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I'm looking for a way to evaluate the expectation

$$ \mathbb{E}\left[ \Phi\left(\frac{X}{c}\right) \Phi\left(\frac{Y}{c}\right) \right]$$

where $\Phi(x)$ denotes the cdf of a standard normal distribution and $X$ and $Y$ follow a standard bivariate normal distribution, i.e. a bivariate normal distribution with means equal to zero, variances equal to 1 and correlation coefficient equal to $\rho$.

I have found this proof but I'm having some trouble understanding it so could someone please explain to me how we get from the first equality to the second? I can take over from there.

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  • $\begingroup$ you mean $(X,Y) \sim \mathcal{N}(\mu,\Sigma)$ follows a $2$-dimensional normal distribution with mean $\mu = (0,0)$ and covariance matrix $\Sigma = \left( \begin{array}{ll} \sigma^2 & \rho \\ \rho & \sigma^2 \end{array}\right)$ (X,Y) \sim \mathcal{N}(\mu,\Sigma), \Sigma = \left( \ begin{array}{ll} \sigma^2 & \rho \\ \rho & \sigma^2 \end{array}\right) $\endgroup$ – reuns Mar 16 '16 at 20:05
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\begin{align} E[P^X(X-cZ_1>0)P^Y(Y-cZ_2>0)] &= E[P(X-cZ_1>0|X=x)P(Y-cZ_2>0|Y=y)]\\ &= E[E[\mathbb{I}_{[X-cZ_1>0]}|X=x]E[\mathbb{I}_{[Y-cZ_2>0]}|Y=y]]\\ &= E[E[\mathbb{I}_{[X-cZ_1>0, \,\, Y-cZ_2>0 ]}|X=x\,, Y=y]]\\ &=P(X-cZ_1>0, \, Y-cZ_2>0 ) \end{align}

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  • $\begingroup$ Thanks, where does the third line follow from? What is it that allows us to merge these expectations? $\endgroup$ – JohnK Mar 16 '16 at 20:32
  • $\begingroup$ $E1_{A\cap B}=E1_{A}1_{B}=E1_{A}E1_{B}$. First equality stems from properties of indicator functions, and the second from independence of $Z_1$ and $Z_2$. $\endgroup$ – V. Vancak Mar 16 '16 at 20:38
  • $\begingroup$ But how do you know that they are independent? I don't think it's given. $\endgroup$ – JohnK Mar 16 '16 at 20:38
  • $\begingroup$ Perhaps you are right. I'm not sure about the formal/correct justification. Maybe someone else can help to clarify this point. $\endgroup$ – V. Vancak Mar 16 '16 at 21:12
  • $\begingroup$ I thought you are right but someone else told me that we have no need for the independence. Maybe there is a trick I am ignoring. $\endgroup$ – JohnK Mar 16 '16 at 21:24

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