TITLE: Integral of Wiener Squared process
QUESTION [7 upvotes]: I don't have a background of stochastic calculus. 
It is known fact that definite integral of standard Wiener process from $0$ to $t$ results in another Gaussian process with slice distribution that is normal distributed with mean equal to $0$ and variance $\frac{T^3}{3}$ i-e
$$ \int_0^{t} W_s ds \sim \mathcal{N}(0,\frac{t^3}{3})    $$
Question:
 What if we square the standard Wiener process and then integrate i-e $$ \int_0^{t} W_s^2 ds \sim ?    $$
Would that be scaled Chi-square distributed ?

REPLY [7 votes]: The distribution of $\int_0^t W^2_s\,ds$ was found by Cameron and Martin in the 1940s. Mark Kac (1949, Transactions of the AMS) applied his method to find the Laplace transform of this integral:
$$
\Bbb E\left[\exp(-u\int_0^t W^2_s\,ds)\right]={1\over\sqrt{\cosh(t\sqrt{2u})}}.
$$