4
$\begingroup$

Consider the process,

$$ dX_t=(-aX_t+b(1-X_t))dt + \sqrt{X_t(1-X_t)}dW_t $$

How do I show that the stationary distribution for the transition density is a beta distribution?

I tried expanding the corresponding Kolmogorov Forward Equation but it seems too difficult to solve the equation.

$\endgroup$
4
$\begingroup$

There is a shortcut around the Forward Equation when you are looking for the stationary distribution. Let me write $$ dX = \mu(X)dt +\sigma(X)dW $$ for $$ \mu(x)=b(1-x)-ax\ \text{ and }\ \sigma^2(x)=x(1-x) $$

The Forward Equation indeed states that the stationary distribution $p(x)$ will be satisfied for $\partial p/\partial t = 0$, therefore $$ \frac{1}{2}\frac{d^2}{dx^2}\left[\sigma^2(x)p(x)\right] - \frac{d}{dx}\left[\mu(x)p(x)\right] = 0 $$

The trick is to take one differential as a common factor and write $$ \frac{d}{dx} \left\{ \frac{1}{2}\frac{d}{dx}\left[\sigma^2(x)p(x)\right] - \mu(x)p(x) \right\}= 0 $$ Then, the term in the braces will be a constant (it's derivative is zero), and we can take it to be zero. Then we are facing the first order ODE $$ \frac{1}{2}\frac{d}{dx}\left[\sigma^2(x)p(x)\right] = \mu(x)p(x) $$ Solving this yields the stationary distribution up to the normalization constant. The solution is actually given by $$ p(x) \propto \sigma^{-2}(x) \exp\left( \int^x \frac{2\mu(u)}{\sigma^2(u)} du \right) $$

The above holds for any process. In your particular case, the integral becomes $$ \int^x \frac{2b(1-u)-2au}{u(1-u)} du = 2b \int^x \frac{du}{u} -2a\int^x \frac{du}{1-u} = \log \left( x^{2b} (1-x)^{2a} \right) $$

Hence, overall the stationary distribution is Beta with parameters $(\alpha,\beta)=(2b,2a)$ $$ p(x) \propto x^{2b-1} (1-x)^{2a-1} $$

$\endgroup$
  • $\begingroup$ Very nice solution. $\endgroup$ – Gordon May 13 '15 at 18:19
0
$\begingroup$

try to calculate the moment generating function and show that it correspond to the one of a beta distributed random variable

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.