Consider the stochastic differential equation (SDE) \begin{align} dX_t = \alpha \left( \theta - X_t + \sqrt{2\mu} \left(\int_0^t e^{\mu(s-t) }dW_s \right) \right) dt \end{align} where $W_t$ is a standard Brownian motion, $\alpha>0$ and $\theta>0$. I am interested in studying the mean and the variance of a process $X_t$ that satisfies the SDE above as $t\to\infty$.

For the mean, by a known property of the Ito's integral, $\int_0^t e^{\mu(s-t) }dW_s$ has a normal distribution with mean 0 and variance $\frac{1-e^{-2t \mu}}{2\mu}$ given by Ito's isometry. This should give $\mathbb{E}[X_t]\to \theta$ as $t\to\infty$.

For the variance, I tried to compute the second moment but the calculation gets very messy, and I was wondering whether there is a known result for the specific SDE above or what would be a good approach?


1 Answer 1


Helping with the essential steps:

Writing $$\tag{1} Y_t=\alpha\,\theta+\alpha\sqrt{2\mu}\left(\int_0^t e^{\mu(s-t) }\,dW_s \right) $$ this SDE is more an inhomogeneous ODE of the form $$\tag{2} \dot X_t=-\alpha X_t+Y_t $$ which has the solution $$\tag{3} X_t=\textstyle X_0\,e^{-\alpha t}+\int_0^tY_u\,e^{\alpha(u-t)}\,du\,. $$ The mean of $Y_t$ is $\mathbb E[Y_t]=\alpha\,\theta\,.$ Therefore, the mean of $X_t$ is $$\tag{4} \mathbb E[X_t]=\textstyle X_0\,e^{-\alpha t}+\int_0^t\alpha\,\theta\,e^{\alpha(u-t)}\,du=X_0\,e^{-\alpha t}+\theta\,(1-e^{-\alpha t})\,. $$ Further \begin{align}\tag{5} \mathbb E[X_t^2]&=\textstyle X_0^2\,e^{-2\alpha t}+2\,X_0\,e^{-\alpha t}\,\mathbb E\Big[\int_0^tY_u\,e^{\alpha(u-t)}\,du\Big]+\mathbb E\Big[\Big(\int_0^tY_u\,e^{\alpha(u-t)}\,du\Big)^2\Big]\,. \end{align} The second term in (5) is (similar to the calculation for the mean) $$\tag{6} 2\,X_0\,e^{-\alpha t}\,\theta\,(1-e^{-\alpha t})=2\,X_0\,\theta\,(e^{-\alpha t}-1)\,. $$ The last term in (5) is \begin{align}\tag{7} \textstyle\mathbb E\Big[\int_0^t\int_0^tY_u\,Y_v\,e^{\alpha(u+v-2t)}\,du\,dv\Big] =\int_0^t\int_0^t\mathbb E[Y_u\,Y_v]\,e^{\alpha(u+v-2t)}\,du\,dv\,. \end{align} Using (1) we get \begin{align} \mathbb E[Y_uY_v]&=\alpha^2\,\theta^2\,+\sigma^2\,\mathbb E\Big[\Big(\textstyle\int_0^u e^{\mu(s-u)}\,dW_s\Big)\Big(\textstyle\int_0^v e^{\mu(s-v)}\,dW_s\Big)\Big]\\[2mm] &=\alpha^2\,\theta^2\,+\sigma^2\,e^{-\mu(u+v)}\mathbb E\Big[\Big(\textstyle\int_0^u e^{\mu s}\,dW_s\Big)\Big(\textstyle\int_0^v e^{\mu s}\,dW_s\Big)\Big]\,.\tag{8} \end{align} Because $W_t$ has independent increments and the integrands in the stochastic integrals are deterministic the last expectation in (8) is $$\tag{9} \textstyle\int_0^{\min(u,v)} e^{2\mu s}\,ds=\displaystyle\frac{e^{2\mu\min(u,v)}-1}{2\mu}\,. $$ (This is a relatively simple generalization of the covariance function $\mathbb E[W_uW_v]=\min(u,v)$ and of the Ito isometry $\mathbb E[(\int_0^ue^{\mu s}\,dW_s)^2]=\int_0^u e^{2\mu s}\,ds\,.$)

Therefore \begin{align} \mathbb E[Y_uY_v] &=\alpha^2\,\theta^2\,+\,\sigma^2\,e^{-\mu(u+v)}\,\frac{e^{2\mu\min(u,v)}-1}{2\mu}\\ &= \alpha^2\,\theta^2\,+\,\sigma^2\,\frac{e^{-\mu|u-v|}-e^{-\mu(u+v)}}{2\mu}\,. \tag{10} \end{align} The last expression is known as the covariance function of an Ornstein-Uhlenbeck process.

Though some work is left to do it should be straightforward to perform the integral on the right hand side of (7) now.

  • $\begingroup$ Thank you this helps. I don't understand the second equality in (8), how did you get it? $\endgroup$
    – user52227
    Mar 24 at 17:04
  • $\begingroup$ @user52227 . Let me check. It could be slightly wrong. $\endgroup$
    – Kurt G.
    Mar 24 at 17:29
  • $\begingroup$ @user52227 should be fixed now. $\endgroup$
    – Kurt G.
    Mar 24 at 17:49
  • $\begingroup$ Yes, thank you! The rest is straightforward calculus (just a bit boring eheh). Just want to notice that the ode is of the form $\dot X_t= - \alpha X_t+Y_t$ but this does not change much things. $\endgroup$
    – user52227
    Mar 26 at 8:56
  • $\begingroup$ @user52227 Thank you too. Will fix that answer a little bit again. When $\alpha>0$ the mean of $X_t$ converges to $\alpha\theta$ as $t\to\infty\,.$ Would you reveal your findings of the variance? (We don't need the boring details. Just the result.) $\endgroup$
    – Kurt G.
    Mar 26 at 9:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .