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I was studying an example given in Variance of a time integral with respect to a Brownian Motion function

There we need to calculate the Variance of $I_t = \int\limits_{0}^{t} f \left(s \right) W_s ds$.

Then, it said that $E \left[ I_t^2 \right] = \int\limits_{0}^{t}\int\limits_{0}^{t} f \left(s \right) f \left(u \right) \min \left(s,u\right) dsdu$

Can someone please help to understand the detailed calculation on how it can be arrived?

It is also stated that $I_t$ follows a Normal distribution. What is the reason for that?

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This uses the autocorrelation of the Weiner process (proved in this post), ${\mathbb E}[W_s W_u] = \min(s,u)$

From your expression, \begin{align} {\mathbb E}[I^2_t] &= {\mathbb E}[\int_0^t f(s)W_sds \int_0^t f(u)W_u du]\\ &= {\mathbb E}[\int_0^t \int_0^t f(s)W_s f(u)W_u ds du]\\ &= \int_0^t \int_0^t f(s) f(u) {\mathbb E}[W_s W_u] ds du \end{align} where we moved the expectation into the integral as the Weiner terms are the only non-deterministic things

Then we substitute in the autocorrelation, and bingo! \begin{align} {\mathbb E}[I^2_t] &= \int_0^t \int_0^t f(s) f(u) {\mathbb E}[W_s W_u] ds du\\ &= \int_0^t \int_0^t f(s) f(u) \min(s,u) ds du \end{align}

as required

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  • $\begingroup$ Ahh and as for the normality of $I_t$ - remember that the sum of any number of normal variables is still normally distributed, even if they're correlated. An integration is the limit of a sum, so hopefully it's easy enough to believe that the sum of these brownians (weighted at each step by the local value of $f(s)$) must also be brownian $\endgroup$ – StackG Aug 4 '20 at 11:37

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