Distribution of a squared standard Brownian motion

Question

During a self study I encountered the following issue:

I got a standard Brownian motion $B(t)$ on the interval $[0,1]$ which is $N(0,t)$ distributed by definition, but now I want to figure out the distribution of $B^2(t)$. I assume its a $\chi^2$ - distribution, but im not sure how to show it.

One approach I see is using the fact that $B(1) \sim N(0,1)$ and thus $B^2(1) \sim \chi^2(1)$, but I dont think its correct to rewrite $B^2(t) \stackrel{d}{=} t B(1)$ and receiving a $\chi^2$ - distribution this way.

Can anyone point me in the right direction? Thank you!

Answer 1

Since $B(t)\sim N(0,t)$, we have that $B(t)=t^{1/2}\cdot t^{-1/2}B(t)$, where $t^{-1/2}B(t)\sim N(0,1)$. Hence, $B^2(t)=tQ$, where $Q$ is the chi-squared distribution with $1$ degree of freedom. Indeed, it is correct to write $B^2(t)\stackrel{d}{=}tB^2(1)$ since $B^2(1)$ has the chi-squared distribution with $1$ degree of freedom.