cumulative binomial equals integral Can somebody explain why the cumulative binomial equals an integral expression?
Thanks!${}$
$$
\sum_{j=0}^{k-1}\binom{n}{j}\theta^j(1-\theta)^{n-j} = 1- \binom{n}{k}k\int_0^\theta t^{k-1}(1-t)^{n-k}dt\,.
$$
 A: 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 $f=g$.
A: This is essentially the same as the solution by T. Amdeberhan, but I find that it is sometimes easier to prove such identities simultaneously for all $k$. I'll use the standard formula
$$ \sum_{k=0}^n k\binom{n}{k}U^{k-1} = n(1+U)^{n-1}\quad(*) $$
obtained by differentiating the binomial formula. Let
$$ I(k) = \binom{n}{k}k\int_0^\theta t^{k-1}(1-t)^{n-k}\,dt $$
that you want to compute, and consider the sum
$$ \begin{aligned}
\sum_{k=0}^n I(k)X^{k-1}
&= \int_0^\theta \left\{\sum_{k=0}^n\binom{n}{k}k t^{k-1}(1-t)^{n-k}X^{k-1}\right\}\,dt \\
&= \int_0^\theta \left\{\sum_{k=0}^n\binom{n}{k}k \left(\frac{tX}{1-t}\right)^{k-1}\right\}(1-t)^{n-1}\,dt \\
&=  \int_0^\theta n\left(1+\frac{tX}{1-t}\right)^{n-1}(1-t)^{n-1}\,dt, \quad\text{using $(*)$,}\\
&= n\int_0^\theta (1-t-tX)^{n-1}\,dt\\
&= n\int_0^\theta \sum_{k=0}^{n-1} \binom{n}{k} t^k(-1-X)^k\,dt \\
&= n \sum_{k=0}^{n-1} \binom{n}{k} (-1)^k(1+X)^k \frac{\theta^{k+1}}{k+1}.
\end{aligned}
$$
Now expand $(1+X)^k$ using the binomial theorem, flip the order of the sums, and equate the coefficients of the powers of $X$.
