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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).

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    $\begingroup$ 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). $\endgroup$
    – kolobokish
    Commented Nov 10, 2021 at 9:44
  • $\begingroup$ 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? $\endgroup$
    – broken_urn
    Commented Nov 10, 2021 at 9:59
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    $\begingroup$ 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.) $\endgroup$
    – kolobokish
    Commented Nov 10, 2021 at 10:10
  • $\begingroup$ 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. $f(t,s)=e^{\frac{-(t-s)^2}{2\sigma^2}}$)? $\endgroup$
    – broken_urn
    Commented Nov 10, 2021 at 10:37
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    $\begingroup$ 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 $t.$ $\endgroup$
    – UBM
    Commented Nov 10, 2021 at 19:06

1 Answer 1

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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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  • $\begingroup$ Many thanks, this is exactly what I was looking for! Does the standard Leibniz rule justify bringing the derivative inside the integral here? $\endgroup$
    – broken_urn
    Commented Nov 10, 2021 at 12:25
  • $\begingroup$ What is the natural filtration of this "process"? $\endgroup$
    – UBM
    Commented Nov 10, 2021 at 13:09
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    $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Commented Nov 10, 2021 at 13:48
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    $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Commented Nov 11, 2021 at 12:29

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