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I am looking for an example of two random variables $X,Y$ such that

(a) $X,Y$ are not independent.

(b) At least one of $X,Y$ is not normal.

(c) $E(X|y)$ (expected value of $X$ given $Y=y$) is linear in $y$, i.e. of the form $a+by$, and $E(Y|x)$ is linear in $x$.

(d) The correlation coefficient $\rho\neq \pm 1$.

4 Answers 4

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Note that if Y is a binary random variable, then any function of Y is linear in Y. For example $ \sin(Y)\;=\;\sin(0)\;+\;(\sin(1)-\sin(0))\cdot Y. $ This means that any generic distribution over $\{0,1\}^2$ satisfies your conditions.

By the way, in general, if $E(X|Y)$ is linear in $Y$, then it must be necessarily of the form $ E(X|Y)\;=\;E(X)\;+\;\frac{{\rm cov}(X,Y)}{{\rm var}(Y)}(Y-E(Y)). $

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Let $Y$ be any discrete random variable, choose constants $a$ and $b$, and just set $X=aY+b$. They are not independent, not Gaussian, and the conditional expectations are linear as you wanted.

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    You are right. I just added an additional condition.2011-04-15
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$Y, Z$ are r.v.'s. Let $X=aY+bZ$. Then the correlation should not be $\pm1$.

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    @TCL You are right. It's more complicated tha$n$ I thought.2011-04-15
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A general method to get such $X, Y$ is as follows:

Let $Y,Z$ be independent identically distributed random variables. Let $X=Y+Z$. Then as noted above $E(X|y)=y+E(Z)$ is linear in $y$.

And intuitively, $E(Y|x)=E(Z|x)=x/2$. This can be proved rigorously (for continuous case) as follows:

Suppose the density function for $Y,Z$ is $f(u)$. Then the joint pdf for $X,Y$ is $f(y)f(x-y)$, and the density function for $X$ is the convolution of $f$ with itself. Then $ E(Y|x)=\frac{\int y f(y)f(x-y) dy}{\int f(y)f(x-y) dy} .$ This is equal to $x/2$ as shown by Eric Naslund in [this post].

Proving an integral identity: $\int\nolimits_{-\infty}^\infty x f(x)f(t-x) dx =\frac{t}{2} \int_{-\infty}^\infty f(x)f(t-x) dx $

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    If $X=2Y+Z$ and $Y,Z$ are independently uniform on $[0,1]$. Then $E(Y|x)=x/4$ for 0, $=(2x-1)/4$ for 1 and $=(x+1)/4$ for 2, showing that it is not linear.2011-04-25