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).

  • 1
    $\begingroup$ 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 $X+Y$ $\endgroup$ – Henry Dec 5 '19 at 19:28
  • $\begingroup$ thanks for your comment, I clarified in the question. $\endgroup$ – mexmex Dec 5 '19 at 19:40
  • $\begingroup$ $X \geq 1, X+Y \geq 2, Y \geq 0$ 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 $\endgroup$ – Henry Dec 5 '19 at 19:47
  • 2
    $\begingroup$ 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. $\endgroup$ – mexmex Dec 5 '19 at 19:58
  • 1
    $\begingroup$ 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 $U(1/2,3/2)$. $\endgroup$ – Xi'an Dec 5 '19 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.