Correct formulas for the mean and variance of negative binomial distribution

Question

According to ScienceDirect and StatTrek, a negative binomial distribution where:

$x$ number of trials, $x = \textrm{1, 2, ...}$

$r$ number of failures, $r = \textrm{1, 2, ... }x$

$k$ number of successes, $k = \textrm{0, 1, ... }(x-r)$

$p$ probability of success, $0<p<1$

The the mean and variance are calculated by:

$$\mathbf E[X_k] = \frac{k}{p}$$

$$\sigma_{X_k}^2 = \frac{k(1-p)}{p^2}$$

However, Wikipedia and this question say they are:

$$\mathbf E[X_k] = \frac{pr}{1-p}$$

$$\sigma_{X_k}^2 = \frac{pr}{(1-p)^2}$$

I am completely lost here. Can someone please help?

Answer 1

The negative binomial distribution has many different parameterizations, because it arose multiple times in many different contexts. Hilbe's Negative Binomial Regression gives a good overview in case you are interested.

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I'll concentrate on tying the Wikipedia (W) and ScienceDirect (SD) articles together. The StatTrek one is a bit hard for me to parse.

In the present case, there are two sources of confusion:

• On the one hand, the W article defines the negbin as counting the number of failures until we have a certain number of successes, whereas the SD article defines it as the number of trials (so, failures plus successes).
• On the other hand, the SD article does not explicitly define what $p$ is. It turns out that if SD's $p$ is the probability of failure, whereas W's $p$ is the probability of success, everything falls into place.

Of course, we also have that W denotes the number of successes by $r$ and SD by $k$.

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So, let's unify things. Here is our common nomenclature:

• $r$ is the number of successes (following W rather than SD)
• $p_W$ is the probability of success
• $p_{SD}$ is the probability of failure, so $p_{SD}=1-p_W$

Now, for our random variables: let

• $X_W$ denote the number of failures until we have $r$ successes
• $X_{SD}$ denote the number of trials until we have $r$ successes

So obviously, we have

$$ X_{SD} = X_W+r. $$

Now, do the formulas for the expectation match?

$$ \begin{align*} E X_{SD} = & EX_W+r \quad\text{by additivity of the expectation} \\ = & \frac{p_Wr}{1-p_W}+r \quad\text{from W} \\ = & \frac{(1-p_{SD})r}{p_{SD}}+r \quad\text{because $p_W=1-p_{SD}$} \\ = & \frac{r}{p_{SD}} \\ = & E X_{SD} \quad\text{from SD.} \end{align*} $$

So the formulas the expectation match.

For the variance, $$ \begin{align*} \sigma^2_{X_{SD}} = & \sigma^2_{X_W} \quad\text{by additivity of the expectation} \\ = & \frac{p_Wr}{(1-p_W)^2} \quad\text{from W} \\ = & \frac{(1-p_{SD})r}{p_{SD}^2} \quad\text{because $p_W=1-p_{SD}$} \\ = & \sigma^2_{X_{SD}} \quad\text{from SD.} \end{align*} $$

So the formulas for the variance also match.