# Stochastic process as an Ito integral over time-dependent integrand (without $t$ in upper limit)

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).

This post here (Stochastic process as an Ito integral with time-dependent integrand) seems to imply that (1) may be true, but doesn't answer (2).

• Seems that your integral does not define process in general. (I mean it defines a random variables evaluated at point t. i.e. It defines $Y(t)$ for some fixed $t$, you took in integral). In the cited question t is both parameter and time-variable of stochastic process defined by its integral. What I mean, $dY_{t}$ would probably be something like this $\int_{0}^{\infty} \frac{\partial f}{\partial t} (t,s) dW_s$. (as it seems to be parameter of non-random function). Commented Nov 10, 2021 at 9:44
• Thanks for your comment. Do you not think that it would define a stochastic process in the case of a suitably nice deterministic function $f(t,s)$? For example, if $f(t,s)=1$ for $s\in [t-\frac{1}{2},t+\frac{1}{2}]$ and $0$ otherwise, the integral reduces to $\int_{t-\frac{1}{2}}^{t+\frac{1}{2}} dW_s=W_{t+\frac{1}{2}}-W_{t-\frac{1}{2}}$. Is this not a continuous stochastic process? Commented Nov 10, 2021 at 9:59
• It is indeed. (Though it is some sort of cheating )) when you take parameter to act like that.). But even for that case, for fixed t you get just random variable. But in your case you have something more. It is hard to define $\frac{\partial f}{\partial t}$, without $\delta$ (Dirac) function. But with the use of $\delta$ function you'll indeed get correct answer. (I mean differential of found difference of Wiener process, is indeed the integral of two $\delta$ functions (at given points) with respect to Wiener. It seems to me.) Commented Nov 10, 2021 at 10:10
• I see what you mean in this case with the Dirac $\delta$. Do you think then that $dY_t = \int_0 ^\infty \frac{\partial f}{\partial t}(t,s) d W_s$ is well-defined for nicer functions than step functions (e.g. $f(t,s)=e^{\frac{-(t-s)^2}{2\sigma^2}}$)? Commented Nov 10, 2021 at 10:37
• To me the infinity in the integral seems strange and it makes this question different from the link you added. If the natural filtration is $\{ \mathcal F_t\}$, then $Y_t$ is $\mathcal F_0$-measurable for every $t.$
– UBM
Commented Nov 10, 2021 at 19:06

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 $$dY_t=\left(\int_0^\infty \partial_t f(t,s)\,dW_s\right)\,dt\,.$$
• The natural filtration of this process is simply ${\cal F}_t=\sigma(Y_s;s\le t)\,.$ The formula for $dY_t$ can also be viewed as a statement of the stochastic Fubini theorem. In other words, we recover $Y_t$ when we integrate $dY_t$ from $0$ to $t$ and exchange $dt$ and $dW_s\,.$ The question by @broken_urn thus boils down to 'when does the stochastic Fubini theorem hold?'. Looking at this paper I don't think we are missing anything. Commented Nov 10, 2021 at 13:48
• One more remark on $Y_t$ being a "process" : It's more natural to view $Y_t$ as a random variable that depends on a parameter $t$ and is differentiable w.r.t. $t\,.$ Likewise, one could view $t\mapsto Y_t$ as a $C^1$-curve that is random. The natural filtration of $Y_t$ clearly exists but is far larger than that of $W_t\,.$ The times of $Y$ and $W$ are very different. Commented Nov 11, 2021 at 12:29