# Distribution of a squared standard Brownian motion

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!

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.