# Square of arithmetic brownian motion process

We have an arithmetic Brownian motion process $X_t$ that follows $dX_t=\mu dt + \sigma dZ_t$ and we define the asset price $S_t=X_t^2$ and we are asked to find the stochastic differential equation that $S_t$ satisfies, as well as the density and distribution functions of $S_t$. The first part I accomplished through a straightforward application of Ito's lemma: \begin{align*} dS_t&=\left(\frac{\partial S_t}{\partial t}+\mu\frac{\partial S_t}{\partial X_t}+\frac{1}{2}\sigma^2\frac{\partial^2 S_t}{\partial X_t^2}\right)dt+\sigma\frac{\partial S_t}{\partial X_t}dZ_t\\ &=\left(2\mu X_t+\sigma^2\right)dt+2\sigma X_tdZ_t\\ \end{align*}

but the second part I am having more trouble on. We know that $X_t$ is normally distributed in that $X_t-X_0\sim N(\mu t,\sigma\sqrt{t})$, so I know from other classes that $S_t=X_t^2$ will follow some sort of $\chi^2$ distribution, however because $X_t$ has neither zero mean nor unit variance, $S_t$ ends up being a scaled noncentral $\chi^2$ distribution and the CDF and PDF are ugly at best (not to mention I'm pretty sure that the knowledge that the square of normal is $\chi^2$ is beyond the scope of this class).

Is there something wrong with my logic in going from $X_t\sim$ Normal to the distribution of $S_t$?

## 1 Answer

Your logic is fine $$X_t \sim \mathcal {N}(X_0+\mu t, \sigma^2 t)$$

Thus, $\left (\frac {X_t}{\sigma\sqrt {t}}\right)^2$ indeed exhibits a non central chi-squared distribution

$$\left (\frac {X_t}{\sigma\sqrt {t}}\right)^2 \sim \chi^2\left(k=1,\lambda=\left (\frac {X_0+\mu t}{\sigma\sqrt {t}}\right)^2\right)$$

whence the law of $S_t := X_t^2$.

As regards the pdf/cdf of $S_t$, the key here is that there is only one degree of freedom, so no need to know the non central $\chi^2$ pdf/cdf by heart.

Actually, there is no need to know that its a chi-squared distribution to begin with.

Indeed, for any variable $X^2$ with support on $[0,\infty [$, the cumulative distribution function $F_{X^2}$ writes:

$$F_{X^2}(x) := P [X^2 \leq x]$$ which is equivalent to writing \begin{align} F_{X^2}(x) &= P [\vert X \vert \leq \sqrt {x}] \\ &= P [-\sqrt {x} \leq X \leq +\sqrt {x}] \\ &= F_X (\sqrt {x}) - F_X (-\sqrt {x}) \end{align}

Hence the relationship between the cumulative distribution function of $X^2$ and that of $X$.

Now differentiate the above equation with respect to $x$ to obtain the relationship between the probability density functions.

• Very good answer! – Richard Mar 29 '16 at 8:20