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Suppose $X, Y$ are non-independent random variables with $X\sim NB(r,p)$ and $Y\sim NB(s,p)$; correlation between $X$ and $Y$ is $<$ 1.

Is sum $X + Y$ distributed as negative binomial? Or not?

den2042
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  • How can X+Y be distributed as negative-binomial? Negative binomial has positive probability for arbitrarily large numbers, where as $P(X+Y>r+s)=0$ – Meet Taraviya Jun 23 '17 at 06:57
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    Maybe this can help https://stats.stackexchange.com/questions/21549/how-to-add-two-dependent-random-variables – Deep North Jun 23 '17 at 07:04
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    @Meet There are two kinds of negative binomials, one for "number of trials to the $r$-th success" and one for "number of *failures* to the $r$-th success". Under independence and given the same $p$, the sum of two negative binomials of the second type is negative binomial of that same type. See the two different sets of support [here](https://en.wikipedia.org/wiki/Negative_binomial_distribution#Alternative_formulations) – Glen_b Jun 24 '17 at 03:59

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As stated your question doesn't specify enough information to give a definite answer.

In some situations the answer is "yes". In general the answer is "no".

An example of a situation where the answer is "yes" is as follows:

Let $W\sim \text{negbin}(n_w,p)$, $U\sim \text{negbin}(n_u,p)$, $V\sim \text{negbin}(n_v,p)$, where $W,U$ and $V$ are all independent -- here we're talking about the parameterization in the sidebar to the right in the Wikipedia page on the negative binomial.

Now let $X=W+U$ and $Y=W+V$. Then $X$ and $Y$ are each negative binomial but they're correlated (since $\text{cov}(X,Y) = \text{cov}(W+U,W+V) = \text{var}(W)+0+0+0$).

In general the distribution of $X+Y$ depends on their joint distribution, which you did not specify.

Glen_b
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