Expected Discounted Integral of Drifted Brownian Motion

Summary

The document asks for the expectation of a time integral of Brownian motion with drift, weighted by exponential discounting. It specifies drift μ, volatility σ, and discount rate k, then applies Fubini’s theorem to move expectation inside the integral. Under the stated convention that the process has expected value μs at time s, the expectation reduces to integrating μs times the discount factor over the interval.

The resulting expression is μ times a function of k and the horizon t, divided by k squared. Volatility does not appear because the expectation depends on the process’s mean, while its zero-mean Brownian component contributes no expected value. The result assumes the stated drift convention and a valid exchange of expectation and integration. The source does not discuss special handling when k equals zero, nor does it provide further derivation or applications.

Key ideas

• Fubini’s theorem allows the expectation and time integral to be interchanged under suitable conditions.
• The expected value of the drifted Brownian motion at time s is μs under the convention used.
• The discounted expectation is obtained by integrating that mean against the exponential discount factor.
• The volatility parameter does not affect this expectation because the Brownian component has zero mean.

Tags

# Expectation of an Integral of a function of a Brownian Motion


# Expectation of an Integral of a function of a Brownian Motion












I would really appreciate some guidance on how to calculate the expectation of an integral of a function of a Brownian Motion.

> Let $B(t)$ be a Brownian motion with drift $\mu$ and standard deviation $\sigma$. At time $t$, $e^{-kt}$ represents time discounting with a time discount factor of k. I need to evaluate the following: $$ \mathbb{E}\left[\int_0^t B(s)e^{-ks} \,\mathrm{d}s\right]$$

## Answer by user217285 (score 3)

https://quant.stackexchange.com/a/40333

By Fubini's theorem, $$\newcommand\bbE{\mathbb{E}} \bbE\left[ \int_0^t B(s)e^{-ks} \,\mathrm{d}s \right] = \int_0^t \bbE[B(s)e^{-ks}] \,\mathrm{d}s = \int_0^t \mu se^{-ks} \,\mathrm{d}s = \frac{\mu (1 - e^{-kt} (1+kt))}{k^2}.$$

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