For the stochastic integral, where $W_t$ is a Wiener process, I am trying to find the mean of $X_t = \int_0^t sW_sdW_s$. I have read before that any stochastic integral with $dWt$ has mean zero, but I dont know if it extends to cases where I have a random variable in the integrand as well. My approach is to decompose the integral $X_t$ into:

$$ \int_0^t sW_sdW_s = \lim_{n \to \infty}\sum_{j=0}^{n-1}t_jW_{t_i}(W_{t_{i+1}}-W_{t_i}) $$


$$ E\left(\int_0^t sW_sdW_s\right) = \lim_{n \to \infty}\sum_{j=0}^{n-1}E\left(t_jW_{t_i}(W_{t_{i+1}}-W_{t_i})\right) $$

I believe that I can separate the terms in the expectation into:

$$ E\left(t_jW_{t_i}(W_{t_{i+1}}-W_{t_i})\right) = E\left(t_jW_{t_i}\right)E\left(W_{t_{i+1}}-W_{t_i}\right) $$

I am not sure how to find $E\left(t_jW_{t_i}\right)$, although I know it is finite so it doesn't matter because $E\left(W_{t_{i+1}}-W_{t_i}\right) = 0$

  • 1
    $\begingroup$ Once the integrand is adapted, the expectation is zero. $\endgroup$
    – zhoraster
    Apr 8, 2016 at 20:08
  • 1
    $\begingroup$ Also you simply have $E[t_jW_{t_i}] = t_jE[W_{t_i}] = 0 $ $\endgroup$
    – BGM
    Apr 9, 2016 at 2:32


Your Answer

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

Browse other questions tagged or ask your own question.