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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:
$$ \mathbb{E}\left[\int_0^t B(s)e^{-ks} \,\mathrm{d}s\right]$$

user217285
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Effie Stavrulaki
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1 Answers1

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

user217285
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