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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?

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  • $\begingroup$ 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). $\endgroup$
    – Alex
    Jun 21 at 11:38
  • $\begingroup$ @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 $V$? $\endgroup$ Jun 21 at 15:23
  • $\begingroup$ @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? $\endgroup$ Jun 21 at 16:01
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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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