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The following is a problem from my probability practice final:

Let $X$ and $Y$ be discrete random variables with joint mass function

$$f(x,y) = \frac{C}{(x+y-1)(x+y)(x+y+1)} $$ for $x, y = 1, 2, 3...$

Calculate $C$ and find the marginal mass functions of $X$ and $Y$.

Now, to find the marginal distributions, I've tried to sum over all possible values of either variable (which is a sum that looks like it converges), but it's not a very pleasant task, and I can't find a way to do it nicely. Is there some shortcut here that I'm missing?

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1 Answer 1

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The constant $C$ is determined by the normalization condition, i.e. $f(x,y) \geqslant 0$ for all $x,y=1,2,3,\ldots$ and $\sum_{x=1}^\infty \sum_{y=1}^\infty f(x,y) = 1$. $$ \begin{eqnarray} \frac{1}{C} = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{(n+m)((n+m)^2-1)} &=& \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{(n+m)((n+m)^2-1)} \sum_{p=1}^\infty \delta_{p,n+m-1} \\ &=& \sum_{p=1}^\infty \frac{1}{(p+1)p(p+2)}\sum_{n=1}^\infty \sum_{m=1}^\infty \delta_{p,n+m-1} \\ &=& \sum_{p=1}^\infty \frac{1}{(p+1)p(p+2)} \sum_{n=1}^{p} \sum_{m=1}^{p}\delta_{p+1,n+m} \\ &=& \sum_{p=1}^\infty \frac{1}{(p+1)p(p+2)} [z]^{p+1} \left( z\frac{1-z^{p}}{1-z}\right)^2 \\ &=&\sum_{p=1}^\infty \frac{1}{(p+1)p(p+2)} [z]^{p+1} \left( \frac{z}{1-z}\right)^2 \\ &=&\sum_{p=1}^\infty \frac{1}{(p+1)p(p+2)} p \\ &=& \sum_{p=1}^\infty \frac{1}{(p+1)(p+2)} = \sum_{p=1}^\infty\left( \frac{1}{p+1} - \frac{1}{p+2}\right) \\ &\stackrel{\color\maroon{\text{telescopes}}}{=}& \frac{1}{2} \end{eqnarray} $$ Hence $C=2$.

Given that $f(x,y) = f(y,x)$ is suffices to determine marginal in one direction. Similarly, applying the telescoping idea: $$\begin{eqnarray} f_Y(y) &=& \sum_{x=1}^\infty \frac{2}{(x+y)(x+y-1)(x+y+1)} = \sum_{x=1}^\infty \left( \frac{1}{x+y-1} - \frac{2}{x+y} + \frac{1}{x+y+1}\right) \\ &=& \sum_{x=1}^\infty\left( \left( \frac{1}{x+y-1} - \frac{1}{x+y}\right) - \left(\frac{1}{x+y}- \frac{1}{x+y+1}\right)\right) = \\ &=& \frac{1}{y(1+y)} \end{eqnarray} $$

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@jake It might be simpler to first find the marginal, and then, requiring that marginal be normalized to $1$ find the constant $C$. This would be simpler. –  Sasha May 9 '12 at 21:48
    
Thanks. Forgive my ignorance, but I have no idea what you're doing with the series in the first five steps. –  jake May 9 '12 at 21:50
    
In the first line I inserted identity, valid for $m,n \geqslant 1$, i.e. $1 = \sum_{p=1}^\infty \delta_{p,n+m-1}$, then exchanged the order of summation. Then, since $n+m$ is required to be $p-1$, simplified the joint mass function, then I needed to evaluate $\sum_{n=1}^\infty \sum_{m=1}^\infty \delta_{p,n+m-1}$. You could evaluate it by noting that for each $n$, there is a unique $m$, such that $n+m-1=p$, hence $\sum_{n=1}^\infty \sum_{m=1}^\infty \delta_{p,n+m-1} = \sum_{n=1}^p 1 = p$. I used generating functions, and series coefficient thereof. –  Sasha May 9 '12 at 21:59
    
Thanks again, though as you noted, that is all unnecessary. I can just find the marginals in terms of $C$, then sum through to solve for $C$, which is another telescoping series. Just pointing out: I think you need to switch the last two terms in the sum on the second to last line. –  jake May 9 '12 at 22:11

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