This might be a stupid question but I'm going to ask it anyway because I can't find a way to do it.

I'm trying to find the expectation (and -- if possible -- higher moments) of the solution of the SDE

$$\mathrm d u = -u^3\mathrm d t + u^2 \mathrm d W.$$

Now it is easy to see that

$$\mathbb E u(t) = -\mathbb E \int_0^tu^3(s)\mathrm d s, $$


$$\frac{\mathrm d \mathbb E u(t)}{\mathrm d t} = -\mathbb E u^3(t), $$

but I don't know how to solve this: The third moment of $u$ is in turn related to higher moments and so on ...

  • $\begingroup$ Why do you think the expectation of the stochastic integral is $0$? You don't know whether that integral defines a martingale. $\endgroup$
    – Calculon
    Jul 4, 2016 at 8:50
  • $\begingroup$ Sorry, I forgot to mention that $W$ is Brownian motion, so the Ito integral has zero expectation. $\endgroup$ Jul 4, 2016 at 8:58
  • $\begingroup$ You should actually also add that the process $u(t)$ is adapted to the filtration of the Brownian motion $W(t)$ in order to make sure that the integral has zero expectation. Otherwise that integral can be whatever. $\endgroup$
    – RandomGuy
    Jul 4, 2016 at 8:59
  • $\begingroup$ @FasEtNefas That is still no guarantee that the stochastic integral is a martingale. You also don't know whether there is a process that satisfies the SDE you have given. So there may be no $u$ to compute the expectation of. $\endgroup$
    – Calculon
    Jul 4, 2016 at 9:00
  • $\begingroup$ Even if the adaptation property that @RandomGuy mentioned holds? Also, in that case I am also grateful for hints on how to prove existence (standard results for Lipschitz-continuous coefficients obviously does not hold). $\endgroup$ Jul 4, 2016 at 10:27

2 Answers 2


We assume that all integrability conditions are satisfied. Since \begin{align*} du(t) = -u(t)^3dt + u(t)^2 dW_t,\tag{1} \end{align*} we obtain that \begin{align*} du(t)^3 &= 3u(t)^2 du(t) + 3u(t)\,d\langle u, u\rangle_t\\ &=3u(t)^4dW_t. \end{align*} That is, \begin{align*} u(t)^3 = u(0)^3 + \int_0^t 3u(s)^4dW_s. \end{align*} Consequently, \begin{align*} E(u(t)^3) = u(0)^3. \end{align*} Moreover, from $(1)$, \begin{align*} u(t) = u(0) - \int_0^t u(s)^3 ds + \int_0^t u(s)^2 dW_s, \end{align*} Then \begin{align*} E\left(u(t)\right) &= u(0) -\int_0^t E(u(s)^3) ds\\ &=u(0)- u(0)^3t. \end{align*}

  • $\begingroup$ But now $\mathbb E(u(0)) = 0$ and $\mathbb E(u(1)) = - u(0)$? That doesn't look right... $\endgroup$ Jul 5, 2016 at 5:00
  • $\begingroup$ @FasEtNefas: See revision. If $u(0)=0$, then $E(u(t))=0$. $\endgroup$
    – Gordon
    Jul 5, 2016 at 12:50

Sorry for posting this as an answer instead of a comment (but I need to upload a picture). I made a simulation. The slightly thicker blue line is the empiric average of the sample paths shown. To me, it doesn't look like $\mathbb E u(t) = u(0) - u(0)^3t$ (a linear function).Simulation

  • $\begingroup$ Where did you get this question? $\endgroup$
    – Gordon
    Jul 5, 2016 at 13:51
  • $\begingroup$ It's a simplified version of a part of a more complex problem I have. $\endgroup$ Jul 5, 2016 at 13:58

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