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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 it?

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

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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    Good. FYI, it is usually denoted through $\Phi$2012-06-02

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So... after much discussions in the comments, in the end it seems that the letter $\mathcal N$ is used here to denote $\Phi$ the CDF of the standard normal distribution, defined by $\Phi(x)=\mathrm P(X\leqslant x)$, 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 $Z$ is symmetric and $\mathrm E(\Phi(W_t))=\Phi(0)=\frac12$ for every $t$.

By the same decomposition, $\mathrm E(\Phi(W_t+a))=\mathrm P(Z\leqslant a)$. Since $Z$ is centered normal with variance $1+t$, $\mathrm P(Z\leqslant a)=\mathrm P(\sqrt{1+t}\cdot X\leqslant a)$, hence, for every $a$ and every nonnegative $t$,

$\color{red}{\mathrm E(\Phi(W_t+a))=\Phi\left(\frac{a}{\sqrt{1+t}}\right)}$

Exercise: Use this approach to completely and readily solve this nearly duplicate question of yours.