Calculate $E[e^{2 \sqrt{2}(W_T-W_t)}]$, where $W_s$ is the Wiener process, t is fixed, T>t

Question

Let $X_T = W_t - W_t$, so that $dX_T = dW_T$, and let $Z_T = e^{2 \sqrt{2}(W_T-W_t)}$. By Ito's lemma $$dZ_T =2\sqrt{2}Z_Td_T + 4Z_TdW_T $$ Integrating from $t$ to $T$ and taking expectations we get $E(Z_T) = 1 + 2\sqrt{2}\int_t^TE(Z_s)ds $

We then get the ODE in $T$, for $m(T) = E(Z_T)$

$$m'(T) = 2\sqrt{2}m(t)$$ $$m(t) = 1$$

which has solution $E(Z_T) = E(X_T^2) = m(T) = e^{2\sqrt{2}(T-t)}$

Is this correct?

Answer 1

First let me say that I don't find any logical error. To me it seems correct. You could give more details, but that's okay.

However, in your first equaton you mean: $X_T = W_T - W_t$. In second line it is a $8$ not $4$, when I get the second derivation of $e^{2\sqrt{2} \zeta}$ right and I don't really understand that you write: $\Bbb{E}[Z_T] = \Bbb{E}[X_T^2]$. I don't see a equation here, especially because: $\Bbb{E}[X_T^2] = T- t$