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This equation solution is a geometric brownian motion $$dx_t=r.x_t.dt+\sigma .x_t.dB_t \to \\ x_t=x_0.e^{(r-\frac{1}{2}\sigma^2)t+B_t}$$ now I am asking for $E[x_t]$
I saw wikipedia ...It said that $E[x_t=x_0.e^{(r-\frac{1}{2}\sigma^2)t+B_t}]=x_0.e^{rt}$ https://en.wikipedia.org/wiki/Geometric_Brownian_motion

But I think this is is wrong ...!

I tried this :$$E[x_t=x_0.e^{(r-\frac{1}{2}\sigma^2)t+B_t}]=\\E[x_0].e^{E[(r-\frac{1}{2}\sigma^2)t+B_t]}=\\E[x_0].e^{E[(r-\frac{1}{2}\sigma^2)t]+E[B_t]}=\\E[x_0].e^{E[(r-\frac{1}{2}\sigma^2)t]}$$ because $E[B_t]=0$ Please help me to understand ,that mine is correct or not ? And what is the right answer

Khosrotash
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1 Answers1

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The (big) problem in your calculation is that you seem to think that

$$ E(e^X) = e^{E(X)}$$

which is wrong.

If you search for characteristic function of a normal and remember that $B_t \sim N(0,t)$ you should be able to compute the correct result

Ant
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  • :Thank you.So I had mistake . can you show me more ? how to obtain expectation ?(for a standard brownian motion ) – Khosrotash Nov 08 '16 at 16:56
  • See here: https://math.stackexchange.com/questions/40826/expectation-of-geometric-brownian-motion – xiA May 02 '19 at 04:45