Using Ito Isometry formula

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How would I use the Ito Isometry formula to compute,

$$E\left [ \left ( \int_{0}^{t}g(s,W_s)dW_s \right )^2 \right ]$$

for the case $g(t,W_t)=(1+t^2+W_t)$ with $W_t$ being Brownian motion.

I would really appreciate any help on this. Thanks

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1 Answer

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I write a solution, also as an exercise for me...

Step 1: apply Ito isometry

$I=E[\left(\int_0^t g(s,W_s) dW_s\right)^2]=E[(\int_0^t g(s,W_s)^2 ds]$

Step 2: Insert and expand

$I=E[\int_0^t (1+s^4+W_s^2 +2s^2+2sW_s+2W_s )ds]$

Step 3: Exchange expectation with integral

This way we can prove that:

$E[\int_0^t W_s ds]=\int_0^t E[W_t]ds=0$

$E[\int_0^t sW_s ds]=\int_0^t sE[W_t]ds=0$

$E[\int_0^t W^2_s ds]=\int_0^t E[W^2_s] ds=\int_0^t s ds=t^2/2$

The rest is just deterministic integrals:

$I=t+t^5/5+t^2/2+2t^3/3$

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