I am stuck in finding the sum of $\sum_{n=0}^{\infty} {n \choose y} p^{y+1}(1-p)^{2n-y}$.
The sum looks quite similar to a negative binomial sum but I can't really find the exact form. Can anyone help?
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$\begingroup$ This can be rewritten as $$p^{y+1}(1-p)^{-y}\sum_n \binom ny (1-p)^{2n}$$ This means you really want to solve $$\sum_n \binom ny x^n$$ for $|x|<1.$ $\endgroup$– Thomas AndrewsCommented Nov 8, 2021 at 3:48
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$\begingroup$ hint: $\binom{n}{y} = \left.\frac1{y!}\frac{d^y}{dt^y}(t^n) \right|_{t=1}$ $\endgroup$– achille huiCommented Nov 8, 2021 at 3:49
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$\begingroup$ Please try to avoid math-only titles. See Guidelines for good use of MathJax on question titles for more information. $\endgroup$– souplessCommented Nov 8, 2021 at 6:09
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1 Answer
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It is a negative binomial sum. In a sequence of iid Bernoulli trials with success probability $\theta$, suppose we are interested in counting the total number of trials $N$ just before we observe the $(y+1)^{\rm th}$ success; that is to say, we exclude the final success from the count. Then $N$ is negative binomial with parameters $y + 1$ and $\theta$ with parametrization $$\Pr[N = n] = \binom{n}{y} \theta^{y+1} (1-\theta)^{n-y}, \quad n \in \{y, y+1, \ldots \}.$$ Now this looks a lot like your sum except you have $(1-p)^{2n-y}$ instead. So this suggests we let $(1-\theta)^n = (1-p)^{2n}$, or $$\theta = 1 - (1-p)^2 = p(2-p),$$ and the rest is trivial.
