# 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 :)

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 $\kappa>0$, $$\lim_{t\to \infty}\text{Var}(X_t)=\frac{\sigma^2}{2\kappa}\tag 5$$