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Finding the Beta Stationary Distribution of a Bounded Diffusion

Article Quant Q&A · Author: Danny
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Summary

The document shows how to obtain a stationary density for a diffusion whose state remains between zero and one, with drift pulling the process according to two parameters and volatility proportional to the square root of the product of the state and its complement. Instead of solving the full second-order stationary forward equation directly, it integrates once and sets the resulting probability current to zero, reducing the task to a first-order differential equation.

A general density formula in terms of drift and squared volatility then yields a density proportional to the state raised to one parameter minus one and its complement raised to another parameter minus one. This identifies a beta distribution with parameters twice the model’s drift coefficients. The argument assumes zero stationary probability current and omits boundary-condition details; the resulting density must also be normalizable for the chosen parameters. A second answer merely suggests using a moment-generating function and provides no derivation.

Key ideas

  • Integrating the stationary forward equation once reduces it to a first-order equation when probability current is set to zero.
  • The stationary density is proportional to inverse squared volatility times an exponential integral of drift over variance.
  • For the bounded square-root diffusion, the density has the form of a beta distribution.
  • The beta parameters are twice the two drift coefficients in the stated model.
  • Boundary behavior and normalizability conditions are not developed in the excerpt.

Tags

StatisticsMean reversion
Full text
# Stationary distribution for square root process


# Stationary distribution for square root process












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.

## Answer by Kiwiakos (score 4)

https://quant.stackexchange.com/a/17781

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 $(\alpha,\beta)=(2b,2a)$ $$ p(x) \propto x^{2b-1} (1-x)^{2a-1} $$

## Answer by Phun (score 0)

https://quant.stackexchange.com/a/17355

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

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.