Looking for answers on the site I came across this answer with 3 upvotes where I am having trouble to understand an integral. Having not enough reputation (SE requires 50 reputation at least and I did not have access to chat until after editing) I cannot comment the post and ask Sasha, the answerer for clarification.

The said user stated: If $X := \int_1^2 W_s^2 \, dW_s$ then $E[X]=0$ where $W_s$ is a Brownian motion.

Yet I cannot understand how he came to this conclusion.

I followed the next steps and started from a more general question starting with $X=\int_0^t W_s^2dW_s$ : 1)Ito's isometry tells me that :$$ E\left[ \left(\int_0^t W_s \, dW_s \right)^2 \right] = E \left[\int_0^t W_s^2 \, ds \right]$$

2) Fubini's theorem (if I did unsterdand correctly from this another answer by Byron Schmuland) that i have: $$ E \left[\int_0^t W_s^2 \, ds \right] = \int_0^t E[W_s^2] \, ds = \int_0^t s \, ds = \frac{t^2}{2}$$

3) applying this to the particular case i would end up with this primitive evaluated in the following bounds : $$\left[\frac{s^2}{2}\right]^2_1=1.5$$

Which is different than $E[X] =0$

I cannot see what I am doing wrong. On the form on the question are the \left \right brackets necessary for readability?

  • 2
    $\begingroup$ What you are doing is to calculate $\mathbb{E}(X^{\color{red}{2}})$. Itô's isometry allows you to compute $\mathbb{E}(X^{\color{red}{2}})$; you are interested in $\mathbb{E}(X)$! $\endgroup$
    – saz
    Aug 13, 2017 at 8:28
  • $\begingroup$ Thank you very much for pointing that out! I will redo the maths and editing $\endgroup$
    – RandowMalk
    Aug 13, 2017 at 8:32
  • $\begingroup$ Asking the OP in a comment to their answer was not an option because? (You did not even signalled them you were discussing the merits of their answer elsewhere, so much for courtesy.) $\endgroup$
    – Did
    Aug 13, 2017 at 9:01
  • 1
    $\begingroup$ The site tells me i need 50 reputation to comment a post. Also the user Sasha (the one who came with the answer) did seem to come sporadically on the site when i looked at the frequency of his answers and questions. (1 every ~ month) and did not want to bother him for a 4 year old answer. And i am also unaware on the best practices of the site. The introducing tour doesn't tell you every step and etiquette. When writing any doc, I do put references but i do not send a notice to the author. I assumed it was the same here. Sorry if I was wrong. $\endgroup$
    – RandowMalk
    Aug 13, 2017 at 9:14
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    $\begingroup$ @Did I politely invite you to criticize my behaviour / mistakes in the chat (e.g. C.R.U.D.E where I found you or a private one). But this debate is irrelevant to the current answer or the comment section and has its place in the meta (again from my understanding of the mechanics of the site). $\endgroup$
    – RandowMalk
    Aug 13, 2017 at 10:43

1 Answer 1


What you are doing is to calculate $\mathbb{E}(X^{\color{red}{2}})$. Itô's isometry allows you to compute $\mathbb{E}(X^{\color{red}{2}})$; you are interested in $\mathbb{E}(X)$!

It is well-known the stochastic integral $$\int_0^t H_s \, dW_s$$ has expectation zero for any (nicely measurable) function $H$ such that $$\mathbb{E} \left( \int_0^t H_s^2 \, ds \right)<\infty$$ for any $t \geq 0$. To prove this, recall that

$$M_t := \int_0^t H_s \, dW_s$$

is a martingale which implies that $$\mathbb{E}(M_t)= \mathbb{E}(M_0)=0, \qquad t \geq 0$$ since martingales have constant expectation.

  • $\begingroup$ thanks for showing me how to edit, but why did you delete the equality and replace with :=? Because the integral is random and i can't put a strong equality? $\endgroup$
    – RandowMalk
    Aug 13, 2017 at 9:08
  • 3
    $\begingroup$ @RandowMalk What do you mean by "strong inequality"? For me ":=" reads as "defined as" $\endgroup$
    – saz
    Aug 13, 2017 at 11:24

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