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I'm sitting with a special case of GBM where $B(t)$ is Brownian motion with drift $\mu$ and variance $\sigma^2$. I want to find the expected value of $A(t)$, where $A(t)$ is given by $$A(t)=exp(B(t))B(t)+C$$ and C is a constant. Has anyone worked with such a model before?

I'm thinking that $$E[A(t)]=E[B(t) \exp(\mu t)]+C=\mu t\exp(\mu t)+C$$ is the answer to the expected value (not 100% sure), but how do I determine the variance of $A(t)$?

Thanks in advance.

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up vote 0 down vote accepted

Since $B(t)$ has a normal distribution with mean $\mu t$ and variance $\sigma^2 t$, you know that its density function is $$p_t(x) = \frac{1}{\sqrt{2 \pi \sigma^2 t}} e^{-(x-\mu t)^2/2\sigma^2 t}.$$ So you can find the expected value of any function $f(B(t))$ by computing the integral $\int_{-\infty}^\infty f(x) p_t(x)\,dx$. For $E[A(t)]$, you'll compute $$\int_{-\infty}^{\infty} (e^x x + C) \frac{1}{\sqrt{2 \pi \sigma^2 t}} e^{-(x-\mu t)^2/2\sigma^2 t}\,dx = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} x e^{x -(x-\mu t)^2/2\sigma^2t}\,dx + C.$$ This should be easily computed by a change of variables. Maple says we get $$e^{\mu t + \sigma^2 t/2} (\sigma^2 t + \mu t). $$ The variance should be computable by a similar technique; I think there may be an integration by parts needed as well.

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Thanks, I see p_t(x) = \frac{1}{\sqrt{2 \pi \sigma^2 t}} e^{-(x-\mu t)^2/2\sigma^2 t} right? – user117142 Dec 21 '13 at 20:25
@Jimmy: Yes, fixed. – Nate Eldredge Dec 21 '13 at 21:55

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