I was deriving the solution to the stochastic differential equation $$dX_t = \mu X_tdt + \sigma X_tdB_t$$ where $B_t$ is a brownian motion. After finding $$X_t = x_0\exp((\mu - \frac{\sigma^2}{2})t + \mu B_t)$$ I wanted to calculate the expectation of $X_t$. However I think I'm not quite getting it. I thought that I'd just calculate $$E(x_0\exp((\mu - \frac{\sigma^2}{2})t + \mu B_t) = x_0\exp((\mu - \frac{\sigma^2}{2})t)E(\exp(\mu B_t))$$ but the book I'm using gives as answer $E(X_t) = x_0\exp(\mu t)$. I found this quite surprising as I don't quite see how $\sigma$ could just disappear. After reading Wikipedia I see that the result could be either $E(X_t) = x_0\exp((\mu + \frac{\sigma^2}{2})t)$ or $E(X_t) = x_0\exp(\mu t)$, depending on whether you use the Itô interpretation or the Stratanovich interpretation.

Since the book I use only considers the Itô formulation of stochastic integration I am interested in the latter result. But how do I obtain this? Do I just fail in calculating $E(\exp(\mu B_t))$? Thanks in advance.

  • $\begingroup$ Hint: The solution you've given for $X_t$ is incorrect and, based on the subsequent remarks, not simply a typo. $\endgroup$ – cardinal May 23 '11 at 12:51

The answer is that $E(X_t)=x_0e^{\mu t}$. The easiest way to see it is to start from the SDE and to note that $$\mathrm{d}E(X_t)=\mu E(X_t)\mathrm{d}t,\qquad E(X_0)=x_0.$$ Hence $a(t)=E(X_t)$ solves $a'(t)=\mu a(t)$ and $a(0)=x_0$, that is, $a(t)=x_0e^{\mu t}$ as claimed above.

Your solution goes astray when you solve the SDE, the factor of $B_t$ is wrong and, in fact, $$ X_t=x_0e^{(\mu-\sigma^2/2)t+\sigma B_t}. $$ Hence $$ E(X_t)=x_0e^{(\mu-\sigma^2/2)t}E(e^{\sigma B_t}). $$ Since $E(e^{uZ})=e^{u^2/2}$ for every real number $u$ and every standard normal random variable $Z$, the identity $E(e^{\sigma B_t})=e^{\sigma^2 t/2}$ follows from the fact that $\sigma B_t$ is distributed like $\sigma\sqrt{t}Z$. Simplifying, one gets the same expression of $E(X_t)$ than by the direct route, namely, $$E(X_t)=E(X_0)e^{\mu t}.$$

  • $\begingroup$ I haz a stupid once again. Thanks. $\endgroup$ – Stijn May 23 '11 at 13:41
  • 1
    $\begingroup$ @Stijn, note that, when accepting an answer you can also simultaneously upvote it as well. :) $\endgroup$ – cardinal May 23 '11 at 14:47
  • $\begingroup$ Indeed, thanks for reminding me. $\endgroup$ – Stijn May 23 '11 at 15:27
  • 5
    $\begingroup$ @Ievgenii Recall that $$dX_t = \mu X_tdt + \sigma X_tdB_t$$ is actually a shorthand for $$X_t = X_0+\mu\int_0^t X_sds + \sigma \int_0^tX_sdB_s$$ which implies $$E(X_t)=E(X_0)+\mu E\left(\int_0^t X_sds\right)+\sigma E\left(\int_0^t X_sdB_s\right)=E(X_0)+\mu \int_0^t E(X_s)ds+0,$$ which implies that $$\frac{d E(X_t)}{dt}=\mu E(X_t).$$ More generally, if $$dX_t=a(t,X_t)dB_t+b(t,X_t)dt,$$ then, under suitable integrability conditions, $$\frac{d E(X_t)}{dt}=E(b(t,X_t)).$$ $\endgroup$ – Did Nov 28 '15 at 21:54
  • 1
    $\begingroup$ After you write $E(X_t) = ... + \mu\int E(X_s)ds$ we again go back into differential form( phrase "which implies that"). This is only possible way, yes? Below, @Ilya wrote that there 2 options for calculating expectations: 1) taking expectation of Stochastic process; 2) taking expectation of SDE. Is that correct and these 2 options are equivalent? $\endgroup$ – Ievgenii Nov 28 '15 at 22:12

An easy way to calculate an expected value of SDE's solutions is to find an equation on $m(t) = \mathsf{E}[X_t]$. More precisely, from $$ dX_t = \mu X_tdt+\sigma X_t dB_t $$ taking expectations of both sides (while using that $\mathsf{E}[dB_t] = 0$) we obtain $$ dm(t) = \mu m(t)dt, $$ which gives a desired answer. Indeed, from the latter equation follows that $m(t) = m_0 \mathrm{e}^{\mu t}$. @Did has already described such method, but it may be useful for you to apply such method for other SDEs.

You should be aware of taking the expectation in SDE which is quite informal. On the other hand, you can equivalently write $$ X_t = X_0+ \int\limits_0^t\mu X_sds+\int\limits_0^t\sigma X_s dB_s. $$

Here you can take an expectation since on both sides there are just random variables (at each fixed moment of time). The latter integral has zero expectation since it is Ito integral.

  • $\begingroup$ can you explain more about your second option for calculating expectation straight from SDE? What I conclude from your answer: I can do both ways: 1) taking expectation of Stochastic process; 2) taking expectation of SDE. Are these 2 options equivalent? Are you implicitly using Fubini theorem in second case? (when you are taking $\mathbb{E} \int \sigma X_s dB_s$). And the last question, how do you compute: $\mathbb{E} \int \mu X_s ds$, I am confused. Thank you in advance! $\endgroup$ – Ievgenii Nov 28 '15 at 21:49

Even though it's pretty late I think this still may be helpful for a few of you.

First of all notice as $B_t$ is a geometric Brownian motion, by definition it is normally distributed with mean $0$ and variance $t$.

I.e. $B_t$ has the moment-generating function

\begin{equation}\label{moment}\tag{1} \mathbb{E}[\exp(u B_t)]=\exp(\frac{1}{2}u^2t),\qquad u\in\mathbb{R}. \end{equation}

Now we have for $X_t$ being a geometric Brownian motion

$$X_t= x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t+\sigma B_t\Big).$$

Therefore, applying the expectation value yields

\begin{align} \mathbb{E}[X_t]&= \mathbb{E}\Big[x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t+\sigma B_t\Big)\Big]\\ &=x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t\Big)\mathbb{E}\Big[ \exp\big(\sigma B_t\big)\Big]. \end{align}

So, by using equality \eqref{moment} with $u=\sigma$ we get

\begin{align} \mathbb{E}[X_t]&= x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t\Big)\mathbb{E}\Big[ \exp\big(\sigma B_t\big)\Big]\\ &=x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t\Big)\exp\big(\frac{1}{2}\sigma^2 t\big)\\ &=x_0 \exp\big(\mu t\big). \end{align}

This gives you the desired result.


Identify $E(\exp (\mu B_t ))$ with the moment-generating function of the normal distribution with mean zero and variance $t$.

  • 1
    $\begingroup$ I'm not downvoting, but it's worth noting that, while you've provided a good hint, applying it to the OP's development still does not yield the stated result. $\endgroup$ – cardinal May 23 '11 at 12:53
  • $\begingroup$ @cardinal: I noticed that; the answer addresses the OP's question "Do I just fail in calculating $E(\exp (\mu B_t ))$?" -- providing a quick way to calculate it. $\endgroup$ – Shai Covo May 23 '11 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.