I'm trying to understand Negative Binomial Random Variables and have across the following:

$ Z\sim \mathrm{NegBin}(n,p)$ if $Z = X_1 +\cdots+ X_r $ where $X_i's$ are independent identically distributed variables, $\mathrm{Geo}(p)$.

Apparently, $Z$ refers to the time when the $r^\text{th}$ success occurs.

Q.1 However, I don't understand the following:

$$ Z \in \mathbb \{r, r+1, r+2,\ldots\}$$

Q.2 And this bit: $$P(Z=k) = p \binom{n-1}{k-1}p^{k-1}(1-p)^{n-k} $$

Q.3 I also got why this is the case: $ \sum\limits_{k=0}^n k\binom{n}{k} p^k (1-p)^{n-k} = np $ from $\mathrm{Bin}(n,p)$

Thanks a lot

  • $\begingroup$ Each of the geometric random variables is $\ge 1$, so their sum is $\ge r$. $\endgroup$ – André Nicolas Sep 3 '14 at 23:29
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    $\begingroup$ @user131983 : Apparently my copy-editing of your post interfered with an attempt by you to edit it. Go ahead and do it again; I'll stay away from it for now. $\endgroup$ – Michael Hardy Sep 3 '14 at 23:34
  • $\begingroup$ @AndréNicolas Thanks, that helps me understand the first bit. Would you be able to possibly help on the next two bold sentences, or is it related to the answer you just gave? $\endgroup$ – user131983 Sep 3 '14 at 23:36

Each of the geometric random variables $X_1,X_2,\dots,X_r$ can take on any integer value $\ge 1$, so their sum $Z$ can take on any integer value $\ge r$.

For the formula for $\Pr(Z=k)$, we will assume that you are familiar with the binomial distribution.

The random variable $Z$ is the sum of independent identically distributed random variables with geometric distribution. So $Z$ is the "waiting time" (total number of trials) until the $r$-th success.

The $r$-th success occurs on the $k$-th trial if and only if (i) we had exactly $r-1$ successes in the first $k-1$ trials and (ii) we have a success on the $k$-th trial.

By properties of the binomial distribution, the probability of (i) is $\binom{k-1}{r-1}p^{r-1} (1-p)^{(k-1)-(r-1)}$.

And of course the probability of (ii) is $p$.

Multiply, and clean up a bit. We obtain the formula quoted in the post.

Remark: The third question has to do with the binomial, not the negative binomial. We want to evaluate the sum $$\sum_{k=0}^n k\binom{n}{k}p^k(1-p)^{n-k},\tag{1}$$ which is the expression for the mean of the binomial. Alternately we could sum from $k=1$, since the $k=0$ term makes no contribution.

For $k\ne 0$, we have the identity $$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}.$$ This identity can be verified by expressing the binomial coefficients on each side in terms of factorials. Using this identity, we find that the sum (1) is equal to $$n\sum_{k=1}^n n p^k(1-p)^{n-k}.$$ This is $$np\sum_{i=0}^{n-1} p^i q^{n-1-i},$$ which is $np$.

I don't like the above argument, it is messy. More importantly, it is computational, not *conceptual.

I would much rather say that the binomial is a sum $B_1+\cdots+B_n$ of independent Bernoulli. Each of the Bernoulli has expectation $p$. So by the linearity of expectation $E(B_1+\cdots+B_n)=E(B_1)+\cdots +E(B_n)=np$.

  • $\begingroup$ Thank You very much! I just have 2 further questions, Q1) how did you go from $(1)$ to $\sum_{k=1}^n n p^k(1-p)^{n-k}.$ Wouldn't that imply $n = n\binom{n-1}{k-1}$ and Q2) how do we know $\sum_{i=0}^{n-1} p^i q^{n-1-i} = 1$? Thanks again! $\endgroup$ – user131983 Sep 4 '14 at 1:32
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    $\begingroup$ The 2) is easiest to explain. By the Binomial Theorem, the sum is $[p+(1-p)]^{n-1}$, that is, $1^{n-1}$. another way of seeing it is that it is the sum of $\Pr(W=i)$, where $W$ is binomial random variable, $n-1$ trials, probability of success $p$. The sum of all probabilities is $1$. I will use a separate comment for the other one, in case I run out of room. $\endgroup$ – André Nicolas Sep 4 '14 at 3:03
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    $\begingroup$ For your first question, by the identity I had mentioned, the sum (1) is equal to $\sum_{k=1}^n k\cdot\frac{n}{k}\cdot\binom{n-1}{k-1}p^k (1-p)^{n-k}$. The $k$'s in front cancel. $\endgroup$ – André Nicolas Sep 4 '14 at 3:05
  • $\begingroup$ @AndreNicols Thank You! Its clear now. $\endgroup$ – user131983 Sep 5 '14 at 2:24
  • $\begingroup$ You are welcome. It is good to continue until things are clear. $\endgroup$ – André Nicolas Sep 5 '14 at 5:19

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