1
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you! This was much easier than I thought $\endgroup$ – Bazzan Dec 5 '18 at 13:16
  • $\begingroup$ @Bazzan You're welcome! $\endgroup$ – Cm7F7Bb Dec 5 '18 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.