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I encounterred a question on Stochastic Calculus as following, but I don't understand the meaning of $\mathcal{N}$ here, can any expert explain me a little bit? Thank you very much in advance!

$\mathbb{E}[\mathcal{N}(W_t)]=?$ where $W_t$ is standard Brownian Motion.

Besides, a follow-up is $\mathbb{E}[\mathcal{N}(W_t+a)]=?$ I was also blocked by the meaning of $\mathcal{N}$.

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I don't know what is $\mathscr N$, could you put a source for this problem? –  Ilya May 26 '12 at 16:06
    
Hi Ilya, I am also not sure about it, it seems it's understandable to many stochastic analysis experts but I am not. –  pidig May 27 '12 at 4:26
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Hi Ilya, I know how to handle it, $\mathcal{N}$ is just the cdf of standard normal distribution. –  pidig Jun 2 '12 at 7:48
    
Good. FYI, it is usually denoted through $\Phi$ –  Ilya Jun 2 '12 at 8:28
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1 Answer

So... it seems that $\mathcal N$ is $\Phi$ the CDF of standard normal distribution, that is $\Phi(x)=\mathrm P(X\leqslant x)$, where $X$ is any standard normal random variable.

Since $W_t=\sqrt{t}Y$ where $Y$ is standard normal, $\mathrm E(\Phi(W_t))=\mathrm P(X\leqslant\sqrt{t}Y)=\mathrm P(Z\leqslant0)$ with $Z=X-\sqrt{t}Y$. Since $X$ and $Y$ are independent and centered normal, $Z$ is centered normal hence $\mathrm E(\Phi(W_t))=\Phi(0)=\frac12$.

By the same decomposition, $\mathrm E(\Phi(W_t+a))=\mathrm P(Z\leqslant a)$. Since $Z$ is centered normal with variance $1+t$, this yields $\mathrm E(\Phi(W_t+a))=\Phi(a/\sqrt{1+t})$.

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