Can somebody explain why the cumulative binomial equals an integral expression?
Thanks!${}$


$$
\sum\_{j=0}^{k-1}\binom{n}{j}\theta^j(1-\theta)^{n-j} = 1- \binom{n}{k}k\int\_0^\theta t^{k-1}(1-t)^{n-k}dt\,.
$$



cumulative binomial equals integral
**Key players:** $m\binom{n}m=n\binom{n-1}{m-1}$, derivative $\frac{d}{d\theta}$ and re-indexing $j-1\rightarrow j$.


Denote the LHS by $f(\theta)$ and the RHS by $g(\theta)$. Take derivatives w.r.t. $\theta$ to get
$$g'(\theta)=-k\binom{n}k\theta^{k-1}(1-\theta)^{n-k}=-n\binom{n-1}{k-1}\theta^{k-1}(1-\theta)^{n-k},$$
\begin{align} f'(\theta)
&=\sum\_{j=0}^{k-1}j\binom{n}j\theta^{j-1}(1-\theta)^{n-j}
-\sum\_{j=0}^{k-1}(n-j)\binom{n}{n-j}\theta^j(1-\theta)^{n-j-1} \\
&=n\sum\_{j=1}^{k-1}\binom{n-1}{j-1}\theta^{j-1}(1-\theta)^{n-j}
-n\sum\_{j=0}^{k-1}\binom{n-1}{n-j-1}\theta^j(1-\theta)^{n-j-1} \\
&=n\sum\_{j=0}^{k-2}\binom{n-1}j\theta^j(1-\theta)^{n-j-1}
-n\sum\_{j=0}^{k-1}\binom{n-1}j\theta^j(1-\theta)^{n-j-1} \\
&=-n\binom{n-1}{k-1}\theta^{k-1}(1-\theta)^{n-k}.
\end{align}
Therefore, $f'(\theta)=g'(\theta)$. Since $f(0)=g(0)=1$, we conclude $f=g$.


