Differentiating a Brownian Integral with the Product Rule
Summary
This exchange explains how to differentiate a stochastic expression that contains an exponential time factor multiplied by an integral with respect to Brownian motion. The key step is to treat the product as two processes: the deterministic, finite-variation exponential term and the stochastic integral. Applying the product rule yields a drift contribution from differentiating the exponential and a Brownian increment from differentiating the integral. Because the exponential factor has finite variation, there is no quadratic covariation term between it and the stochastic integral.
The answer confirms the proposed reasoning and identifies an extra factor of time in an intermediate displayed equation as a typo. It presents no market data or trading application; its value is in clarifying an Itô-calculus manipulation used in stochastic modeling. The derivation assumes the integrands and processes are sufficiently well behaved for the stated stochastic integral and product rule, and the original expression’s surrounding model is not developed in the exchange.
Key ideas
- A product of a finite-variation process and a Brownian integral can be differentiated with the product rule.
- The finite-variation factor contributes no quadratic covariation term.
- Differentiating the upper limit of the Brownian integral produces the current integrand times the Brownian increment.
- An extra time factor in the displayed intermediate equation is a typo, not part of the corrected differential.
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# Ito's Lemma, differentiating an integral with Brownian motion
# Ito's Lemma, differentiating an integral with Brownian motion
In How were these SDE derived? I don't understand one part of Gordon's answer, specifically:
$$\ln S_t=\ln F_{0,t}-\frac{\sigma^2}{4\lambda}(1-e^{-2\lambda t})+\sigma e^{-\lambda t}\int_0^t e^{\lambda s}dB_s$$ $$d\ln S_t=\bigg(\frac{\partial \ln F_{0,t}}{\partial t}-\frac{\sigma^2 }{2}e^{-2\lambda t}-\lambda \sigma e^{-\lambda t}\int_0^t e^{\lambda s}dB_s\bigg)dt+\sigma dB_t$$
My question is, how to get from the first line to the second. Evidently, Ito's Lemma was applied, however it has been used in a way in which I have not encountered. It seems that the integral with the Brownian motion term was treated as a constant, but I don't see how that is allowed when there is $t$ in the upper limit.
Any help is appreciated.
Edit: I have come up with a solution, as long as the following is valid. Can anyone please confirm? The first line uses the product rule. \begin{align} d\bigg(\sigma e^{-\lambda t}\int_0^t e^{-\lambda s}dB_s\bigg)&=d(\sigma e^{-\lambda t})\int_0^t e^{-\lambda s}dB_s+\sigma e^{-\lambda t}d\bigg(\int_0^t e^{\lambda s}dB_s\bigg)\\ &=-\lambda\sigma e^{-\lambda t}dt \int_0^t e^{-\lambda s}dB_s+\sigma e^{-\lambda t}e^{\lambda t}dB_t\\ &=-\lambda\sigma e^{-\lambda t}dt \int_0^t e^{-\lambda s}dB_s+\sigma dB_t \end{align}
## Answer by M. Jeunesse (score 1, accepted)
https://quant.stackexchange.com/a/30608
You are right in your use of product rule.
Since $\sigma e^{-\lambda t}$ is a finite variation process, you do not have crochet term and this is the standard product rule.
In your second line, there is an extra $t$ (which is a typo I presume)
$$\ln S_t=\ln F_{0,t}-\frac{\sigma^2}{4\lambda}(1-e^{-2\lambda t})+\sigma e^{-\lambda t}\int_0^t e^{\lambda s}dB_s$$ $$d\ln S_t=\bigg(\frac{\partial \ln F_{0,t}}{\partial t}-\frac{\sigma^2 }{2}e^{-2\lambda t}-\lambda \color{red}{t}\sigma e^{-\lambda t}\int_0^t e^{\lambda s}dB_s\bigg)dt+\sigma dB_t$$ the correct equation is: $$d\ln S_t=\bigg(\frac{\partial \ln F_{0,t}}{\partial t}-\frac{\sigma^2 }{2}e^{-2\lambda t}-\lambda \sigma e^{-\lambda t}\int_0^t e^{\lambda s}dB_s\bigg)dt+\sigma dB_t$$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.