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I am trying to compute this quantity:

$\frac{d}{dt}\int_{0}^{t} W_s ds $

Where $W_t$ is a Wiener process. Is there a theorem which tells how this can be computed?

I have tried https://en.wikipedia.org/wiki/Leibniz_integral_rule , however here the functional form is $\frac{d}{dx}\int_{a(x)}^{b(x)} (f(x,t) dt) $, that is the function $f$ takes 2 inputs - $x$ and $t$.

Using the above rule,

$\frac{d}{dt}\int_{0}^{t} W_s ds = W_t\frac{d}{dt}t - W_0\frac{d}{dt}0 + \int_{0}^{t}\frac{\partial{}}{\partial{t}}W_sds$
I am not sure what the term $\int_{0}^{t}\frac{\partial{}}{\partial{t}}W_sds$ means.

Any help is appreciated.

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    $\begingroup$ You are confusing the inputs for the Wiener process. The process is a function of the underlying state of the world and time, which you indicate as $s$, so the Wiener process is $W(\omega,s)$. So basically $t$ is a constant and you don't have that integral issue. So the answer should be $W_t$. So basically think of $t$ as a user input time that we want to see how the Wiener process will act up until. But the time variable for the process itself is different, which you wrote as $s$. Hope that makes sense. $\endgroup$ – Slade Feb 28 at 14:36
  • $\begingroup$ Please consider adding the tag "Homework" or "Self-study" $\endgroup$ – Sanjay Feb 28 at 15:11
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    $\begingroup$ Thanks for your reply, however your answer is not entirely clear to me. So $W(w,s)$ is the weiner process, where $w$ denotes the paths and $s$ denotes the time. If I fix $w$, then the path is fixed. Then, for that particular path, the integral from $0$ to $t$ should exist, i.e. $s$ will vary from $0$ to $t$. $\endgroup$ – SPaul Feb 28 at 15:13
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    $\begingroup$ Yes the integral exists, but once you take its derivative with respect to $t$, it follows the integral rule as you said. I was pointing out that the variable $t$ is different than the variable $s$, and therefore the Wiener process does not have any dependence on $t$. So for example, a student wants to simulate a Wiener Process from time 0 to time $450$. So here $t$ is $450$, but this does not affect the Wiener process's ($W(\omega,s)$) values for any $s$. So we'd have values for $W$ from 0 to 450, but the fact that $t = 450$ does not affect the Weiner process. $\endgroup$ – Slade Feb 28 at 15:30
  • $\begingroup$ I am not sure why you think you need a theorem to show the result. Since you are integrating with respect to ${ds}$, you are integrating (pathwise) continuous functions which enables you to use regular calculus rules. $\endgroup$ – phantagarow Feb 28 at 19:21
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You can actually use Leibniz here. And your computation above is correct. Just be aware that $\frac{d}{dt} W_s=0$. $W_s$ is a function of $s$ differentiating wrt $t$ is zero.

To sum up: $$\frac{d}{dt}\int_{0}^{t} W_s ds =W_t $$

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