What is the integral of a family of diffusion processes?

Question

Let $S$ be an infinite subset of $[0,1]$. For all $s \in S$, let W_s(t) be a standard Wiener process.

Definite P(s)_t = \mu(P,s,t) dt + \sigma(P,s,t) dW^s_t

Can we characterize? $$F_t= \int_S P(s)_t ds$$

Answer 1

Here's my take on this,

Let $S_t(s) = \int_0^t \mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)dW_t$

Then, $S_t(s)$ is a normally distributed random variable parametrized by it's expectation and variance.

To see that $S_t(s)$ is normal,

$S_t(s) = \int_0^t \mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)dW_t$,

which is equal to a constant $\int_0^t \mu(t,S_t,s)dt = A + \text{an infinite sum of normal random variables since } dW_t \text{ is distributed }$ $\mathcal N(0,1).$ The sum of normal random variables is again normally distributed, hence, the limit of the sum, the integral, is normally distributed by weak convergence.

$\Bbb E[S_t(s)] = \Bbb E[\int_0^t\mu(t,S_t,s)dt] + \Bbb E[\int_0^t \sigma(t,S_t,s)dW_t]$

$= \int_0^t\mu(t,S_t,s)dt + \int_0^t \sigma(t,S_t,s)\Bbb E[dW_t]$

Where we are able to pass the Expectation operator under the integral by dominated convergence because the integral is a limit of sums, so we can pass the expectation operator through the integral. Also assuming that $\sigma(t,S_t,s)$ is deterministic.

$\Rightarrow \Bbb E[S_t(s)] = \int_0^t\mu(t,S_t,s)dt$ Because $\Bbb E[dW_t] = 0$

Similarly,

$\Bbb E[S_t(s)^2] = \Bbb E[(\int_0^t \sigma(t,S_t,s)dW_t)^2] = \int_0^t \Bbb E[\sigma(t,S_t,s)^2dW_t^2]$ By Ito Isometry.

$\Rightarrow \Bbb E[S_t(s)^2] = \int_0^t \Bbb E[\sigma(t,S_t,s)^2dt] = \int_0^t \sigma(t,S_t,s)^2dt \Rightarrow Var(S_t(s)) = \Bbb E[S_t(s)^2] - \Bbb E[S_t(s)]^2$

I hope that's what you were asking.