2

Let $n\geq 0$ be a natural number. Consider the probability distribution on $\{0,\ldots,n\}$ whose probability mass function is given by $$ \mathbb{P}(X=k) := \binom{k-\frac{1}{2}}{k}\binom{n-k-\frac{1}{2}}{n-k} $$ where as usual $\binom{x}{k} := \frac{1}{k!}x(x-1)\cdots(x-k+1)$.

Equivalently, consider $X$ and $Y$ two independent negative binomial distributed variables with parameters $p$ (arbitrary) and $r=\frac{1}{2}$, so that their sum $X+Y$ follows a geometric distribution with parameter $p$; then the distribution of $X$ conditional to $X+Y=n$ is given by the distribution above (this does not depend on $p$).

I'd be very surprised if this distribution didn't have a standard name, so:

Question: What is this called?

Edit: I just realized that this is a particular case of the negative hypergeometric distribution (if we allow for non-integer $r$, here $\frac{1}{2}$), where $K$ and $N$ in Wikipedia's notation are both equal to (my) $n$. Nevertheless, I think this isn't a very satisfactory answer to my question, because it's a very special case which I'm hoping has a more specific name, and it's not clear whether the term “negative hypergeometric” is appropriate for non-integer $r$ (also, Wikipedia's description of the distribution by drawing without replacement makes absolutely no sense if we try to apply it here). So I'm hoping for a better name or, at least:

Alternate question: What is a natural scenario in which this distribution might occur?

Gro-Tsen
  • 5,541

1 Answers1

1

Answering my own question, it turns out that this is a beta-binomial distribution with parameters $\alpha=\beta=\frac{1}{2}$. I was confused by the fact that Wikipedia has an article called “negative hypergeometric” distribution and one called “beta-binomial” while the two are really the same (or at least, the former is a special case of the latter) with different labeling of the parameters, and no link from one to the other (I made a minimal edit to point out the connection). In the particular case I was taking about, this involves the identity $$ \binom{k-\frac{1}{2}}{k}\binom{n-k-\frac{1}{2}}{n-k} = \binom{n}{k}\frac{\mathrm{B}(k+\frac{1}{2},n+k+\frac{1}{2})}{\pi} $$ which is itself easy to see: $$ \begin{aligned} \binom{k-\frac{1}{2}}{k} &= \frac{(k-\frac{1}{2})(k-\frac{3}{2})\cdots(\frac{1}{2})}{k!} = \frac{(2k-1)!!}{2^k k!}\\ \binom{n-k-\frac{1}{2}}{n-k} &= \frac{(n-k-\frac{1}{2})(n-k-\frac{3}{2})\cdots(\frac{1}{2})}{(n-k)!} = \frac{(2(n-k)-1)!!}{2^{n-k} (n-k)!}\\ \Gamma(k+\frac{1}{2}) &= \big(k-\frac{1}{2}\big)\big(k-\frac{3}{2}\big)\cdots\big(\frac{1}{2}\big) \sqrt{\pi} = \frac{(2k-1)!!}{2^k}\sqrt{\pi}\\ \Gamma(n-k+\frac{1}{2}) &= \big(n-k-\frac{1}{2}\big)\big(n-k-\frac{3}{2}\big)\cdots\big(\frac{1}{2}\big) \sqrt{\pi} = \frac{(2(n-k)-1)!!}{2^{n-k}}\sqrt{\pi}\\ \frac{\mathrm{B}(k+\frac{1}{2},n+k+\frac{1}{2})}{\pi} &= \frac{(k-\frac{1}{2})!\,(n-k-\frac{1}{2})!}{n!\,\pi} = \frac{(2k-1)!!\,(2(n-k)-1)!!}{2^n\,n!}\\ \binom{k-\frac{1}{2}}{k} \binom{n-k-\frac{1}{2}}{n-k} &= \frac{(2k-1)!!\,(2(n-k)-1)!!}{2^n\,k!\,(n-k)!} = \binom{n}{k} \frac{\mathrm{B}(k+\frac{1}{2},n+k+\frac{1}{2})}{\pi} \end{aligned} $$

Gro-Tsen
  • 5,541