# Distribution of Brownian Motion squared

Could anyone please help me resolve the following paradox?

(i) Brownian Motion squared (i.e. $(B(t)^2$) can be written as:

$$2*[Ito Integral(B(t))] - t$$

(simple, using Ito's lemma). The distribution of Ito integral of a process is Normal, where the mean is zero and the variance can be computed using Ito Isometry. Therefore, using this argumentation, the distribution of Brownian motion squared should be normal with mean zero and variance equal to $2t^2$.

(ii) Standard Brownian motion $B(t)$ at a point in time is, by definition, normally distributed with mean zero and variance $t$. Therefore, it would seem logical that Brownian Motion squared has a (non-centered) Chi-squared distribution with one degree of freedom.

Which one is correct? Or am I making a logical error in my reasoning somewhere?

Thank you so much for your time and help,

Best regards,

J.

## 1 Answer

Your second reasoning is correct, and the first one is wrong. The stochastic integral

$$\int_0^t f(s) \, dB_s$$

is Gaussian if $f$ is a deterministic function. You are interested in the stochastic integral

$$\int_0^t B_s \, dB_s = \frac{B_t^2}{2}-\frac{t}{2}$$

i.e. $f(s) = B_s$ which is clearly not deterministic.

Remark: Note that $B_t^2$ is a non-negative random variable - this means, in particular, that it cannot be Gaussian!