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Suppose that $X\sim N(0,1)$, $Y\sim N(0,1)$ with correlation $(X, Y) =ρ$ where $ρ ∈ (−1, 1)$.

Show the following,

Correlation $(Φ(X),Φ(Y))=\dfrac6π \arcsin\dfrac ρ2 $.

Here $Φ(X)$, and $Φ(Y)$ denote the CDF of Random variables $X$ and $Y$ respectively.

What I know so far:

$$\text{Cov}( Φ(X),Φ(Y)) =E(Φ(X)Φ(Y)) - E(Φ(X)) ×E(Φ(Y)) $$

We also know since $X \sim N(0,1)$ and $Y\sim N(0,1)$; $Φ(X)\sim\text{unif}(0,1)$, $Φ(Y)\sim\text{unif}(0,1)$. Hence $$E(Φ(X)) =E(Φ(Y)) =\frac12.$$

I am stuck in finding $E(Φ(X)Φ(Y))$ [I tried to find it using double expectation].

What I have written may be the wrong way of attempting the question. So I am thankful for any help.

amWhy
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Amelia
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1 Answers1

4

Assuming $(X,Y)$ is jointly normal with zero means, unit variances and $\text{Corr}(X,Y)=\rho$, having joint density $f_{X,Y}$.

Define $$(X',Y')\stackrel{\text{i.i.d}}{\sim} N(0,1)$$

such that $(X',Y')$ is independent of $(X,Y)$.

So $(X',Y')$ is (trivially) jointly normal with correlation zero.

This implies $$(U,V)=\left(\frac{X-X'}{\sqrt 2},\frac{Y-Y'}{\sqrt 2}\right)$$ is also jointly normal with zero means and unit variances and $\text{Corr}(U,V)=\rho/2$.

Then,

\begin{align} E(\Phi(X)\Phi(Y))&=\int_{\mathbb R}\int_{\mathbb R}\Phi(x)\Phi(y)f_{X,Y}(x,y)\,dx\,dy \\&=\int_{\mathbb R}\int_{\mathbb R}P(X'\leqslant x,Y'\leqslant y)f_{X,Y}(x,y)\,dx\,dy \\&=\int_{\mathbb R}\int_{\mathbb R}P(X'\leqslant x,Y'\leqslant y\mid X=x,Y=y)f_{X,Y}(x,y)\,dx\,dy \\\\&=P(X'\leqslant X,Y'\leqslant Y) \\\\&=P(X-X'\geqslant 0,Y-Y'\geqslant 0) \\\\&=P\left(\frac{X-X'}{\sqrt 2}\geqslant 0,\frac{Y-Y'}{\sqrt 2}\geqslant 0\right) \\\\&=P(U\geqslant 0,V\geqslant 0) \\\\&=\frac{1}{4}+\frac{1}{2\pi}\arcsin\left(\frac{\rho}{2}\right) \end{align}

In the last line, we used a popular result for the probability that two jointly normal variables both lie in the first quadrant. Proofs can be found here and here.

Finally,

\begin{align} \text{Corr}(\Phi(X),\Phi(Y))&=\frac{\frac{1}{4}+\frac{1}{2\pi}\arcsin\left(\frac{\rho}{2}\right)-\frac{1}{4}}{\frac{1}{12}} \\\\&=\frac{6}{\pi}\arcsin\left(\frac{\rho}{2}\right) \end{align}

StubbornAtom
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  • While finding E(Φ(X)Φ(Y)), @StubbornAtom how did u go from step 1 to step 2 that is ;how can u say Φ(x)Φ(y)=P(X'⩽x, Y'⩽y) – Amelia Oct 14 '18 at 15:39
  • @Amelia I have taken $X'$ and $Y'$ to be independent standard normal variables.

    So,

    $$\Phi(x)\Phi(y)=P(X'\leqslant x)P(Y'\leqslant y)=P(X'\leqslant x,Y'\leqslant y)$$

     – StubbornAtom Oct 14 '18 at 15:45
  • No I meant to ask, is the following right Φ(x)Φ(y)=P(X⩽x) P(Y⩽y) and P(X⩽x)= P(X'⩽x) ,P(Y⩽y)=P(Y'⩽y) Thus Φ(x)Φ(y)=P(X'⩽x, Y'⩽y) – Amelia Oct 14 '18 at 16:18
  • @Amelia Yes, but that is a joint probability in your last equation. So we need independence of $X'$ and $Y'$. – StubbornAtom Oct 14 '18 at 16:25
  • Inorder to avoid misconceptions on my part if u don't mind can u please explain how u went from step 3 to step 4.The rest is all clear. And also Thanks for ur help☺️ – Amelia Oct 14 '18 at 17:07
  • @Amelia It is the familiar total probability theorem for continuous random variables, conditioning on two variables simultaneously:

    $$P(X'\leqslant X,Y'\leqslant Y)=\iint P(X'\leqslant x,Y'\leqslant y\mid X=x,Y=y)f_{X,Y}(x,y),dx,dy$$

     – StubbornAtom Oct 14 '18 at 17:25
  • @Amelia If the answer is helpful, do consider voting it up. – StubbornAtom Oct 15 '18 at 03:48
  • Your answer have been indeed very helpful @StubbornAtom, but the pblm is that I am a new member and thus I don't have enough reputation to do so. I am helpless here – Amelia Oct 15 '18 at 05:18
  • @Amelia Okay, do not worry about that. Glad to help. – StubbornAtom Oct 15 '18 at 05:25