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I have with the benefit of context the following result:

$$\sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{n} \left({m+j-1 \choose j}\left(\frac{\theta_1}{\theta_1+t}\right)^m \left(\frac{t}{\theta_1+t}\right)^{j}\right) \left({m+n-j-1 \choose n-j}\left(\frac{\theta}{\theta+t}\right)^m \left(\frac{t}{\theta+t}\right)^{n-j}\right)=1$$

I have verified this also numerically. Now, how do I go about proving it?


My attempt: I basically collected all terms together and "simplified" this to (calling the summation $\beta$ since it's related to the false negative rate of the Binomial test applied to the negative binomial distribution):

$$\beta = \sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{n} \frac{n \choose j}{2m+n-2 \choose m+j-1}{2m-2\choose m-1} \left(\frac{\theta_1\theta}{(\theta_1+t)(\theta+t)}\right)^m \frac{t^n}{(\theta_1+t)^j(\theta+t)^{n-j}}$$

Taking as many terms out of the summations as possible we get:

$$\beta ={2m-2\choose m-1} \left(\frac{\theta_1\theta}{(\theta_1+t)(\theta+t)}\right)^m \sum\limits_{n=0}^{\infty} {2m+n-2\choose n} t^n\sum\limits_{j=0}^{n} \frac{n \choose j}{2m+n-2 \choose m+j-1}\frac{1}{(\theta_1+t)^j(\theta+t)^{n-j}}$$

Now, the inner summation over $n$ becomes a thorn since it involves a ratio of binomial terms. This has not provided any simplification over the original expression.


Note: I secretly don't care so much about the summation in the question since I know it's $1$ (it's basically the sum over entire domain of the PMF of the density of a sum of two negative binomial distributions). The summation I really care about is:

$$\sum\limits_{n=0}^{\infty} \sum\limits_{j=0}^{[\frac{n}{2}]} \left({m+j-1 \choose j}\left(\frac{\theta_1}{\theta_1+t}\right)^m \left(\frac{t}{\theta_1+t}\right)^{j}\right) \left({m+n-j-1 \choose n-j}\left(\frac{\theta}{\theta+t}\right)^m \left(\frac{t}{\theta+t}\right)^{n-j}\right)$$

But hoping working out the mechanics of the original one (which should be easier and I know the answer) will provide insight into this one.

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Recall the formula \begin{eqnarray*} \sum_{i=0}^{\infty} \binom{K+i-1}{i} x^{i} =\frac{1}{(1-x)^K}. \end{eqnarray*} You also need to invert the order of the sums \begin{eqnarray*} \sum_{n=0}^{\infty} \sum_{j=0}^{n} \cdots = \sum_{j=0}^{\infty} \sum_{n=j}^{\infty} \cdots \end{eqnarray*} The sums will then decouple into the following $2$ products \begin{eqnarray*} \sum_{j=0}^{\infty} \binom{m+j-1}{j} \left( \frac{\theta_1}{\theta_1+t} \right)^m \left( \frac{t}{\theta_1+t} \right)^j \end{eqnarray*} and if you then let $p=n-j$ \begin{eqnarray*} \sum_{p=0}^{\infty} \binom{m+p-1}{p} \left( \frac{\theta}{\theta+t} \right)^m \left( \frac{t}{\theta+t} \right)^p. \end{eqnarray*} Both of these are easily seen to equal $1$.

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  • $\begingroup$ thanks. I don't see how reversing the order of summations and change of indices worked. Could you please elaborate on that part? Also, what would this maneuver look like if the inner sum was from $0$ to $[\frac{n}{2}]$ instead of $0$ to $n$? $\endgroup$ Dec 11, 2019 at 4:46
  • $\begingroup$ Never mind. If we visualize an $n-j$ grid, this becomes quite obvious and the extension becomes possible as well. Much appreciated! $\endgroup$ Dec 11, 2019 at 5:21

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