I have such stochastic process with which I struggle all day, finally I found 2 mistakes, however answer is still unsatisfying. $$X_t = atW_t^2 - \int_0^t(W_s^2+s)ds,$$ I need to check if it is a martingale. I simply write Ito formula for $X_t(t,W_t,S_t)$, where I denote by $S_t = \int_0^tW_s^2ds$. So I apply it, and get: $$dX_t= (aW_t^2 - t -W_t^2 +at )dt + 2atW_tdW_t,$$ I pick function $f_t$, which should be equal to zero if the process is a martingale. $$aW_t^2 - t -W_t^2 +at = 0,$$ so my $a$ is $a = 1$.

BUT! I have decided to check it substituting the $a$ into the $X_t$ and calculating expectation(using martingale property ($\mathbb{E}(X_t \mid \mathcal{F_s}) = X_s$), and I get problems. As far as I understand I should get: $$\mathbb{E} \left(tW_t^2- \int_0^t(W_s^2+s)ds \mid \mathcal{F_s} \right) = sW_s^2 - \int_0^s(W_u^2+u)du$$ But what I get is: $$\mathbb{E}(tW_t^2 \mid \mathcal{F_s})=tW_s^2 + t^2 -ts$$ $$\mathbb{E} \left( \int_0^tW_s^2ds \mid \mathcal{F_s} \right)= \int_0^sW_u^2du +st -s^2$$ $$\mathbb{E} \left(t- \int_0^tsds \mid \mathcal{F_s} \right) = t - t^2/2$$ Substituting: $$\mathbb{E} \left(tW_t^2 - \int_0^t(W_s^2+s)ds \mid \mathcal{F_s} \right) = tW_s^2 + t^2 -ts - \int_0^sW_u^2du - st + s^2 -t^2/2$$ which is does not look like needed: \begin{align} \mathbb{E} \left(tW_t^2- \int_0^t(W_s^2+s)ds \mid \mathcal{F_s} \right) &= sW_s^2 - \int_0^sW_u^2du - s^2/2\\ & = tW_s^2 - \int_0^sW_u^2du+ s^2 - 2st +t^2/2 \end{align}


Without your (detailed) calculations it is hard to say what you did wrong. So let's do it step by step:

It is not difficult to check that $(W_t^2-t)$ is a martingale. Therefore, we have

$$\begin{align*} \mathbb{E}(t W_t^2 \mid \mathcal{F}_s) &= t \mathbb{E}((W_t^2-t)+t) \mid \mathcal{F}_s) \\ &= t (W_s^2-s+t) = t W_s^2 + t^2-ts. \end{align*}$$

That agrees with your result for this term. Now the next one.

$$\begin{align*} \mathbb{E} \left( \int_0^t W_u^2 \, du \mid \mathcal{F}_s \right) &= \int_0^s W_u^2 \, du + \mathbb{E} \left( \int_s^t W_u^2 \, du \mid \mathcal{F}_s \right) \\ &= \int_0^s W_u^2 \, du + \mathbb{E} \left( \int_s^t ((W_u-W_s)+W_s)^2 \, du \mid \mathcal{F}_s \right) \\ &= \int_0^s W_u^2 \, du + \int_s^t \mathbb{E}((W_u-W_s)^2) \, du + 2W_s \int_s^t \underbrace{\mathbb{E}(W_u-W_s)}_{0} \, du \\ &\quad + W_s^2 (t-s) \\ &= \int_0^s W_u^2 \, du + \int_s^t (u-s) \, du + W_s^2 (t-s) \\ &= \int_0^s W_u^2 \, du + \frac{(t-s)^2}{2} + W_s^2 (t-s). \end{align*}$$

This is different from your result. The third one is again correct. Adding all up, we get

$$\begin{align*} \mathbb{E}(X_t \mid \mathcal{F}_s) &= t W_s^2 + t^2-ts - \left( \int_0^s W_u^2 \, du + \frac{(t-s)^2}{2} + W_s^2 (t-s) \right) - \frac{t^2}{2} \\ &= s W_s^2 - \int_0^s (W_u^2-u) \, du = X_s \end{align*}$$

  • $\begingroup$ I thought that we can apply expectation to $\int_s^t W_u^2 du$ with no decomposition into $(W_u-W_s),W_s$ parts as it is obvious that we are going to look at interval $[s,t]$ which is not measurable by $\mathcal{F_s}$, no? This is why I got there such a result. One more question: I have few questions almost about the same, should I extend this post, or I can ask here? $\endgroup$ – Ievgenii Dec 8 '15 at 20:30
  • 1
    $\begingroup$ @Ievgenii Note that $W_u^2$ is not independent of $\mathcal{F}_s$. Therefore, $$\mathbb{E} \left( \int_s^t W_u^2 \, du \mid \mathcal{F}_s \right) \neq \mathbb{E} \left( \int_s^t W_u^2 \, du \right).$$ (And thanks, there very indeed several "du" missing.) $\endgroup$ – saz Dec 8 '15 at 20:32
  • $\begingroup$ Sorry, but this is still unclear. I am afraid, I can get this mistake again in some time. Why is it not independent? We have bounds of integral that strictly say we have $u$ which is from $s$ to $t$. Actually far before I wrote post, I was trying to interpret integral as $\sum_{i=0}^n W^2_{u_i}(u_{i+1}-u_i),$ and I thought that $\mathbb{E} (W^2_{u_i} \big\vert \mathcal{F_s})$ is just variance. If you say that because only increments are independent this will not help me much in understanding, gg. It seems to me, that I have to get used to it $\endgroup$ – Ievgenii Dec 8 '15 at 20:44
  • 1
    $\begingroup$ @Ievgenii No, it is not. The Brownian motion has independent increments, but this does "only" mean that $W_u-W_s$ is independent of $\mathcal{F}_s$ for $u \geq s$ and not that $W_u$ is independent of $\mathcal{F}_s$. (Not sure how to help you understanding this. One can show explicitly that $W_u$ is not independent from $\mathcal{F}_s$, e.g. by calculating $\mathbb{E}(W_s W_u)$...but I guess that also doesn't help much in understanding.) $\endgroup$ – saz Dec 8 '15 at 20:47
  • 1
    $\begingroup$ @Ievgenii The second one looks raster nasty; usually I prefer checking the martingale property directly, but in this case I would go for a different reasoning. But yes, please ask a new question - otherwise this is getting a mess. $\endgroup$ – saz Dec 8 '15 at 20:59

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.