# Stieltjes convolution with white noise

I'm looking for a reference that would discuss a Stieltjes convolution between a wiener process and a function of bounded variation. Additionally, I had a question about this sort of convolution.

Is Stieltjes convolution between a function $F(t)$ with bounded variation and a wiener process, $W(t)$ commutative?

if $F(t) \bigotimes W(t)=\int F(t-x)dW(x)$

then does $W(t) \bigotimes F(t)= \int W(t-x)dF(x) = F(t) \bigotimes W(t)$?

Additionally, I'd like to evaluate this integral numerically when I have no formula for F, but a list of points ${t_0,t_1,...t_{N-1} }$ and the values of $F(t)$ at those points. Any references that would point me in this direction would be appreciated.

• What is the "white noise process" $W$? Is it a Brownian motion? If not, what does $dW$ mean? Also, what is $F$? Any bounded variation function? – George Lowther Jun 28 '12 at 1:26
• Thanks for the comment. I clarified the question. By W I meant Brownian motion, and F is any function with bounded variation. – ncRubert Jun 28 '12 at 15:42
• The answer is yes. But it is just a direct application of Ito's formula, and I wouldn't call it research level. So, I voted to close. It'd fit better at math.stackexchange in my opinion. – George Lowther Jun 29 '12 at 0:12
• But I'll probably answer it here anyway, as it is better to have an answer than just left open with no responses. – George Lowther Jun 29 '12 at 0:13

This is just an application of Ito's formula. Normally, Brownian motion $W(t)$ is only defined for nonnegative times $t$, so I'm assuming that $W(t)=0$ for $t\le0$, although it doesn't really matter much to the answer.
For a fixed $t$, set $V_s=F(t-s)$, which is a deterministic finite variation process. As $V$ is of finite variation, the quadratic covariation $[V,W]$ vanishes. So, by Ito's formula,
$$\int_{-\infty}^TF(t-s)\,dW(s)=\int_{-\infty}^TV_s\,dW(s) =V_TW(T)-\int_{-\infty}^TW(s)dV_s.$$ Using the change of variables $u=t-s$ in the final (Stieltjes) integral, $$\int_0^TF(t-s)\,dW(s)=F(t-T)W(T)+\int_{t-T}^\infty W(t-u)dF(u)$$ Using Ito's isometry, the integral on the left converges as $T\to\infty$ iff $\int_{-\infty}^tF(s)^2\,ds$ is finite for all $t$ (convergence is in $L^2$, although it can be shown to be in $L^p$ for all $1\le p <\infty$ and almost sure). The first term on the right vanishes in $L^2$ as $T\to\infty$ iff $F(-t)^2t\to0$ as $t\to\infty$. So, I'm assuming that these conditions hold. Taking the limit in $L^2$ as $T\to\infty$ gives $F\otimes W=W\otimes F$.
To compute this, $F\otimes W$ is just a standard stochastic integral, and a Riemann sum approximation method should work fine. By Ito's Isometry, $F\otimes W(t)$ is normally distributed with mean $0$ and variance $\int_0^t F(s)^2\,ds$. If you want to get faster convergence in distribution for $F\otimes W$, you could do something like use an improved approximation to the integral for the variance (e.g., Simpson's rule) and then use the corresponding weights to adjust the Riemann sum approximation to obtain the correct variance.
• @ George: I agree in general, but have a small nitpick. $\int_{-\infty}^t F(s)^2 ds$ finite for all $t$ is not enough, it has to be uniformly bounded (or equally $\int_{-\infty}^{\infty}F(s)^2 ds< \infty$). Otherwise the indicator function on the nonnegative reals provides an easy counterexample. – Stephan Sturm Jun 29 '12 at 17:06
• In that example you get $F\otimes W(t)=W\otimes F(t)$, which seems fine to me. – George Lowther Jun 29 '12 at 19:16
• Should have added $=W(t)$ in that equality. – George Lowther Jun 29 '12 at 19:16