# Origin of strange formula for equilibrium standard deviation

In the paper

M. Avellaneda and J. H. Lee, Statistical arbitrage in the U.S. equities market, July 2008,

in the Appendix on page 46, how does he get equilibrium standard deviation as following:

$$\sigma_{eq} = \sqrt{\frac{\text{Variance}(\zeta)}{1 − b^2}}$$

If anyone knows the paper, please explain.

Much appreciated,

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I posted another question on the same paper - stats.stackexchange.com/questions/9396/…. Please look at it if possible. Thanks! – user862 Apr 9 '11 at 23:26

The authors are providing a simple means for estimating the parameters of a mean-reverting Orstein-Uhlenbeck process via a regression on returns at discretized points in time.

The model they are considering has a representation as a stochastic differential equation of the form [pg. 16, Eq. (12)] $$\newcommand{\rd}{\mathrm{d}} \rd X(t) = \kappa (m - X(t)) \rd t + \sigma \rd W(t)$$ where $W(t)$ is a standard Brownian motion.

The solution to this SDE is well-known and easy to find via Ito's lemma and an analogous technique to integrating constants in ODEs. The solution is [pg. 17, Eq. (13)] $$X(t_0 + \Delta t) = e^{-\kappa \Delta t} X(t_0) + (1-e^{-\kappa \Delta t}) m + \sigma \int_{t_0}^{\,t_0 + \Delta t} e^{-\kappa(t_0+\Delta t - s)} \, \rd W(s) .$$

This is a Gaussian process and so is characterized by its mean and covariance as a function of time. Letting "time go to infinity" (i.e., $\Delta t \to \infty$), we get an equilibrium mean and variance of \begin{aligned} \mathbb{E} X(t) &= m \\ \mathbb{V}\mathrm{ar}(X(t)) &= \frac{\sigma^2}{2 \kappa} \end{aligned}

Now, skipping to the appendix [bottom of page 45], the authors are trying to estimate the parameters by doing a regression using the discrete values of the process and model $$X_{n+1} = a + b X_n + \zeta_{n+1} .$$

Matching up the parameters $a$ and $b$ with the portions from above, we get that \begin{aligned} a &= m (1 - e^{-\kappa \Delta t}) \\ b &= e^{-\kappa \Delta t} \\ \mathbb{V}\mathrm{ar}(\zeta) &= \sigma^2 \frac{1-e^{-2\kappa \Delta t}}{2\kappa} \end{aligned}

Substituting the second equation into the first and solving for $m$ gives $m = a / (1-b)$. Use the same substitution in the third equation and rearrange to get $$\sigma^2 = \frac{\mathbb{V}\mathrm{ar}(\zeta) \cdot 2 \kappa}{1 - b^2} \>,$$ but, recall that the variance of the equilibrium distribution (by looking far into the future) for $X(t)$ is just $\sigma^2 / 2 \kappa$ and so this gives your result.

Addendum: If you're wondering how the expression $$\mathbb{V}\mathrm{ar}(\zeta) = \sigma^2 \frac{1-e^{-2\kappa \Delta t}}{2\kappa}$$ was obtained, it is via the (remarkable and beautiful!) Ito isometry and the fact that an Ito integral is a zero-mean martingale; namely, in this instance, $$\mathbb{E}\Big(\sigma \int_{t_0}^{\,t_0 + \Delta t} e^{-\kappa(t_0+\Delta t - s)} \, \rd W(s)\Big)^2 = \sigma^2 \int_{t_0}^{\,t_0 + \Delta t} e^{-2 \kappa(t_0 +\Delta t - s)} \, \rd s$$ where we note that the integrand has been squared on the right-hand side and we "get to replace" $\rd W(s)$ with $\rd s$, converting the problem into one of solving a standard Riemann integral.

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 Oh I just got confused with the substitution as I got caught up with how did they set $\kappa=1$ to get that, which is not case at all and it is a simple substitution from previous equation. I did understand the derivation of variance of mean reverting process using ito-isometry. Thanks for a nice description. – user862 Apr 9 '11 at 22:21 @user862: I figured it might just be a simple situation like this. But, I thought the background of the question might be interesting enough to others that I'd put together a quick exposition. That way the question and answer are more self-contained, especially in case that link ever goes down. – cardinal Apr 9 '11 at 23:06 Completely agree. Thanks again. – user862 Apr 9 '11 at 23:07