Is a Beta distribution a continuous version of the Binomial Theorem?

Question

The visual appearance of the PDF for a Beta distribution resembles that for the terms in the Binomial Theorem. Is the former a continuous variant of the discrete terms of the latter? Are they related?

Here's what I have so far:

$$ \begin{align} \textrm{PDF Beta Distribution} &= \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha, \beta)} \\ &= \frac{\Gamma(\alpha+\beta)x^{\alpha-1}(1-x)^{\beta-1}}{\Gamma(\alpha)\Gamma(\beta)} \\ &= \frac{(\alpha+\beta-1)!x^{\alpha-1}(1-x)^{\beta-1}}{(\alpha-1)!(\beta-1)!} \\ &= \alpha \binom{\alpha+\beta-1}{\alpha} x^{\alpha-1}(1-x)^{\beta-1} \end{align} $$ compared with the k-th term of the expansion of $(x + y)^n$:

$$ \binom{n}{k}x^{n-k}y^k $$ now setting $y=(1-x)$, $k=\alpha$, and $n=\alpha+\beta-1$ gives:

$$ \binom{\alpha+\beta-1}{\alpha}x^{\beta-1}(1-x)^\alpha $$ The dots aren't quite connected but do suggest a connection.

So are the terms of a binomial expansion related to the discrete steps for a Beta- distribution where $\alpha$ and $\beta$ are natural numbers?

Answer 1

Conceptually, the two distributions are different views of the same model. The binomial distribution is the PMF of k successes given n independent events each with a probability p of success. The beta distribution is the PDF for p given n independent events with k successes.

Mathematically, when $\alpha=k+1$ and $\beta=n - k + 1$, the beta distribution and the binomial distribution are related by a constant factor: $$ Beta(p; \alpha, \beta) = (n+1) Binom(k; n, p) $$

This resource proves the relationship to the binomial distribution:

Suppose $\mathit{X}$ is a random variable having a Beta distribution with parameters $\alpha$ and $\beta$. Let $\mathit{Y}$ be another random variable such that its distribution conditional on $\mathit{X}$ is a binomial distribution with parameters $n$ and $\mathit{X}$. Then, the conditional distribution of $\mathit{X}$ given $\mathit{Y}=y$ is a Beta distribution with parameters $\alpha+y$ and $\beta+n-y$.