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I have been reading a paper by Desmond J. Higham titled "An Algorithmic Introduction to Numerical Simulation of Stochastic Differential Equations." Here is the link:

https://epubs.siam.org/doi/pdf/10.1137/S0036144500378302

Suppose $W(t) , t \geq 0$ is a Brownian motion. The author wants to evaluate the function $u\big(W(t)\big) = \exp\big(t + \frac{1}{2}W(t)\big)$ along $1000$ discretized Brownian paths. He writes some MATLAB code which I 'm clear with. He then wishes to compare the mean of the discretised paths with the expected value of $u\big(W(t)\big).$

On page $528,$ fourth line from the last, he mentions that the expected value of $u\big(W(t)\big)$ is $\exp (9t/8).$ I'm not clear as to how he got this. Any help highly appreciated.

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2 Answers 2

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Write $$E[\exp\big(t + \frac{1}{2}W(t)\big)]=e^{t} \cdot E[\exp\big( \frac{1}{2}W(t)\big)]$$ at this point remember the fact that $W(t)$ is a centered Gaussian r.v. with variance equal to $t$, and use the moment generating function to obtain $$=e^t\cdot e^{1/2 \cdot (1/2)^2 \cdot t}=e^{9t/8 }$$

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By Ito you get $$ X_t=F(t,W_t)=\exp(t+\frac12W_t) \\ \implies dX_t=(F_t+\frac12F_{ww})dt+F_wdW_t =(X_t+\frac12\frac1{2^2}X_t)dt+\frac12X_tdW_t $$ and the expectation of $X$ follows the equation $$\newcommand{\E}{\mathbb{E}} d\E(X_t)=\frac98\E(X_t)dt\implies \E(X_t)=e^{\frac98t}. $$

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  • $\begingroup$ I think you are missing $X_t$ with $\frac{1}{2}dW_t$. Could you elaborate how the expectation of last term is $0$? I know Brownian Motion increments have mean $0$ but we have an extra $X_t$ term as well. And are you solving ODE in the last line? $\endgroup$
    – user715112
    Jan 14, 2021 at 13:37
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    $\begingroup$ Thank you, corrected. $X_t$ is independent of $dW_t$, or in other words, $E(X_tdW_t|{\cal F}_t)=X_tE(dW_t|{\cal F}_t)=0$. And yes, for the expectation it is only an ODE that is solved as such. $\endgroup$
    – Lutz Lehmann
    Jan 14, 2021 at 14:12
  • $\begingroup$ @Lehmann In the paper, the author defines $dW(t_j) = W(t_j) - W(t_{j-1}).$ How do we know that $dW(t_j)$ is not $\mathcal{F}(t_j)$-measurable? $\endgroup$
    – user715112
    Jan 15, 2021 at 0:54
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    $\begingroup$ That is not Ito compatible. For that it has to be $dW_t=W_{t+dt}-W_t$. $\endgroup$
    – Lutz Lehmann
    Jan 15, 2021 at 4:56

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