$ I_t = \int_0^t e^{i W_s} dWs $ where $W_s$ is the standard brownian motion and $i$ is the complex number. Any help will be appreciated!

  • 3
    $\begingroup$ Evaluate sounds too vague for me. You want a close formula or you want to simulate it... it would be nice to have more details. $\endgroup$ Jan 26, 2017 at 6:08
  • $\begingroup$ What do you mean "it looks like"? What does the "It" refer to? $\endgroup$
    – Gordon
    Jan 26, 2017 at 16:24
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    $\begingroup$ @Pandaaaaaaa you changed the question pretty drastically, how can you be sure this is what he wanted originally? $\endgroup$
    – SRKX
    Feb 20, 2017 at 1:08
  • $\begingroup$ @SRKX I have seen the same problem somewhere. It asks for variance and mean. $\endgroup$ Feb 20, 2017 at 1:13

2 Answers 2


This process is martingale and we have

$$ E[I_t|t=0]=0 $$

To find the variance, let's write it into differential form $$ dI_t =e^{iW_t}dW_t $$ Apply Ito's isometry $$ Var(I_t)=\int_0^tE[e^{2iW_s}]ds $$ Apply MGF of normal $$ Var(I_t)=\int_0^te^{\frac{1}{2}(2i)^2s}ds=\int_0^te^{-2s}ds=\frac{1-e^{-2t}}{2} $$ Please let me know if anything is incorrect.


The expectation of an Itô integral (and a Wiener integral) is always zero.

  • $\begingroup$ This does not answer the question and rather has the quality of a comment. $\endgroup$ Feb 20, 2017 at 22:36

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