Long-Run Variance of a Mean-Reverting Ornstein–Uhlenbeck Process

Summary

The document examines an Ornstein–Uhlenbeck process whose drift targets a level equal to the process’s limiting variance. It asks whether that apparent relationship is mathematically consistent, using a process with positive mean-reversion speed, constant diffusion, and a specified initial value.

The answer applies an integrating factor, equivalent to solving the linear stochastic differential equation with Itô calculus. The resulting expression separates the decaying contribution from the initial value, the drift’s long-run level, and a stochastic integral. The variance follows from the Itô isometry and converges to the stated level as time grows. This establishes that the equality can hold for this parameterization; it does not mean that a process’s mean and variance are generally identical, nor does the discussion address estimation from market data or broader model suitability.

Key ideas

• An Ornstein–Uhlenbeck process can have a drift target numerically equal to its limiting variance under the stated parameterization.
• An integrating factor gives an explicit solution to the linear stochastic differential equation.
• The stochastic integral determines the time-dependent variance through the Itô isometry.
• The variance converges to its long-run value when the mean-reversion speed is positive.
• The result is a property of this model setup, not a general identity between mean and variance.

Tags

# Mean Reverting to its own variance?


# 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 :)

## Answer by user16651 (score 1, accepted)

https://quant.stackexchange.com/a/31507

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

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