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Why the Time Integral of Brownian Motion Has Increment Wₜ dt

Article Quant Q&A · Author: rarnor
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Summary

The document addresses a stochastic calculus question about the differential of the time integral of a Brownian motion. Its accepted explanation treats the integral as a definite integral with a moving upper limit. Extending that limit by a small time increment adds the integrand's current value, multiplied by the length of the extension, giving an increment of Wₜ dt.

This clarifies that the integral is accumulated with respect to ordinary time, rather than directly with respect to Brownian motion. The questioner's proposed partial derivative with respect to Wₜ is not the applicable way to differentiate this accumulated quantity. A second response asks the reader to specify the function needed to apply Itô's lemma, but does not develop that analysis. The treatment is brief and addresses the differential intuition; it does not discuss stochastic integration with respect to Brownian motion or applications to pricing and trading.

Key ideas

  • A definite time integral of Brownian motion changes by the current integrand times the added time interval.
  • The moving upper integration limit determines the differential of the accumulated quantity.
  • Differentiating with respect to the Brownian value is not the method used in the accepted explanation.
  • The discussion distinguishes a time integral from a stochastic integral with respect to Brownian motion only implicitly.

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# Integral of Wiener process over time


# Integral of Wiener process over time












This should hopefully be an easy question to answer, but I am new to Stochastic Calculus and am gapping as to why the following is true, for a brownian motion $W_t$:

$$d(\int W_t dt ) = W_t dt$$

I have seen many of the relevant linked posts about the integral, but all of them use this as a basic fact.

I have tried just applying Ito's lemma, which yields the first term when taking the partial derivative w.r.t. $t$, but I am wondering why there is no $dW_t$ term. In particular, why don't we have:

$$\frac{\partial}{\partial W_t} \int W_t dt = \int \frac{\partial}{\partial W_t} W_t dt = \int dt = t $$.

## Answer by nbbo2 (score 1, accepted)

https://quant.stackexchange.com/a/43175

Compare $\int_o^t W_t dt$ and $\int_o^{t+dt} W_t dt$

The increment between the first integral and the second is equal to $W_t dt$ (i.e. the value of the integrand at the upper limit of integration ($W_t$) multiplied by the length of time by which the integral has been extended to the right ($dt$).

That is what we mean when we write $$d(\int W_t dt ) = W_t dt$$

(The limits of integration have been left out, but it is a definite integral that we are talking about here).

## Answer by Ezy (score 0)

https://quant.stackexchange.com/a/43168

The ito lemma applies to a function $f(t,W_t)$. To help you understand why you are confused i would ask: since you believe that you can apply the ito lemma, can you explictly provide the function $f$ you believe is applicable in this case ?

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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