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A negative binomial distribution is given by $$P(X=x) = \binom{x-1}{k-1}p^{k}(1-p)^{(x-k)},$$ where $p$ is probability of a success and $x \in \{k , k+1 , k+2 , \ldots\}$.

Its mean is given by $$ E(X) = \sum_{x=k}^{\infty} \binom{x-1}{k-1}p^{k}(1-p)^{(x-k)} \cdot x $$ So what I tried $$ \sum_{x=k}^{\infty} \binom{x-1}{k-1}p^{k}(1-p)^{(x-k)}\cdot x = \sum_{x=k}^{\infty} \binom{x}{k}p^{k}(1-p)^{(x-k)} \cdot k, $$ as $\binom{x-1}{k-1} \cdot x$ = $\binom{x}{k} \cdot k$

Now, $$ \sum_{x=k}^{\infty} \binom{x}{k}p^{k}(1-p)^{(x-k)} \cdot k = k \sum_{x=k}^{\infty} \binom{x}{k}\frac{p^{k+1}}{p}(1-p)^{(x-k)} = \frac{k}{p} \sum_{x=k}^{\infty} \binom{x}{k}p^{k+1}(1-p)^{(x-k)}. $$ So, $$ \sum_{x=k}^{\infty} \binom{x}{k}p^{k+1}(1-p)^{(x-k)}=1 $$ (Sum of all probabilities). This implies $$ \frac{k}{p}\sum_{x=k}^{\infty} \binom{x}{k}p^{k+1}(1-p)^{(x-k)} =\frac{k}{p}. $$ Thus, Mean = $\frac{k}{p}$ But this isn't the correct answer, could anyone tell what am I doing wrong ?

Sha Vuklia
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User9523
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  • Look in Wiki for the definition $P(X=x)$. – user64494 Aug 06 '15 at 18:43
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    There are several definitions of the negative binomial. For the one you are using, the mean is $\frac{k}{p}$. What makes you think it is wrong? – André Nicolas Aug 06 '15 at 18:47
  • Your answer is okay. You are looking at the number of trials needed to come to the point that you have $k$ successes, right? That means that $X$ takes the values $k,k+1,\dots$ etc. In other definitions one looks at an $X$ taking the values $0,1,\dots$ etc. There $X$ denotes the number of failures only. – drhab Aug 06 '15 at 18:50
  • Why didn't you accept the answer? – ViktorStein Jun 22 '19 at 10:18

1 Answers1

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You didn't ask for it, but I cannot withold myself from offering you an alternative route to find $\mathbb EX$. One that happily confirms that you are correct.

Write: $$X=G_1+G_2+\cdots+G_k$$

If e.g. $k=3$ then sequence is $FFSFFFFSS$ results in $G_1=3$, $G_2=5$ and $G_3=1$.

Here $G_1$ corresponds with the number of trials needed to come to the first success, $G_2$ after that likewise for the second success, et cetera.

The $G_i$ are iid and geometrically distributed with parameter $p$.

Then: $$\mathbb EX=\mathbb EG_1+\cdots\mathbb EG_k=k\mathbb EG_1=\frac{k}{p}$$

edit:

Even calculation of $\mathbb EG_1$ can be avoided.

If $S$ denotes the event that the first trial is a success then:

$\mathbb{E}X=\mathbb{E}\left(X\mid S\right)p+\mathbb{E}\left(X\mid S^{c}\right)q$.

Here $\mathbb{E}\left(X\mid S\right)=1+\frac{k-1}{k}\mathbb{E}X$ and $\mathbb{E}\left(X\mid S^{c}\right)=1+\mathbb{E}X$.

To understand the expression for $\mathbb E(X|S)$ realize that a success in the first trial gives $X$ the form $1+G_2+\cdots+G_k$.

This gives you the equality:

$$\mathbb{E}X=\left[1+\frac{k-1}{k}\mathbb{E}X\right]p+\left[1+\mathbb{E}X\right]q$$

Easy to solve and resulting in $\mathbb{E}X=\frac{k}{p}$.

drhab
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