# 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 $$\bar{V}$$.

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

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

• 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).
– Alex
Jun 21 at 11:38
• @Alex Thanks for the comment, Alex. If you want to formalize your response as an answer, I will choose it! Also, did you mean to have $t$ as a subscript to $V$? Jun 21 at 15:23
• @Alex, in its asymptotic behavior under the conditional expectation, we see the volatility term plays no role, so I understand the magnitude of the volatility would not impact how "jiggly" it will converge to the long-term mean asymptotically. Would you agree? Jun 21 at 16:01

## 1 Answer

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 $$\kappa_V,\bar{V},\sigma$$.