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I am wondering why for a Brownian Motion $B_t$ and where $s < t$:

$$ E(sin(B_{t-s})) = 0 $$ ?

I can't understand why, can anyone see?

1 Answers 1

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$\sin$ is an odd, bounded, function. If the distribution of $X$ is symmetric about zero, then $E[\sin(X)]=0$.

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    What is the CPV sense and why do you think the assertion E(sin(X))=0 has to be qualified? – 2017-01-13
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    @Did CPV usually stands for the Cauchy principal value, but I do not understand why we need it here since I think $E[\sin X]$ exists since it is bounded by one. – 2017-01-14
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    In general, there are odd functions $g(X)$ for which the convergence of the integral for $E[g(X)]$ might be in question, For example, it's easy to construct a probability distribution symmetric about 0 such that $E[X^{3}]$ converges only in the Cauchy principal value sense. As user148692 points out, $| \sin(x) | \leq 1$, so this particular expected value will converge without having to consider the CPV. That's what I get for answering questions late at night... – 2017-01-14
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    @user148692 My point, indeed. – 2017-01-14