Proving the strict stationary of $X(t)=e^{-t/2}W(e^t)$

36 Views Asked by At

Let $W(t)$ be a Wiener process with a parameter $\alpha=1$ and a process $X(t)=e^{-t/2}W(e^t)$.

Show that X(t) is stationary in the strict sense.

Resolution attempt:

$$E(X(t))=E(e^{-t/2}W(e^t))=e^{-t/2} E(W(e^t))=e^{-t/2}.0=0$$

$$Var(X(t))=E(X(t)^2)-E(X(t))^2=E(e^{-t} W^2(e^t))-0=e^{-t}Var(W^2(e^t))=e^{-t}.e^{t}=1$$

So $X(t)\sim N(0,1)$.

Since the variance is a constant then $F(X(t+\tau))=F(X(t))$ hence $X(t)$ is stationary in the strict sense.

Questions:

Is my resolution right? If not how should I prove the claim?

Thanks in advance!