Is there a way to get the Binomial coefficients as terms that normalize a Binomial distribution?

Consider a Binomial distribution. We toss a coin $n$ times and model the number of heads, $X$ we'll see. If the probability of heads is $p$, it is clear that:

$$P(X=x) \propto p^x (1-p)^{n-x} \tag{1}$$

For continuous distributions, we typically go from here and integrate to get the normalizing factor. For the Binomial distribution, we go from first principals.

I'm wondering if we can instead work from just equation (1) and get the binomial coefficient by trying to ensure the distribution sums to $1$ for $x \in \{0,1,\dots n\}$?

I'm looking for something similar to what we do with the Beta where we integrate equation (1) over $p$ to get the Beta distribution.

Solution 1:

Your question has some confused premises which is probably messing you up a bit. Firstly, the initial result $(1)$ you give is only valid for proportionality in $p$ --- it is not proportionate with respect to the argument variable $x$, because you removed a term that depends on $x$. So the proportionality result you have given (interpreted as $\overset{p}{\propto}$) is perfectly fine for getting the likelihood function (and the resulting posterior in Bayesian analysis), but it is not helpful for getting the other term in the binomial distribution.

It is not possible to go from the part you have created to get the binomial distribution (because you have removed a term that depends on $x$). If you were to normalise the function you have used here (over the values $x=0,...,n$) you would actually get a truncated geometric distribution instead:

$$\begin{align} p(x) &\overset{x}{\propto} p^x (1-p)^{n-x} \cdot \mathbb{I}(x \in \{ 0,...,n \}) \\[12pt] &= (1-p)^n \Big( \frac{p}{1-p} \Big)^x \cdot \mathbb{I}(x \in \{ 0,...,n \}) \\[8pt] &= (1-p)^n \Big( 1 - \frac{1}{1-p} \Big)^x \cdot \mathbb{I}(x \in \{ 0,...,n \}) \\[8pt] &\overset{x}{\propto} \text{TruncGeom}(x | \tfrac{1}{1-p}, n). \\[6pt] \end{align}$$