2
$\begingroup$

In a paper I am reading, they define the negative binomial as the following: random variable $X$ has a negative binomial distribution with parameters $p \in (0,1),k \in \mathbb{N}$ if

$$\mathbb{P}[X=t]=\begin{pmatrix}t-k \\ k-1\end{pmatrix}p^{k}(1-p)^{t-k},$$

for all $t \in \mathbb{N}$.

I haven't seen this form for the negative binomial before, and I am a little confused. My questions are

  • What is the interpretation of this form for the negative binomial?
  • It's possible that we can have $k>t$ in which case the binomial coefficient isn't well-defined, surely?

I feel like I am missing something. Thanks.

$\endgroup$
2
  • 2
    $\begingroup$ Can you add a link to the paper? This looks strange indeed. $\endgroup$ Commented Sep 30, 2022 at 9:10
  • $\begingroup$ Some fonts, especially in PostScript, occasionally do not show the vertical crossbar in the "+" symbol well. Most likely the numerator of the Binomial coefficient is $t+k,$ not $t=k.$ But that can't explain the appearance of "$t-k$" in the exponent of $1-p.$ $\endgroup$
    – whuber
    Commented Sep 30, 2022 at 15:59

1 Answer 1

1
$\begingroup$

This formula makes no sense, there must be an error somewhere. For instance, if we have parameters $p=0.2$ and $k=3$ (so the negbin, in the standard interpretation, would give the probability of $t$ failures before we see $k=3$ successes in a sequence of Bernoulli trials with success probability $p=0.2$ - but the interpretation here may be different), we get the following PMF:

negbin probs

This is the wrong way around, the probabilities should (potentially) first rise, then fall.

R code:

pp <- 0.2
kk <- 3
tt <- seq(0,3*kk)
(probs <- choose(tt-kk,kk-1)*pp^kk*(1-pp)^(tt-kk))
plot(tt,probs,pch=19)
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.