# When (if ever) is the sum of two dependent geometric RVs negative binominal?

Imagine you have two random variables $$X$$ and $$Y$$, you know $$X \sim \text{Geometric}(p) \\ X + Y \sim \text{Negative Binomial}(2, p)$$ I am interested in what if anything can be said about the distribution of $$Y$$ under these conditions. I now that a possible solution for Y that fits the above system is $$Y \sim \text{Geometric}(p)\\ Y \perp X$$ If no assumptions are made, it seems plausible that there might be infinite solutions for $$Y$$.

I am interested whether something more specific can be said in the case where $$Y$$ is required to have a geometric distribution given $$X$$ (parameter may depend on X). Is the trivial solution for that case (just assuming independence as above) the only solution in that case or are there others too?

EDIT: For clarity, if all variables are supported on the non-negatives. (i.e. $$X \geq 0, X+Y \geq 0, Y \geq 0$$), then it can be shown (see comments below) that there is only one solution (namely the independence solution).

But in my particular use case, I have a slightly weirder scenario: $$X$$ is supported only on the positives, $$Y$$ is supported only on the positives if $$X=1$$ and on the non-negatives otherwise. (As a result X + Y is always at least two).

• Are your $X$ and $Y$ distributed on the non-negative integers or on the positive integers? Either way, I suspect you can show that in your final paragraph the conditional distribution of $Y \mid X=x$ is $\text{Geometric}(p)$ (i.e. independence) by considering the probabilities of small values for $X+Y$ – Henry Dec 5 '19 at 19:28
• thanks for your comment, I clarified in the question. – mexmex Dec 5 '19 at 19:40
• $X \geq 1, X+Y \geq 2, Y \geq 0$ is a little strange if $X$ has a geometric distribution, and given $X=x$ you require $Y$ to have a geometric distribution. Ithink you need to choose between $X \geq 1, X+Y \geq 2, Y \geq 1$ and $X \geq 0, X+Y \geq 0, Y \geq 0$ for the supports – Henry Dec 5 '19 at 19:47
• I figured that with the universal non-negative support, there is only one unique solution, which is the independence solution. I found this using the filling-in method you described. – mexmex Dec 5 '19 at 19:58
• A funny coincidence since this week puzzle in our coffee room was to find the joint such that $X,Y$ are $U(0,1)$ and $X+Y$ is $U(1/2,3/2)$. – Xi'an Dec 5 '19 at 21:13