This is in the context of Ito Calculus. Here, $W_t$ is a $P$-Brownian Motion. My book says that "... $(W_t)^2$ has mean $t$, because of the variance structure of Brownian motion.."
I understood that the value of $(W_t)$ is distributed, under $P$, as a normal random variable $N(0,t)$. So, if we square the process $(W_t)$ wouldn't it be distributed as a random normal $N(0,t^2)$, and thus have mean $0$?
Since $Var(W_t) = E(W_t^2) - E(W_t)^2 = t$
And $E(W_t) = 0$, you get $$E(W_t^2) = t$$
Note that saying that $W_t^2$ should be normally distributed with variance $t^2$ is very wrong! It almost never happens that applying continuos, non linear function to random variable preserves how they are distributed. In this specific case $W_t^2 / \sqrt t$ is distrubuted as a chi-square distribution with one degree of freedom $$\frac{W_t^2}{\sqrt t} \sim \chi^2(1)$$
So our $W_t^2$ follows a gamma distribution:
$$W_t^2 \sim \Gamma(1/2, 2\sqrt t)$$
So as you see determining which distribution a (transformation of) a random variable follows is not immediate, and certainly you can't just assume that it preserves the original distribution and just apply the transformation to the relevant parameters.