Itô Integration of an Exponential Brownian Motion

Summary

The document asks how to evaluate the Itô integral of an exponential function of Brownian motion. Its answer uses the exponential Brownian martingale, which includes a time correction in its exponent, and applies the martingale’s differential relationship to express the corresponding stochastic integral in terms of the process at the endpoint. This provides a closed-form result for that corrected integrand.

The original integrand lacks the time correction, so the displayed martingale identity does not directly evaluate the integral as posed. The answer suggests that a change of measure through Girsanov’s theorem could remove the correction while introducing drift into the Brownian integrator. It does not carry out that transformation or give a full expression for the original integral, and concludes that an unevaluated integral may remain. The material is a partial derivation, not a complete solution to the question.

Key ideas

• The exponential Brownian martingale includes a time-dependent correction in its exponent.
• Its differential identifies a stochastic integral that can be written using the terminal martingale value.
• The identity applies to the corrected integrand, rather than directly to the original expression.
• A Girsanov change of measure may shift the correction into drift in the integrator.
• The answer leaves the original integral without a complete closed-form evaluation.

Tags

# Integration of exponential raised with Brownian Motion wrt the Brownian Motion


# Integration of exponential raised with Brownian Motion wrt the Brownian Motion












I have to derive several things for my thesis, however, I have the following expression:

$$ \int^{t}_{0} \exp\{\sigma W_{t}\}.dW_{t} $$

Does anyone know what the solution for this is?

Kind regards.

## Answer by Kurt G. (score 2, accepted)

https://quant.stackexchange.com/a/70053

Unlike for many ordinary integrals in calculus there is not always a solution in Ito calculus.

Partial Answer and Hints

The process $$ M_t=\exp\{\sigma\,W_t-\sigma^2t/2\} $$ is a martingale that satisfies $$ dM_t=\sigma\,M_t\,dW_t\,,\quad\text{ or in integral form }M_t=1+\int_0^t\sigma\,M_s\,dW_s\,. $$ Therefore, $$ \int_0^t\sigma\,M_s\,dW_s=\sigma\int_0^t\exp\{\sigma\,W_s-\sigma^2s/2\}\,dW_s=M_t-1=\exp\{\sigma\,W_t-\sigma^2t/2\}-1\,. $$ If you want to get rid of the $-\sigma^2s/2$ term in the exponential you could apply Girsanov's theorem which will introduce a drift in the integrating BM $dW_t$.

I think some integral will always remain "unsolved".

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