Differentiating the Time Integral of a Wiener Process

Summary

The document asks how to differentiate with respect to the upper limit in the time integral of a Wiener process. The key distinction is between the integration variable and the differentiation variable: within the integral, the process is indexed by the integration variable, so it has no separate dependence on the upper-limit variable.

Applying the Leibniz rule therefore leaves the integrand evaluated at the moving upper endpoint, while the partial derivative term vanishes. The resulting derivative is the Wiener process at that endpoint. The answer gives this rule directly but does not discuss regularity or distinguish ordinary pathwise integration from stochastic integration; here the integral is an ordinary time integral of the process.

Key ideas

• The integral is an ordinary time integral of the Wiener process.
• Leibniz differentiation with respect to the upper limit evaluates the integrand at that limit.
• The integrand’s partial derivative with respect to the upper-limit variable is zero.
• The derivative of the integral is the Wiener process evaluated at the endpoint.

Tags

# Differential of integral of Wiener process over time


# Differential of integral of Wiener process over time












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.

## Answer by Sanjay (score 3)

https://quant.stackexchange.com/a/44335

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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