Find the correlation coefficient $\rho_g$ of $(G(x), H(y))$ where $G(x),H(y)$ are marginal C.D.Fs of $X,Y$ and $X,Y$ follow bivariate normal distribution $\text{B.N.}(\mu_1,\mu_2,\sigma_1,\sigma_2,\rho)$. I am finding it difficult to calculate $E(G(x)H(y))$, otherwise I can find all the other terms. Please help! The answer given is $\frac{6}{\pi} \arcsin{\frac{\rho}{2}}$
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I provided an answer to the rephrased question. – RRL Dec 04 '17 at 18:06
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Good one! Inspiring – Dec 04 '17 at 18:17
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Thanks -- actually quite useful result if you are working with copulas. – RRL Dec 04 '17 at 18:24
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I am just a beginner in probability theory , can you suggest some good books for tough problems? – Dec 04 '17 at 18:26
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Do you want basic discrete probability or continuous probability? Have you learned real analysis and measure theory? – RRL Dec 04 '17 at 18:33
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I would like the continuous one ..more or less yes – Dec 04 '17 at 18:33
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Perhaps Probability and Martingales by Williams, A Probability Path by Resnick, the classic An Introduction to Probability Theory and Applications Vol. 1 by Feller. There should also be some books specifically devoted to problems with answers -- I can't think of one right now. – RRL Dec 04 '17 at 18:38
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By a suitable transformation we reduce this to the case where $X$ and $Y$ have a bivariate normal distribution $\text{BN}(0,0,1,1,\rho)$ with joint density function $f(x,y;\rho)$.
We have
$$\tag{*}E(\Phi(X) \Phi(Y)) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \left(\int_{-\infty}^x \phi(u)\, du\right) \left( \int_{-\infty}^y \phi(v)\, dv\right)f(x,y;\rho)\, dx \, dy,$$
where $\phi$ is the standard normal density function.
The key to solving this problem is to observe that (*) is equivalent to the joint probability that $U \leqslant X$ and $V \leqslant Y$ where $U$ and $V$ are standard normally distributed random variables that are uncorrelated with each other and where each is uncorrelated with $X$ and $Y$:
$$E(\Phi(X)\Phi(Y)) = P(U \leqslant X, V\leqslant Y) = P(X-U \geqslant 0, Y-V \geqslant 0).$$
Now $X-U$ and $Y-V$ are both normally distributed with mean $0$, standard deviation $\sqrt{2}$, and with correlation
$$\text{corr}(X-U,Y-V) = \frac{E((X-U)(Y-V))}{\sqrt{\text{var}(X-U)}\sqrt{\text{var}(Y-V)}} \\= \frac{E(XY)-E(XV) - E(YU) + E(UV)}{\sqrt{2}\sqrt{2}} \\ = \frac{\rho}{2}$$
It is shown here that if $X_1$ and $X_2$ have a standard bivariate normal distribution with correlation $\rho$, then
$$P(X_1 \geqslant 0, X_2 \geqslant 0) = \frac{1}{4} + \frac{\arcsin \rho}{2 \pi}.$$
Thus,
$$E(\Phi(X)\Phi(Y)) = \frac{1}{4} + \frac{1}{2\pi} \arcsin \frac{\rho}{2}.$$
Some further manipulation yields
$$\text{corr}(\Phi(X), \Phi(Y)) = \frac{6}{\pi} \arcsin \frac{\rho}{2}$$
RRL
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I will pose the actual question now: If $(X,Y) \sim B.N(\mu_{1},\mu_{2},\sigma_{1},\sigma_{2},\rho)$.Prove that $\rho_g=corr(G(X),H(Y))=\frac{6}{\pi} \arcsin{\frac{\rho}{2}}$ where $G(x),H(y)$ are the marginal distribution functions of $X,Y$ – Dec 04 '17 at 05:58
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I recall something like that now. Would have saved some time stating it up front. – RRL Dec 04 '17 at 06:05
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