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Consider an arithmetic Brownian motion $X_t$ which follows $dX_t=\mu dt+\sigma dZ_t$ where $\mu$ and $\sigma$ are constants and $r$ is the discount rate. Assume an asset price $S_t=X_t^2$. I need to find the stochastic differential equation satisfied by the process $S_t$ and the density and distribution functions of $S_t$.

I first find the SDE using Ito's lemma on $S_t$ and get \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(0+2\mu X_t+\sigma^2\right)dt+2\sigma X_tdZ_t\\ &=\left(2\mu\sqrt{S_t}+\sigma^2\right)dt+2\sigma\sqrt{S_t}dZ_t \end{align*}

My question is how to find the distribution of $S_t$ now. I know that $X_t$ is normally distributed in that $X_T-X_0\sim N(\mu T,\sigma^2T)$, but I'm not exactly sure how to pull the distribution of $S_t$ out of the SDE.

I recognize from other classes that a square normal is going to be a generalized $\chi^2$ distribution, however I think that is out of the scope of this class so I think I must be reasoning wrongly about something else.

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    $\begingroup$ Actually, using the SDE to determine the distribution of $S_t$ is one of the most impractical ideas one can have. Since one knows the distribution of $X_t$, simply use it to deduce the distribution of $S_t=X_t^2$. (Additionally, note that in your computations, replacing $X_t$ by $\sqrt{S_t}$ is wrong since $P(X_t\geqslant0)\ne1$.) $\endgroup$ – Did Mar 28 '16 at 20:35
  • $\begingroup$ Something does seem odd here, and I wonder if there isn't a mistake in the statement of the problem. Expanding on Did's comment, assuming $\mu \neq 0$, we can't know the drift of $S_t$ without knowing the sign of $X_t$. So $S_t$ alone isn't Markov, and can't be a solution to an SDE. $\endgroup$ – John M Apr 1 '16 at 23:09

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