Conditional Probability Distribution for two Discrete Uniform Random Variables with given Correlation Coefficient

Question

I consider a problem with two random variables $X, Y \sim Unif\{a,b\}$, for which I want to set a correlation coefficient $Corr(X,Y)=\rho$.

Now, I am interested in the conditional probability mass function for (let's say) $Y$, given $X=a$ (that is: $P(Y|X=a)$) as a function of $\rho$.

Answer 1

I don't think so.

$\begin{align} \rho & = \mathsf{Corr}(X,Y) \\[1ex] & = \dfrac{\mathsf E\Big(\big(X-\mathsf E(X)\big)\big(Y-\mathsf E(Y)\big)\Big)}{\mathsf E\Big(\big(X-\mathsf E(X)\big)^2\Big)^{1/2} \;\mathsf E\Big(\big(Y-\mathsf E(Y)\big)^2\Big)^{1/2}} \\[1ex] & =\frac{\mathsf E\Big(\big(X-\frac{a+b}{2}\big)\big(Y-\frac{a+b}{2}\big)\Big)}{\big((b-a+1)^2-1\big)/12} \\[1ex] & =\frac{12\;\mathsf E\Big(XY-\frac{a+b}{2}(X+Y)+\frac{(a+b)^2}{4}\Big)}{(b-a+1)^2-1} \\[1ex] & =\frac{12\;\Big(\mathsf E\big(XY)-\frac{a+b}{2}\mathsf E(X+Y)+\frac{(a+b)^2}{4}\Big)}{(b-a+1)^2-1} \\[1ex] & =\frac{12\;\mathsf E\big(XY)-3(a+b)^2}{(b-a+1)^2-1} \\[3ex] \Big(\big((b-a+1)^2-1\big)\rho+3(a+b)^2\Big)/12 & = \mathsf E(XY) \\[1ex] & =\sum_{x=a}^b \sum_{y=a}^b xy\, \mathsf P(X=x)\mathsf P(Y=y\mid X=x) \\[1ex] & = \frac 1{b-a+1}\sum_{x=a}^b \sum_{y=a}^b x\,y\, \mathsf P(Y=y\mid X=x) \\[3ex] \sum_{x=a}^b x\, \sum_{y=a}^b y\, \mathsf P(Y=y\mid X=x) & = (b-a+1)\Big(\big((b-a+1)^2-1\big)\rho+3(a+b)^2\Big)/12 \end{align}$

That's about as close as I can come.