Assume that $X$ and $Y$ follow the bivariate normal distribution with correlation coeffcient $\rho > 0$, zero means and scale parameters equal to one. I am looking for an elegant way to compute

$$E\left[ \Phi \left(X \right) \Phi \left(Y \right) \right]$$

I can see that the result is

$$E\left[ \Phi \left(X \right) \Phi \left(Y \right) \right] = \frac{1}{4} + \frac{\arcsin\left(\rho/2\right)} {\left(2\pi\right)}$$

but it's not obvious how one gets there. The CDFs are still uniformly distibuted RVs but the dependence complicates things. Had it not been for the dependence we would only have the $\frac{1}{4}$ factor on the RHS but now the shape of the distribution has to be taken into account.

I have experimented with the Law of Iterated Expectations but haven't gone very far. I don't think a trigonometric transformation would be helpful either. I would therefore appreciate it if someone could give some hints on how to approach this.

Thank you.

  • 2
    $\begingroup$ Defining $\Phi$ is a good place to start ... $\endgroup$ – wolfies Jan 24 '16 at 19:45
  • 1
    $\begingroup$ @wolfies I thought it was the universal symbol for the CDF of a standard normal variable... $\endgroup$ – JohnK Jan 24 '16 at 19:46
  • 3
    $\begingroup$ Have to dash, but can't help but notice that your solution is almost identical to the standardised bivariate Normal orthant probability $$P(X<0,Y<0) = \frac{1}{4} + \frac{\arcsin\left(\rho\right)} {\left(2\pi\right)}$$ ... which may (or may not) help, as related integrals may be involved. $\endgroup$ – wolfies Jan 24 '16 at 20:06

That it relates to the standardised bivariate Normal orthant probability pointed out by wolfies is due to the fact that $$\begin{align*} \mathbb{E}^{X,Y}\left[ \Phi \left(X \right) \Phi \left(Y \right) \right] &= \mathbb{E}^{X,Y} \overbrace{\left[\mathbb{E}^Z\{\mathbb{I}(Z\le X)\}\mathbb{E}^W\{\mathbb{I}(W\le Y)\}\right]}^{Z,W\sim\mathcal{N}(0,1)}\\ &=\mathbb{E}^{X,Y,Z,W} \left[\mathbb{I}(Z\le X) \mathbb{I}(W\le Y)\right]\\ &= \mathbb{E}^{X,Y,Z,W} \left[\mathbb{I}(Z-X\le 0) \mathbb{I}(W-Y\le 0)\right]\\ &= \mathbb{E}^{X_1,Y_1} \left[\mathbb{I}(X_1\le 0) \mathbb{I}(Y_1\le 0)\right]\\&= \mathbb{P}^{X_1,Y_1} \left[X_1\le 0,Y_1\le 0\right]\\&= \mathbb{P}^{X_1,Y_1} \left[X_1/\sqrt2\le 0,Y_1/\sqrt2\le 0\right]\\\end{align*} $$ where $(X_1,Y_1)$ is now a bivariate normal vector with correlation $\rho/2$: $$\mathbb{E} \left[X_1Y_1\right]=\mathbb{E} \left[(Z-X)(W-Y)\right]=\mathbb{E} \left[XY \right]=\rho$$ and $$\mathrm{var}(X_1)=\mathrm{var}(Y_1)=\mathrm{var}(Z)+\mathrm{var}(X)=2$$

  • $\begingroup$ I think I got it. You move the expectations with respect to $Z$ and $W$ in front, because they are independent, yes? $\endgroup$ – JohnK Jan 24 '16 at 20:55
  • $\begingroup$ Not really, I simply use a single expectation for all rv's involved, whether they are dependent or independent. The expectation is against the joint distribution over the four rv's, $X,Y,Z,W$. $\endgroup$ – Xi'an Jan 24 '16 at 21:12
  • $\begingroup$ Sorry for being slow but isn't that like claiming that $E\left(X\right) E \left(Y\right) = E\left(XY\right)$? $\endgroup$ – JohnK Jan 24 '16 at 21:22
  • $\begingroup$ No, I am using$$\mathbb{E}^X[\mathbb{E}^Z[h(X,Z)]]=\mathbb{E}^{X,Z}[h(X,Z)]$$ $\endgroup$ – Xi'an Jan 24 '16 at 21:25

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