Then, since $W_t\sim N(0,t)$, the given relationship implies $\mathbb E[e^{\theta X_t}] = \mathbb E [e^{(\theta + \lambda) W_t}] e^{-\frac{1}{2} \lambda^2 t} = e^{\frac{1}{2}\theta^2 t + \theta \lambda t}$; the moment generating function for a random variable distributed as $N(\lambda t, t)$.