There is a claim in my book that there is a connection to the Beta CDF and a Binomial Summation without explaining further.

"Integration by Parts can be used to show that for $0<y<1$, and $\alpha$ and $\beta$ both integers

$$ F(y)=\frac{1}{\beta(\alpha,\beta)}\int_{0}^{y}t^{\alpha-1}(1-t)^{\beta-1}dt=\sum_{i=\alpha}^{n} \binom{n}{i} y^i (1-y)^{n-i}, $$

where $n=\alpha+\beta+1$."

I can see the appeal of using the Binomial tables, algorithms etc. to calculate Beta CDF values instead of having software calculate the integral directly. I am trying to figure out how they would apply the parts to derive the connection. Is it just iterating repetitions of integration by parts? Maybe an inductive derivation? Maybe a trick for connecting using the Binomial Theorem arises? Can anyone see how to make the connection here?


This is basically because we can express the distribution function of a Binomial random variable in terms of the incomplete beta function, which in turn is related to the distribution function of a Beta distributed random variable.

We assume that $k$ and $n-k+1$ are both positive integers for the following 'simplification' to hold.

The survival function of a random variable $X\sim \text{Bin}(n,t)$ is given by

$$P(X\geqslant k)=\sum_{j=k}^n\binom{n}{j}t^j(1-t)^{n-j}$$

Differentiating wrt $t$, we have

\begin{align} \frac{d}{dt}P(X\geqslant k)&=\sum_{j=k}^n\binom{n}{j}\left[jt^{j-1}(1-t)^{n-j}-(n-j)t^j(1-t)^{n-j-1}\right] \\&=\sum_{j=k}^n\left[n\binom{n-1}{j-1}t^{j-1}(1-t)^{n-j}-n\binom{n-1}{j}t^j(1-t)^{n-j-1}\right] \\&=n\binom{n-1}{k-1}t^{k-1}(1-t)^{n-k} \end{align}

Integrating we get,

\begin{align} P(X\geqslant k)&=n\binom{n-1}{k-1}\int_0^t x^{k-1}(1-x)^{n-k}\,dx \\&=\frac{n!}{(k-1)!(n-k)!}\int_0^t x^{k-1}(1-x)^{n-k}\,dx \\\\&=\int_0^t \frac{ x^{k-1}(1-x)^{n-k}}{B(k,n-k+1)}\,dx \end{align}

  • $\begingroup$ I did just work out the integration by parts technique the book suggested to realize you need to repeat parts to obtain each term meaning you need to do by parts n-k+1 times. This solution is far more efficient and I wavered knowing the book I am using tends to suggest inefficient methods. Also I believe there is a way to modify the Bayes Billiards argument to obtain this without the need for the analytic solve. $\endgroup$ – CScholten Oct 14 '18 at 20:08

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