# Calculate Variance of a function of Stochastic Process

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

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

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}
• 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 – StackG Aug 4 '20 at 11:37