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Differentiating a Stochastic Integral with a Time-Dependent Kernel

Article Quant Q&A · Author: Anonymous
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Summary

The document shows how to find the differential of a stochastic integral whose integrand depends on the final time as well as on the integration variable. For the example with an exponential kernel and the sine of Brownian motion, it factors the time-dependent exponential outside the integral, defining a new stochastic process for the remaining integral.

Applying Itô’s lemma to that product yields a stochastic term from the integrand evaluated at the current time and a drift term proportional to the accumulated integral. The response also notes that the stochastic integral is well defined because its integrand is bounded. The displayed derivation appears to omit the differential notation on the final drift term, and its product rule does not require a nonzero quadratic covariation term because the external factor has finite variation. The example illustrates the method but does not discuss more general kernels.

Key ideas

  • Factor the time-dependent part of the integrand outside the stochastic integral when possible.
  • Apply Itô’s product rule to the resulting product of a smooth deterministic factor and a stochastic integral.
  • The instantaneous stochastic term uses the integrand evaluated at the current time.
  • A bounded integrand supports existence of the stochastic integral in the example.

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# Ito's Lemma - Integrand depends on upper limit of integration


# Ito's Lemma - Integrand depends on upper limit of integration












A problem I came across while practicing using Ito's Lemma had a process with an integral whose integrand depends on the upper limit of integration (the goal is to find $dZ_{t}$):

$Z_{t}=\int_{0}^{t}e^{\frac{t-s}{2}}\sin(B_{s})dB_{s}$, where $B$ is a standard Brownian motion

In what way do I need to take this into account in my solving the problem, if at all?

## Answer by Paul (score 1)

https://quant.stackexchange.com/a/7727

Note that the $$dX_t = b_t dt + \sigma_s dB_t$$ notation for a (local) semi-martingale $X = (X_t)_{t \in [ t_0, T]}$ is an abreviation for

$$ X_t = X_{t_0} + \int _{t_0} ^t b_s~ ds + \int _{t_0} ^t \sigma_s ~dB_s$$

where $b$ and $\sigma$ can be for example of the form $b_s = b(\omega, s, X_s)$ and $\sigma_s = \sigma(\omega, s, X_s)$ under condition that they are progressivelly measurable prosses and that $$\ \int _{t_0} ^T b_s ds + \int _{t_0} ^T \sigma_ s^2 ds \ < \infty \quad \mathbb P - as$$

So, since $ Z_t = e^{\frac{t}{2}} Y_t$ where $Y_t:= \int _{0} ^t e^{\frac{-s}{2}} \sin( B_s) ~dB_s$, you have by Itô's Lemma

$$ d Z_t = e^{\frac{t}{2}} ~dY_t + \frac{1}{2}e^{\frac{t}{2}} Y_t ~dt+d\langle e^{\frac{t}{2}} ,Y_t\rangle_t$$

then,

$$ d Z_t = e^{\frac{t}{2}} e^{\frac{-t}{2}} \sin( B_t) ~dB_t + \frac{1}{2}e^{\frac{t}{2}} \int _{0} ^t e^{\frac{-s}{2}} \sin( B_s) ~dB_s$$

for all $ s\in [0, +\infty)$ (note that $Z_0 =0$)

Also note that you must verifie that $Z$ is well defined as an stochastic integral, wich is evidently true since the integrand is bounded in $[0, +\infty)$

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.