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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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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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