Consider the following stochastic integral of a deterministic function $f(t,s)$ with respect to the Wiener process $W_s$:

$$\int_0^\infty f(t,s) d W_s$$

My questions are:

  1. Is such an integral suitably well-defined that it defines a stochastic process $Y_t$?

  2. If so, is there a simple expression for $dY_t$?

I'm aware that the Ito integral with $t$ as the upper limit in the integration defines a stochastic process, but it is unclear what happens in this more general case (we can recover the usual case by $f(t,s)=f(s)(1-\Theta(s-t))$, where $\Theta(x)$ is the Heaviside step function). Apologies in advance if this question has already been answered elsewhere.


1 Answer 1


The process $$ Y_t=\int_0^\infty f(t,s)\,d W_s $$ is well defined when the usual condition $P[\int_0^\infty f^2(t,s) ds<\infty]=1$ holds which in your deterministic case boils down to $\int_0^\infty f^2(t,s) ds<\infty$. When $f(t,s)$ is differentiable in $t$ and $\int_0^\infty \partial_t f^2(t,s) ds<\infty$ then $$ dY_t=\left(\int_0^\infty \partial_t f(t,s)\,dW_s\right)\,dt\,. $$

  • $\begingroup$ Many thanks for your answer. I ended up asking the same question on math.stackexchange here math.stackexchange.com/questions/4302036/… . I would be happy to select this as an answer both here and there if you are willing to post it over? $\endgroup$
    – broken_urn
    Nov 10, 2021 at 10:45
  • $\begingroup$ @broken_urn . I posted it over. $\endgroup$
    – Kurt G.
    Nov 10, 2021 at 11:42

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