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How to calculate sum of negative binomial distribution ? it $ E[X], Var[x]$ and $P\{X = n |\text {first event occurs on the 5-th try}\}$

$P(X=k)= \binom{k-1}{r-1}p^r(1-p)^{k-r}$

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    http://mathworld.wolfram.com/BinomialCoefficient.html2013-10-29

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Suppose that the probability of "success" is $p\ne 0$. Let $r$ be a fixed positive integer, and let $X$ be the number of trials until the $r$-th success. Then $\Pr(X=k)=\binom{k-1}{r-1}p^r(1-p)^{k-r}.$ We want to show that the sum of these probabilities over all $k$ is $1$. In a sense, this is obvious, since with probability $1$ the process will terminate. But probably it is intended that the sum be evaluated "analytically," without the use of probabilistic notions. So we will do it by manipulation of power series.

First note that the smallest possible value of $k$ is $r$, so we want to evaluate $\sum_{k=r}^\infty \binom{k-1}{r-1}p^r(1-p)^{n-r}.$

Let $f(x)=\dfrac{1}{1-x}$. If $|x|\lt 1$, then $f(x)$ has the power series expansion $f(x)=\sum_{k=1}^\infty x^{k-1}.$ Differentiate $f(x)$ a total of $r-1$ times with respect to $x$. We have $f'(x)=(1)(1-x)^{-2}$, $f''(x)=(2)(1)(1-x)^{-3})$, $f'''(x)=(3)(2)(1)(1-x)^{-4}$, and so on. So the $(r-1)$-th derivative of $f(x)$, usually denoted by $f^{(r-1)}(x)$, is given by the formula $f^{(r-1)}(x)=(r-1)!(1-x)^{-r}.$ Now differentiate the power series expansion of $f(x)$ a total of $r-1$ times, term by term. For $k-1\ge r-1$ (i.e.: $k\ge r$), the $(r-1)$-th derivative of $x^{k-1}$ is equal to $(k-1)(k-2)\cdots(k-r+1)x^{k-r}$. It is useful to note that $(k-1)(k-2)\cdots(k-r+1)x^{k-r}=\dfrac{(k-1)!}{(k-r)!}x^{k-r}$. We conclude that $f^{(r-1)}(x)=(r-1)!(1-x)^{-r}=\sum_{k=r}^\infty \frac{(k-1)!}{(k-r)!}x^{k-r}.$ Dividing both sides by $(r-1)!(1-x)^{-r}$, we get $1=\sum_{k=r-1}^\infty \binom{k-1}{r-1}(1-x)^r x^{k-r}.$ Finally, put $x=1-p$. Then $1-x=p$, and we have the desired result.

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    +1, by the way. I was actually searching for a specific answer to my side question, found this, and figured I'd kill two birds with one stone.2013-10-29