distribution function of $X_t = B_t^2 - t$

Question

I have tested that the process $X_t = B_t^2 - t$, where $B_t$ is an standard brownian motion $\mathcal{N}(0,t)$, is a martingal and now need to find is distribuition. I know that $B_t^2$ has a $\mathcal{X}$-squred distribution with 1 degree of freedom, but how can I use this fact to get to the distribution of $X_t$

Answer 1

Just to have this question answered not in the comments.

The Brownian motion respects $B(0)=0$ and $B(t'+t)-B(t') \sim N(\mu=0,\sigma^2=t)$.

This means:

1. $B(t) \sim N(0,t)$. Therefore $B(t) \sim \sqrt{t}Z$, where $Z$ is a standard Gaussian variable.

2. This implies: $B^2(t) \sim tZ^2 \sim t\chi_1^2$, and $B^2(t)-t \sim t\chi_1^2-t$

3. $P(B^2(t)-t\le x)=P(t\chi_1^2-t \le x)=P(\chi_1^2\le \frac{x+t}{t})$

Taking derivatives of the above c.d.f. we get the p.d.f. of our variable:

$f_{B^2-t}(x)=\frac{1}{t}f_{\chi_1^2}(\frac{x+t}{t})$