$(\int_0^uf(r,u)dB_r)_{u \geq 0}$ is Gaussian

Question

Let $f:\mathbb{R}_+\times \mathbb{R}_+\to \mathbb{R}$ such that for all $u>0,\int_0^u(f(r,u))^2dr<\infty.$

Prove that $(\int_0^uf(r,u)dB_r)_{u \geq 0}$ is a Gaussian process.

We note here that $f$ depends of $u$ so that $\int_0^uf(r,u)dB_r$ is not a martingale and we can't apply Ito's formula to prove the exercise.

Perhaps we should find a sequence of Gaussian processes converning in probability to $\int_0^uf(r,u)dB_r$.

Any ideas how to prove the result?

Answer 1

If $h(r)$ is locally square-integrable and $t>0$, then the Ito integal $\int_0^t h(r)\,dB_r$ is a mean zero Gaussian random variable.

For ease of reference define $X_u:=\int_0^u f(r,u)\, dB_r$. Fix $u>0$ and apply the preceding remark with $h(r) = 1_{r\le u}f(r,u)$ and $t=u$ to see that $X_u$ is normally distributed, for each $u>0$. More generally, if $b_1,b_2,\ldots,b_n$ are real numbers and $0<u_1<u_2<\cdots<u_n$, apply the initial remark to $$ h(r) =\sum_{k=1}^n 1_{\{r\le u_k\}}b_k f(r,u_k) $$ to see that $$ \sum_{k=1}^n b_k X_{u_k} $$ is normally distributed. This proves that $\{X_u: u>0\}$ is a Gaussian process.