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I am supposed to find the following ($B_t$ is a Brownian motion, and $\mathcal{F}_s$ the generated filtration):

$$\mathbb{E}[e^{B_t}|\mathcal{F}_s]=?$$ I tried this: shifting by $s$ to the left to get $$\mathbb{E}[e^{B_{t-s}}|\mathcal{F}_0]$$ and then treat $B_{t-s}$ as an RV with normal distribution and $\mathbb{E}=0$ and $Var=t-s$, and then plug this in into $\mathbb{E}[e^X]=e^{\mu+\frac{\sigma^2}{2}}$ to get $$\mathbb{E}=e^\frac{t-s}{2}$$. But then I remembered that for a Brownian motion, we must have $B_0=0$, and simply shifting $B_t$ to the left by $s$ will not give a distribution that is centered at $0$. So what will I do now? Is it enough to add $W_s$ to the expected value, or was the whole shifting a bad idea?

Yours, Marie

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1 Answer 1

up vote 7 down vote accepted

We assume that $s< t$, otherwise $B_t$ is $\mathcal F_s$-measurable and so is $e^{B_t}$, so $E[e^{B_t}\mid\mathcal F_s]=e^{B_t}$. Since $B_s$ and $B_t-B_s$ are independent, $E[e^{B_t}\mid \mathcal F_s]=E[e^{B_t-B_s}e^{B_s}\mid \mathcal F_s]$, and since $B_s$ is $\mathcal F_s$-measurable, $E[e^{B_t}\mid \mathcal F_s]=e^{B_s}E[e^{B_t-B_s}\mid\mathcal F_s]$. $B_t-B_s$ is independent of $\mathcal F_s$ so $E[e^{B_t-B_s}\mid\mathcal F_s]=E[e^{B_t-B_s}]=E[e^X]$, where $X$ is normal random variable of mean $0$ and variance $t-s$. This constant is computable: \begin{align} E[e^X]&=\frac 1{\sqrt{2\pi(t-s)}}\int_{-\infty}^{+\infty}e^xe^{-\frac{x^2}{2(t-s)}}dx\\\ &=\frac 1{\sqrt{2\pi(t-s)}}\int_{-\infty}^{+\infty}\exp\left(-\frac 1{2(t-s)}\left(x^2-2(t-s)x\right)\right)dx\\\ &=\frac 1{\sqrt{2\pi(t-s)}}\int_{-\infty}^{+\infty}\exp\left(-\frac 1{2(t-s)}\left(x-(t-s)\right)^2+\frac{t-s}2\right)\\\ &=\exp\left(\frac{t-s}2\right). \end{align} Conclusion: $$E[e^{B_t}\mid\mathcal F_s]=\begin{cases} e^{B_t}&\mbox{ if }s\geq t\\\ \exp\left(\frac{t-s}2\right)e^{B_s}&\mbox{ if }s<t. \end{cases}$$

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Answered almost 3 years ago - very clear explanation. –  John Smith Feb 5 at 9:49

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