# Sum of hypergeometric distribution

An urn contains $$w$$ white and $$b$$ black balls. $$n$$ extractions without replacement are made (Hypergeometric distribution). The distribution of $$\mathbb{P}(X_i=s)$$ with $$i\geq s$$ ($$s$$ white on $$ith$$ drawn) is:

$$\mathbb{P}(X_i=s)=\frac{\dbinom{w}{s}\dbinom{b}{i-s}}{\dbinom{w+b}{i}}$$

How can I find $$\mathbb{P}(Z=X_i+X_j)$$? I know that $$X_i$$ and $$X_j$$ are not independent.

• "$n$ extractions without replacement are made" - that by itself does not imply a hypergeometric distribution. If you say then that the $Y$ is the number of black balls for example, then you can say $Y$ follows a HG distribution.
– Em.
Jun 19 '16 at 10:32
• @probablyme: I think here $X_i$ is the total number of white balls after $i$ draws and so has a hypergeometric distribution Jun 19 '16 at 10:46
• @Henry Oh, sorry for the confusion. I was making a point about semantics. Like often you will see people say that they roll a die and that this follows a uniform distribution. No, rolling a die does not follow anything. The value that shows on a fair die when you roll it follows a uniform distribution.
– Em.
Jun 19 '16 at 10:50

I suspect that if $i \le j$ and $0 \le z \le i+j$ with $Z=X_i+X_j$ then $$\mathbb{P}(Z=z) = \frac{\displaystyle \sum_{s: \max(0,z-w) \le s \le \min(i,z/2)} \dbinom{w}{s}\dbinom{b}{i-s}\dbinom{w-s}{z-2s}\dbinom{b-i+s}{j-i-z+2s}}{\dbinom{w+b}{i} \dbinom{w+b-i}{j-i}}$$ and I would guess that it might be difficult to simplify this except in special cases