Originally Posted by

**Mathsdog** Hi Harish21,

I wonder if I might bounce another one off you. This time it's three variables, summing one out. It should be similar to the previous one, but again I must be being even dumber than usual.

It is:

Given the joint: $\displaystyle p_{X, Y, Z}(x, y, z) = \frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

Where $\displaystyle x, y, z = 0, 1, 2, 3$ and $\displaystyle 0 \leq x + y + z \leq 3$

Find the joint marginal $\displaystyle p_{X, Y}(x, y)$

Which, I think, is: $\displaystyle p_{X, Y}(x, y) = \sum_{z=0}^{3-x-y}\frac{3!}{x!y!z!(3-x-y-z)!} \left( \frac{1}{2} \right)^x \left( \frac{1}{12} \right)^y \left( \frac{1}{6} \right)^z \left( \frac{1}{4} \right)^{3-x-y-z}$

with the same conditions as above, right?

I did the same trick you did before with the factor I think

so,

$\displaystyle \frac{3!}{x!y!z!(3-x-y-z)!} \cdot \frac{(3- x- y)!}{(3- x- y)!} = \frac{3!}{x!y!(3-x-y)!} \cdot \frac{(3- x- y)!}{z!(3-x-y-z)!}$

Which gives the right factor for a joint multinomial for x and y, n=3, $\displaystyle p_{1} = 1/2$ and $\displaystyle p_{2} = 1/12$

i.e. $\displaystyle p_{X,Y}(x, y)=\frac{3!}{x!y!(3-x-y)!} \cdot (1/2)^x (1/12)^y (5/12)^{3-x-y}$

Which is what the sum over z should reduce to, right?

But, damn it, I can't seem to massage out this distribution from the sum over z in a similar way to before.

Thanks in advance for any help. BW, MD