Mean Reversion Speed in the Square Root Diffusion Process

Summary

The note explains the role of the mean reversion parameter in a square root diffusion, a process often used for nonnegative quantities such as variance. Its conditional expectation is a weighted combination of the initial value and the long-run mean, with weights that change exponentially over time. This gives a direct way to interpret the parameter: when it is positive, a larger value makes the expected process approach its long-run mean more quickly.

The answer distinguishes this average behavior from the full distribution of future values, which also depends on the long-run mean and the volatility parameter. It states that the asymptotic distribution is shaped by these parameters, but does not give its form or discuss constraints required for the process to remain nonnegative. The explanation is therefore about expected reversion speed, not a complete account of path variability or distributional behavior.

Key ideas

• The conditional expectation combines the starting value and long-run mean with time-varying exponential weights.
• A larger positive mean reversion parameter speeds the expected convergence toward the long-run mean.
• Mean reversion speed describes the expectation and does not alone determine individual paths.
• The asymptotic distribution depends on the reversion parameter, long-run mean, and volatility.

Tags

# Boundedness in Square Root Process


# Boundedness in Square Root Process












Consider the following square root diffusion price process:

$$ dV_t = \kappa_V(\bar{V}-V_t)dt+\sigma_V\sqrt{V_t}dW_t $$

It is my understanding that $\kappa_V$ is the rate at which the process reverts back to its long-term mean $\bar{V}$.

- If this rate was bounded, would this mean the process deviates less "intensely" from or to its long-term mean?

- What if $\kappa_V$ could have a large magnitude? What are some implications?

## Answer by Alex (score 1, accepted)

https://quant.stackexchange.com/a/65643

We know that $$\mathbb{E}[V_t|V_0]=V_0e^{-\kappa_Vt}+\bar{V}\left(1-e^{-\kappa_Vt}\right).$$ Thus, as $t\to\infty$, we expect $\mathbb{E}[V_t|V_0]$ to converge to the long-term mean $\bar{V}$. The larger $\kappa_V$ in magnitude, the faster the exponential decay and the faster the convergence to $\bar{V}$. Thus, a large $\kappa_V$ ensures that the process is, on average, quite close to $\bar{V}$ (or returns to $\bar{V}$ very quickly).

The asymptotic distribution (of $V_\infty$) depends on $\kappa_V,\bar{V},\sigma$.

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