Distribution of Standard Brownian Motion squared If the standard Brownian Motion $B_t$ has pdf 
$$f(x)=(2\pi*t)^{-1/2}*\exp (-\frac{x^2}{2t})$$
how would one compute $B_t^2$?
I normally would show a lot of different things that I try myself before posting here, but I don't even know where to start, so if someone has any ideas, I would really appreciate the input!


Hint Let $X$ be a random variable such that $X$ has a density and $\mathbb{P}(X=x) = 0$ for any $x \in \mathbb{R}$. Then, the cumulative distribution function $F_{X^2}$ of $X^2$ equals $$\begin{align*} F_{X^2}(a) &:= \mathbb{P}(X^2 \leq a) = \mathbb{P}(-\sqrt{a} \leq X \leq \sqrt{a}) \\ &= \mathbb{P}(X \leq \sqrt{a})-\mathbb{P}(X<-\sqrt{a}) \\&= F_X(\sqrt{a})-F(-\sqrt{a}) \end{align*}$$ for any $a \geq 0$. Differentiate both sides in order to obtain the density $p_{X^2}$ of $X^2$.
Apply this to $X \sim N(0,t)$, i.e. a normal distributed random variable with mean $0$ and variance $t$.
Remark If $X \sim N(0,t)$, then $X^2$ is a $\chi^2$-distributed random variable.
