Expectations of the Wiener Process

Question

In the Wiener process, by definition, we have that $E(W(t))=0$. What would be the $E(W(e^{2t})$?

My best guess would be that this would equal to 0 again because the expectation with the Wiener process of any t would be equal to 0.

Furthermore, what would be the variance of this Wiener Process?

Answer 1

Since for any $s\ge0$ we have that $W(s) = W(s)-W(0)\sim\mathcal{N}(0,s)$, we have $W(e^{2t})\sim\mathcal{N}(0,e^{2t})$, with mean $0$ and variance $e^{2t}$.

It is not a Wiener process as it has the wrong variance.