# Mean Reverting to its own variance?

Good morning all,

When trying to decipher some documentation I have come across this stochastic process which seems to me much like a Ornstein-Uhlenbeck (or Vasicek) process.

$$dX_t=-\kappa(X_t-\sigma^2/2\kappa)dt+\sigma dW_t$$

However, the long-term mean level coincides with the asymptotic variance of the process: $Var[X_t]=\sigma^2/2\kappa$ as $t\rightarrow \infty$

My question is: Does this make any sense? It certainly does not to me.

Thank you very much, this forum is veryhelful :)

## 1 Answer

Yes, it is true. Let $$dX_t=\kappa\left(\frac {\sigma^2}{2\kappa} -X_t\right)dt+\sigma dW_t\tag 1\\$$ Where $X_0=x$. By application of Ito's lemma we have $$d\left(e^{\kappa t}X_t\right)=\kappa e^{\kappa t}X_tdt+e^{\kappa t}dX_t+\underbrace{d[e^{\kappa t},X_t]}_{0}$$ thus $$d\left(e^{\kappa t}X_t\right)=\frac{1}{2}\sigma^2 e^{\kappa t}dt+\sigma e^{\kappa t}dW_t\tag 2$$ By integration on $[0,t]$, we have $$X_t=x e^{-\kappa t}+\frac{1}{2\kappa}\sigma^2 (1-e^{-\kappa t})+\sigma\int_{0}^{t}e^{-\kappa(t-s)}dW_s\tag 3$$ and $$\text{Var}(X_t)=\mathbb{E}\left[\left(\sigma\int_{0}^{t}e^{-\kappa(t-s)}dW_s\right)^2\right]=\sigma^2\int_{0}^{t}e^{-2\kappa(t-s)}ds=\frac{\sigma^2}{2\kappa} (1-e^{-2\kappa t})\tag 4$$ Since $\kappa>0$, $$\lim_{t\to \infty}\text{Var}(X_t)=\frac{\sigma^2}{2\kappa}\tag 5$$

• Thank you Behrouz Maleki for your answer. But, does it make sense that the process reverts to its asymptotic variance? How do you interpret that as a long-term mean? Thanks a lot! – Pablo Martin Dec 16 '16 at 11:25
• Mean reversion is the theory suggesting that prices and returns eventually move back toward the mean or average. This mean or average can be the historical average of the price or return, or another relevant average such as the growth in the economy or the average return of an industry. – user16651 Dec 16 '16 at 11:31
• – user16651 Dec 16 '16 at 11:33