If X and Y are positively correlated poisson variates, show that X+Y cannot be a poisson variate If X and Y are independent, then I write:
P{X+Y=n}
=P{X=k, Y=n-k} (union of two disjoint events)
=P{X=k}P{Y=n-k} (since they are independent)
=$$\SUM_0^n\$$ exp(-LAMBDA_1) (LAMBDA_1^k/k!) exp(-LAMBDA_2) (LAMBDA_2)^(n-k)/(n-k)!
=exp(-LAMBDA_1-LAMBDA_2) $$\SUM_0^n\$$ ( LAMBDA_1^k LAMBDA_2^(n-k))/(k!(n-k)!)
=exp(-LAMBDA_1-LAMBDA_2)/n! (LAMBDA_1+LAMBDA_2)^n

Thus X+Y have a poisson distribution with parameter LAMBDA_1+LAMBDA_2
But if 
Correlation(X, Y)>0 
=> Cov(X,Y)/sqrt(Var(X)Var(Y)) > 0
=> Cov(X, Y) > 0
=> E[XY] - E[X]E[Y] > 0

But I cant proceed any farther from here. I wanted to include some extra terms while I was breaking P{X=k, Y=n-k} into P{X=k}P(Y=n-k} using E[XY], E[X] and E[Y] and show that the resulting distribution was not a poisson variate. Please can someone guide me? And it would be very kind of you if you can post the entire answer.
 A: Assume that $X+Y$ has Poisson distribution. We will show that the correlation coefficient of $X$ and $Y$ is $0$, so in particular $X$ and $Y$ are not positively correlated.
Note that $E(X+Y)=E(X)+E(Y)$ whether or not $X$ and $Y$ are independent. So if $X+Y$ is Poisson it is Poisson with parameter $\lambda+\mu$, where $\lambda$ and $\mu$ are the parameters of $X$ and $Y$. 
The variance of $X+Y$ is $\lambda+\mu$. Thus 
$$E((X+Y)^2)=(\lambda+\mu)^2+\lambda+\mu=\lambda^2+\mu^2+\lambda+\mu+2\lambda\mu.$$ 
But 
$$E((X+Y)^2)=E(X^2)+2E(XY)+E(Y^2)=\lambda^2+\mu^2+\lambda+\mu+2E(XY).$$ 
It follows that $E(2XY)=2\lambda\mu=2E(X)E(Y)$, so $E(XY)=E(X)E(Y)$ and the correlation coefficient is $0$. 
Remark: If you are accustomed to working with covariance, the calculation can be considerably shortened. For 
$$\operatorname{Var}(X+Y)=\operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X,Y).$$
But $\text{Var}(X+Y)=\lambda+\mu$, and $X$ and $Y$ have variance $\lambda$ and $\mu$ respectively, so $\operatorname{Cov}(X,Y)=0$. It follows that the correlation coefficient of $X$ and $Y$ is $0$.
