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

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 $$N(0, t^2/2)$$?

• have you seen this? math.stackexchange.com/questions/1298661/…
– SBF
Commented Jun 5 at 4:34
• I believe, this paper answers precisely your question samos.univ-paris1.fr/archives/ftp/preprints/samos239.pdf the result itself apparently is given in [6] and [17], but I'm not sure how's your French
– SBF
Commented Jun 5 at 6:08
• @SBF, thank you very much! The first paper is almost exactly what I need; unfortunately, the integral has fixed bounds, and the reference paper is in French. I'll try to read it anyway. Commented Jun 5 at 16:02
• @SBF, the paper trail brought me to en.wikipedia.org/wiki/L%C3%A9vy%27s_stochastic_area Commented Jun 5 at 16:15
• I think you should be able to rescale your original integral to the unit interval. Will that fully answer your question?
– SBF
Commented Jun 7 at 2:53

## 1 Answer

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.