Suppose $\{B_t,t\ge0\}$ be a standard brownian motion and suppose $0\le u\le s Attempts: $E(e^{B(t)}|e^{B(u)},0\le u\le s )=E(e^{B(t-u)+B(u)}|e^{B(u)})=E(e^{B(t-u)}|e^{B(u)})E(e^{B(u)}|e^{B(u)})$ but then not sure how to proceed.
calculate $E(e^{B(t)}|e^{B(u)},0\le u\le s )$
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probability
probability-theory
stochastic-processes
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1$B(t)\ne B(t-u)+B(u)$. – 2012-12-21
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0@did i see $B(t-u)\sim~N(0,\sqrt{t^2+u^2})$ – 2012-12-21
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0No, this is not the distribution of $B(t-u)$. – 2012-12-21
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1OP: What happens with this question? Did you manage to write down a full proof, using @Stefan's hints? – 2013-01-02
1 Answers
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Hint: Write $B_t=B_t-B_s+B_s$ and use that $$ E[XY\mid\mathcal{G}]=XE[Y\mid \mathcal{G}] $$ if $X$ is $\mathcal{G}$-measurable along with the fact that $$ E[X\mid\mathcal{G}]=E[X] $$ if $X$ is independent of $\mathcal{G}$.
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2Also knowing the MGF for a normal random variable will come in handy. – 2012-12-21
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0@RobertIsrael: Indeed! – 2012-12-21