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Say if B is standard Brownian motion then:

$S(t) = S0e^{((𝜇- σ^2)/2)t+σB(t)}$

The mean of this SDE would be

$𝐄[𝑆(𝑡)]=𝑆_0𝑒^{𝜇𝑡}$

I know to do this you use the density function and integrate by parts twice but I keep making a mistake in the integration. Can someone show me how it should look? So I can fix my mistake

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If $S$ is the solution to geometric brownian motion SDE: \begin{equation} dS=\mu S dt + \sigma S dW(t) \end{equation} then \begin{equation} S=S_0e^{(\mu - \sigma^2/2)t + \sigma W(t)} \end{equation} Then if you take expectation \begin{equation} \mathbb{E}[S(t)]=S_0e^{(\mu - \sigma^2/2)t}\mathbb{E}[e^{\sigma W(t)}] \end{equation} Now since $W$ is a wiener process: \begin{equation} W(t) \sim \mathcal{N}(0,t) \end{equation} the expected value is the moment generating function of the normal distribution. This means \begin{equation} \mathbb{E}[e^{\sigma W(t)}]=\int_{-\infty}^{+\infty}e^{\sigma x}\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi t}}e^{-\frac{1}{2t}(x^2 - 2t\sigma x)}dx=\int_{-\infty}^{+\infty}e^{\frac12\sigma^2 t}\frac{1}{\sqrt{2\pi t}}e^{-\frac{1}{2t}(x^2 - 2t\sigma x + \sigma^2t^2)}dx=\int_{-\infty}^{+\infty}e^{\frac12\sigma^2 t}\frac{1}{\sqrt{2\pi t}}e^{-\frac{1}{2t}(x-\sigma t)^2}dx=e^{\frac12\sigma^2 t}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac12y^2}dy=e^{\frac12\sigma^2 t} \end{equation} if you substitute \begin{equation} \mathbb{E}[S(t)]=S_0e^{\mu t} \end{equation}

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  • $\begingroup$ Thank you, I had made minor errors but this has cleared them up for me. $\endgroup$ – JohnOD25 Sep 28 '19 at 18:03

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