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Applying Itô’s Formula to an Exponential of a Path-Dependent Integral

Article Quant Q&A · Author: David Khan
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Summary

The document asks for the stochastic differential of an exponential built from an integral involving a diffusion process. The questioner applies Itô’s formula to the exponential, but assumes the integral has differential equal only to the integrand at the current time. The response points out an additional term arising because the integrand depends on the evolving state variable inside the integral.

It expresses the differential of the integral as the current integrand times time increment plus an integral of the state-dependent differential of the integrand. That inner differential itself includes first- and second-derivative contributions from Itô’s formula. The answer is concise and provides no regularity conditions, derivation details, or discussion of whether the stated integral notation is intended to represent a standard time integral with an evolving state path. Readers should check the precise definition of the process before applying the formula; the response mainly flags why treating the entire integral as having a simple instantaneous derivative may be insufficient.

Key ideas

  • Applying Itô’s formula to an exponential requires the differential of the exponent process.
  • State dependence inside a time integral can add terms beyond the current integrand times time.
  • The response applies Itô’s formula to the integrand with respect to the diffusion state.
  • The resulting expression includes first- and second-order state derivatives.
  • The process definition and regularity assumptions should be clarified before using the formula.

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# Ito formula (lemma) problem


# Ito formula (lemma) problem












I am trying to solve this problem

> Consider the following one-dim. stochastic process $$dX_t = b_t dt + \sigma_t dW_t$$ where $W$ is a one-dim. Brownian motion. The above SDE is well-defined. Consider a smooth and bounded function $g$, and put $$ Z_t := \exp(\int_0^t g(s,X_t)ds).$$ Calculate the stochastic differential $dZ$.

My answer: Put $Y_t = \int_0^t g(s,X_t)ds$ . Then, it follow that $Z_t=e^{Y_t}$ , and from Ito formula, I have $$dZ_t = Z_t(dY_t + \frac{1}{2}(dY_t)^2).$$ Thus, I want to know the stochastic differential $dY$. If I can say that $$dY_t=g(t,X_t)dt$$ then, $$dZ_t = Z_t \bigl(g(t,X_t)dt + \frac{1}{2}(g(t,X_t)dt)^2 \bigl)$$ $$\Leftrightarrow dZ_t = Z_t g(t,X_t)dt .$$ follows. My question: I am not sure if I can say that $$dY_t=g(t,X_t)dt.$$ I suspect that my answer is too simple to be true. Where did I make a mistake?

## Answer by user34971 (score 2)

https://quant.stackexchange.com/a/45794

In your notation, $$ dY_t = g(t,X_t) dt + \int_0^t dg(s,X_t) ds $$ where $$ dg(s, X_t) = \partial_{X_t} g(s,X_t) dX_t + \frac{1}{2} \partial^2_{X_t} g(s,X_t) (dX_t)^2 $$ The rest seems ok.

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