$ 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!
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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. $\endgroup$ – MJ73550 Jan 26 '17 at 6:08
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$\begingroup$ What do you mean "it looks like"? What does the "It" refer to? $\endgroup$ – Gordon Jan 26 '17 at 16:24
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1$\begingroup$ @Pandaaaaaaa you changed the question pretty drastically, how can you be sure this is what he wanted originally? $\endgroup$ – SRKX Feb 20 '17 at 1:08
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$\begingroup$ @SRKX I have seen the same problem somewhere. It asks for variance and mean. $\endgroup$ – Pandaaaaaaa Feb 20 '17 at 1:13
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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.
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The expectation of an Itô integral (and a Wiener integral) is always zero.
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$\begingroup$ This does not answer the question and rather has the quality of a comment. $\endgroup$ – LocalVolatility Feb 20 '17 at 22:36
