Distribution of a Standard Brownian Motion Function

Question

Let $W(t)$ be a standard Brownian motion. What is the distribution of the following function?

$Y(t)=W(t)+W(t^2)$,

Note: We know that $W(t^2)$ is not a Brownian motion, since $R_{W(t^2)}(t_1,t_2)=min(t_1^2,t_2^2)$. $R_{w}(.)$ stands for autocorrelation.

Answer 1

$W(t)$ and $W(s)$ are jointly normal for any $t$ and $s$. Hence $Y(t)$ is normal. You only have to find the mean and the variance. The mean is, of course, $0$. Now $EY(t)^{2}=E[W(t)]^{2}+E[W(t^{2})]^{2}+2EW(t)W(t^{2})=t+t^{2}+2\min \{t, t^{2}\}$.