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If we have $$ X\sim Poisson(\lambda), Y|X = x\sim Binomial(x+1,p) $$ What is the correlation coefficient of X and Y?

So I used $$\rho=\frac{Cov(X,Y)}{\sqrt{Var(x)Var(Y)}} = \frac{E[X[E[Y|X]]-E[X]E[E[Y|X]]}{\sqrt{E[X](E[E[Y|X]^2] -E[E[Y|X]]^2)}}$$ Is this right? Because I got a wacky answer, and is there an easier way?

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I wonder how you come up with your denominator because by Law of Total Variance:

$$\begin{align}\operatorname{var}(Y)&=\operatorname{var}(E[Y|X])+E[\operatorname{var}(Y|X)]=\operatorname{var}((X+1)p)+E[(X+1)p(1-p)]\\&=p^2\lambda+p(1-p)(\lambda+1)=p(\lambda+1-p)\end{align}$$ and $\operatorname{var}(X)=\lambda$. The numerator seems OK.

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  • $\begingroup$ thank you so much!! $\endgroup$
    – Ibrahim B.
    Jun 25, 2019 at 12:59

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