Distribution of a stochastic integral $\int_0^t W_1(s) dW_2(s)$ with independent $W_1$ and $W_2$.

Question

Consider two independent 1-d Wiener processes $W_1(t)$ and $W_2(t)$ with $W_1(0) = W_2(0) = 0$.

I would like to know the distribution of the process $$Y(t) = \int_0^t W_1(s)dW_2(s). \tag{1}$$

I reason that since $W_1(t)$ and $W_2(t)$ are independent, we can treat (1) using the usual calculus. For example, we can use the integration by parts to rewrite (1) as: $$\int_0^t \underbrace{W_1(s)}_{u}\underbrace{dW_2(s)}_{dv} = W_1(t)W_2(t) - \int_0^t W_2(s)dW_1(s).$$ Now, we can clearly see that we have a product $W_1(t) W_2(t)$ of two normally distributed variables. Therefore, (1) can't be normal.

Also, using a partition $P_N = \{t_0, \ldots, t_N\}$ of the interval $[0,t]$, we can show: $$\mathbb{E}\left[ \int_0^t W_1(s)dW_2(s) \right] = \lim_{N \rightarrow \infty}\sum_{j=0}^{N} \mathbb{E}[W_1(t_j)]\mathbb{E}[W_2(t_{j+1})-W_2(t_j)] = 0,$$ and $$\mathbb{E}\left [ \left( \int_0^t W_1(s)dW_2(s) \right)^2 \right] = \int_0^t \mathbb{E}[W_1(s)^2]ds = \int_0^t s \ ds = \frac{t^2}{2}.$$ So, the mean and variance are zero and $t^2/2$, respectively, but I have no idea what the distribution is and if it has a closed form at all.

If it is tough to get an analytical description of the density function of $Y(t)$, would it be reasonable to approximate the distribution with $N(0, t^2/2)$?

Answer 1

For your approximation, the first problem that you have to solve is the following: let $X_i,Y_j $ independent $N(0,1)$, what is the distribution of $$A_N=\sum_{j=1}^N(\sum_{i=1}^j X_i)Y_j$$? Clearly $$E(e^{it A_N})=E(E(e^{itA_N}|X))=E(e^{-\frac{t^2}{2}\sum_{j=1}^N(\sum_{i=1}^j X_i)^2})=E(e^{-\frac{t^2}{2}X^T\Sigma X})$$$$=\frac{1}{(\det (I_N+t^2\Sigma))^{1/2}}=\prod_{i=1}^N\frac{1}{\sqrt{1+t^2\lambda_i}}$$ where $\Sigma$ is the matrix of the quadratic form $X\mapsto \sum_{j=1}^N(\sum_{i=1}^j X_i)^2$ with eigenvalues $\lambda_i.$ I think that $\Sigma$ have its eigenvalues quite explicit, but it takes a bit of time to remember. Anyway since $$\frac{1}{\sqrt{1+t^2}}=E(e^{it{\sqrt{2V}Z}})$$ where $V\sim \Gamma(1/2)$ is independent of $Z\sim N(0,1)$ we get that $$ A_N=\sum_{i=1}^N \lambda_i\sqrt{2V_i}Z_i$$ is a sum of iid rv.

The quick calculation above is probably full of little coefficient mistakes and missing square roots. I will compute the eigenvalues but I am sure that you or somebody of the site will do it before.