Htykuut

I am uncertain how to go about the following problem from the lecture notes on a course in SDE's. We are given the following SDE.

$dX_t=lambdaleft(xi-X_tright) dt+gammasqrt{|X_t|}dB_t$

Where $lambda,xi,gamma$ are positive constants. Show that $X_t$ has Gamma distribution, with a rate parameter $omega=2lambda/gamma^2$ and shape parameter $nu=2lambdaxi/gamma^2$. Finally, what is the mean and variance in stationarity?

I know I have to find the stationary distribution by isolating $phi$ in the following.

$-nablacdotleft( uphi-Dnabla phiright)=0$

Inserting $D=frac{1}{2}gamma^2|X_t|, $ $u=f-nabla D=lambdaleft(xi-X_tright)-frac{1}{2}gamma^2frac{X_t}{|X_t|}$, I get an ugly expression, involving second derivative of absolute value of the stochastic process. Is there a mistake so far? (I take the gradient and divergence w.r.t $x=X_t$)

From the Kolmogorov forward equation,

$$
frac{d^2}{dy^2}left(frac{1}{2}gamma^2yphi(y)right)=frac{d}{dy}left(lambda(xi-y)phi(y)right).
tag{1}label{eq:asdf}
$$
The left-hand side is equal to
$$
frac{d}{dy}left(frac{1}{2}gamma^2phi(y)+frac{1}{2}gamma^2yphi'(y)right).
$$
So integrating eqref{eq:asdf} from $0$ to $y$ we get
$$
frac{1}{2}gamma^2phi(y)+frac{1}{2}gamma^2yphi'(y)=lambda(xi-y)phi(y).
$$
Here, I've used the fact that $phi(0)=0$ (if $X$ hits zero, it'll instantly get reflected). [$phi(0)$ is not necessarily zero. See below.] Rearranging the last equation,
$$
frac{phi'(y)}{phi(y)}
=
frac{lambda(xi-y)-frac{1}{2}gamma^2}{frac{1}{2}gamma^2y}
=
left(frac{2lambdaxi}{gamma^2}-1right)frac{1}{y}-frac{2lambda}{gamma^2}.
$$
Integrating this from $xi$ to $y$,
$$
logfrac{phi(y)}{phi(xi)}
= left(frac{2lambdaxi}{gamma^2}-1right)logfrac{y}{xi} - frac{2lambda}{gamma^2}(y-xi).
$$
So, finally,
$$
phi(y)propto y^{frac{2lambdaxi}{gamma^2}-1}e^{- frac{2lambda}{gamma^2}y},
$$
which is the pdf of a Gamma distribution.

The mean and variance are accordingly given by
$$
frac{2lambdaxi}{gamma^2}frac{gamma^2}{2lambda}=xi
quadtext{and}quad
frac{2lambdaxi}{gamma^2}left(frac{gamma^2}{2lambda}right)^2
=frac{gamma^2xi}{2lambda}.
$$

Addendum

If $frac{2lambdaxi}{gamma^2}le 1$, then $phi(0)ne 0$; and if $frac{2lambdaxi}{gamma^2}< 1$, then $phi'(0)=-infty$, contrary to what I assumed above. In any case, these assumptions are not necessary. Here's the correction.

We still have
$$
frac{1}{2}gamma^2phi(y)+frac{1}{2}gamma^2yphi'(y)
-lambda(xi-y)phi(y)=text{constant}.
$$
Since $phi$ is a pdf defined on $[0,infty)$, $phi(infty)=0$ and $phi'(infty)=0$. Also, if $X_t$ is integrable with respect to the stationary distribution (which I will assume), then $yphi(y)rightarrow 0$ and $yphi'(y)rightarrow 0$ as $yrightarrow infty$. It follows that the constant in the last equation is zero, which brings us back to the derivation above.