Conditions for Stochastic Integrals with Time-Dependent Kernels

Summary

The document explains when integrating a deterministic function against a Wiener process over an unbounded interval defines a stochastic process. For each time value, the kernel must be square-integrable over the integration variable; under that condition, the integral exists as a random variable and can be viewed as a process indexed by time.

It also gives a differential expression when the kernel is differentiable in time: the process has a time derivative formed by integrating the kernel's time derivative against the same Wiener process, multiplied by the time increment. This extends the familiar Itô integral with a moving upper limit. The result is stated with integrability conditions, but the note does not develop regularity requirements for sample paths or justify exchanging differentiation and stochastic integration in greater detail, so those assumptions need care in applications.

Key ideas

• A deterministic kernel must be square-integrable over the integration variable for the Wiener integral to exist at a fixed time.
• The family of such integrals indexed by time defines a stochastic process.
• When the kernel is differentiable in time and suitable integrability holds, its differential is given by integrating the kernel's time derivative against the Wiener process.
• The stated conditions do not elaborate on all regularity assumptions needed to differentiate under the stochastic integral.

Tags

# Stochastic process as integral over window function


# Stochastic process as integral over window function












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:

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

- 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.

## Answer by Kurt G. (score 1, accepted)

https://quant.stackexchange.com/a/68729

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 $$ dY_t=\left(\int_0^\infty \partial_t f(t,s)\,dW_s\right)\,dt\,. $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.