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I saw the following question on another forum:

"Suppose that both height and weight of adult men can be described with normal models, and that the correlation between these variables is 0.65. If a man's height places him at the 60th percentile, at what percentile would you expect his weight to be?"

I see that someone at the forum in question has already pointed out that the question talks about the margins being normal (height and weight ... can be described with normal models), not about bivariate normality and so the question doesn't have a single answer.

Clearly the answer would depend on the actual bivariate dependence relationship (the copula), which left me curious.

My question is:

Given normal margins and a specified population correlation ($\rho$, a Pearson correlation), is there a reasonably straightforward way to find bounds on $E(Y|X=x_q)$ given $X,Y$ both normal, with correlation $\rho$?

If there's an exact largest value and smallest value for the conditional expectation, that (and for preference, the circumstances in which each occurs*) would be good to know about.

* I have some strong suspicions about what those circumstances might be (i.e. the kind of dependence that might be involved; in particular, I expect a specific kind of degenerate distribution will give the bounds) but I haven't yet investigated that thought in any depth. (I figure someone is already likely to know it.)

Failing that, upper or lower bounds on both the largest and smallest values would be interesting.

I don't necessarily require an algebraic answer (some algorithm would do), though an algebraic answer would be nice.

Approximate or partial answers may be useful/helpful.

If nobody has good answers, I may have a go at it myself.

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I think there are no bounds. This conclusion relies on the following construction, which is simplest to describe for arbitrary continuous distributions. As we go along, conditions will be added until we are in the case of Normal marginals.

So, let $X$ be any continuous random variable with distribution function $F$. Given any half-open interval $(a,b]$ (which will eventually become very narrow), define

$$\psi: (a,b] \to (-\infty, c]$$

via

$$\psi(x) = F^{-1}(F(x) - F(a)).$$

This is monotonically increasing and evidently $c = \psi(b) = F^{-1}(F(b)-F(a))$. By construction,

$$\Pr(X \in (a,b]) = \Pr(\psi(X) \le c).$$

Extend $\psi$ to a one-to-one map $\Psi:\mathbb{R} \to \mathbb{R}$ via

$$\eqalign{ \Psi|_{(a,b]} &= \psi, \\ \Psi|_{(-\infty, c]} &= \psi^{-1} }$$

and otherwise $\Psi(x) = x$. The distribution of $\Psi(X)$ is identical to that of $X$, but what it has done is to swap the values between the two intervals $(a,b]$ and $(-\infty, c]$.

Figure 1: graph of Psi

Example of $\Psi$ for $(a,b]=(1.5, 1.75]$.

Let the Pearson correlation of $(X,Y)$ be $\rho \in (-1, 1)$. (Without loss of generality we may now suppose both $X$ and $Y$ have been standardized, because this will change neither $\rho$ nor the continuity of $X$). Let $x_q$ be any real number, as in the question, where the conditional expectation of $Y$ is to be evaluated. Choose $(a,b]$ for which $x_q \in (a,b]$ but make it so narrow that $\Pr(X \in (a,b])$ is tiny. Then the change from $\rho = \mathbb{E}(XY)$ to $\rho^\prime = \mathbb{E}(\Psi(X)Y)$ can be made arbitrarily small. (It takes a little work to show this; it comes down to the fact that the conditional expectation of $Y$ given $X\le c$ increases relatively slowly as $|b-a|$ decreases. If it didn't, $\rho$ would not be defined.) However, applying $\Psi$ changes $\mathbb{E}(Y|X=x_q)$ to

$$\mathbb{E}(Y|\Psi(X) = x_q) = \mathbb{E}(Y|X = \Psi(x_q)),$$

which is a conditional expectation for $Y$ at some value of $X$ less than or equal to $c$.

Figure 2: graph of the PDF of (Psi(X), Y)

Contours of the PDF. Here $(a,b] = (1.5, 1.75]$. The original bivariate normal distribution was given a correlation of $0.85$, which reduced to approximately $0.5$--the target value--when probabilities in the two strips were swapped.

When $(X,Y)$ is a bivariate normal distribution, $c\to -\infty$ as $|b-a|\to 0$. Provided $\rho\ne 0$, the conditional expectation of $Y$ is pushed off to $-\infty$ for $\rho \gt 0$ and to $+\infty$ for $\rho \lt 0$. An analogous construction, swapping the interval $(a,b]$ with $[c, \infty)$, will push the conditional expectation of $Y$ infinitely far in the other direction. By adjusting the original value of $\rho$ slightly we may compensate for the infinitesimal change in $\rho$ that takes place, showing that no matter what the original value of $\rho$ may be, we can say nothing about the conditional expectation of $Y$ at any particular point $X=x_q$.

(The apparent exception $\rho=0$ can be handled by starting with, say, a bivariate distribution with Normal marginals whose support is confined to the lines $y=\pm x$.)

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  • $\begingroup$ +1 This is very interesting. It's somewhat related to the construction I had in mind in writing the question, but it's a better targeted to moving just the conditional in the immediate neighborhood of the quantile and a more thoughtful discussion than I had toyed with. Your conclusion seems at first reading to be correct. Thank you. $\endgroup$ – Glen_b -Reinstate Monica Feb 3 '15 at 23:49
  • $\begingroup$ Actually +1 is inadequate here. $\endgroup$ – Glen_b -Reinstate Monica Feb 3 '15 at 23:57
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If I understand your question correctly the answer depends on the "actual bivariate dependence relationship (the copula)" used.

Well there exist bounds on the value a copula can take right? So why not use the comonotonicity copula and countermonotonicity copula to establish the limits.

enter image description here

Source: Thorsten Schmidt - Coping with copulas

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  • $\begingroup$ The question is more restrictive than the bounds on copulas - you can't attain the co- and counter- monotonicity bounds because of the constraint on $\rho$. $\endgroup$ – Glen_b -Reinstate Monica Feb 5 '15 at 9:19

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