How to prove that the stochastic integral process is gaussian?

Question

I would like to prove that for a $C^1$-function f and a Wiener process W, the integral process defined by $$ Y_t:= \int_0^t f (s)dW_s := f (t)W_t -\int_0^t W_s f'(s)ds $$ Is a centered gaussian process, i.e. , that for any choice of $t_0,...t_n $ the vector $(Y_{t_0},...,Y_{t_n}) $ has multivariate normal distribution.

Since f is $C^1$ (hence f' is continuous) I assume that I could use the already proved fact that for any continuous function h the process $$V_t:= \int_0^t W_s h(s)ds $$ is a centered gaussian process with covariance function $r (s,t) $ (known).

Since for every finite linear combination $$ \sum \alpha_i Y_{t_i} =\sum \alpha_i W_{t_i}f (t_i)-\sum \alpha_i V_{t_i}$$ I've tried using characteristic functions as follows:

$$ E\left( \exp\left(i \sum \alpha_i Y_{t_i} \right)\right) = E\left( \exp\left( i \sum \alpha_i W_{t_i}f (t_i)\right)\exp\left( - i \sum \alpha_i V_{t_i} \right)\right)$$

but I am not sure about how to proceed and how to use the known characteristic function of $\sum \alpha_i V_{t_i}$. Could anyone please give me a hint? What am I missing?

pmo

Answer 1

It's enough that $f$ be Lebesgue measurable and $\int_0^t [f(s)]^2\,ds<\infty$ for each $t$. Indeed, in this case your stochastic integral is well defined. Moreover, if we set $A_t:=\int_0^t[f(s)]^2\,ds$, then $Y$ is a (mean zero) continuous martingale with quadratic variation $\langle Y\rangle_t=A_t$. Let $\tau(t):=\inf\{s:A_s>t\}$. Then $Z_t:=Y_{\tau(t)}$ is a continuous martingale with quadratic variation $t$; by Levy's theorem, $Z$ is Brownian motion. Because $A$ is deterministic, $Y_t=Z_{A_t}$, $t\ge 0$, is a deterministic re-parametrization of a Gaussian process, and is therefore also a Gaussian process.