# Proving an identity of partial binomial sum from a statistical context without using combinatorics

1. The identity is: $$\sum_{i=m}^n {n \choose i}p^i(1-p)^{n-i}=m{n \choose m}\int_0^p {t^{m-1}(1-t)^{n-m}}\,\mathrm dt\quad(0{\le}m{\le}n)$$
2. How I met it: The CDF of $m$-th order statistic is $$P(X_{(m)}<x)=\sum_{i=m}^n {n \choose i}F^i(x)[1-F(x)]^{n-i}$$ as is the left side of the first identity. My text book says it equals to $$m{n \choose m}\int_0^{F(x)}F^{m-1}(x)[1-F(x)]^{n-m}$$(so it's easier to find PDF by differentiating) which is the right side of the first identity. But the proof is left as an excercise.
3. What I know: There is a hint on the book says the two are equal at $p=0$. Differentiate both sides with respect to $p$. The results are equal, then it's proved. I come to this identity after differentiating(the identity I want to proove)$$\sum_{i=0}^{m-1}{n \choose i}p^{i-1}(1-p)^{n-i-1}(np-i) \;?=?\;m{n \choose m}p^{m-1}(1-p)^{n-m}\;(0{\le}m{\le}n)$$ I know it is right, because mathematica proved it.[mathematica output][1]
4. I would be very grateful if you have a more "math" way to solve the identity in 3.(follow the hint) or in 1.(if you have an easier proof not follow the hint)."math" way means not by program like mathematica and not by induction.

Suppose we try to show that

$$\sum_{q=m}^n {n\choose q} p^q (1-p)^{n-q} = m {n\choose m} \int_0^p t^{m-1} (1-t)^{n-m} \; dt.$$

The LHS is clearly a polynomial in $p$ of degree $n$ and the coefficient on $[p^k]$ (here $k\ge m$ by construction) is given by

$$[p^k] \sum_{q=m}^n {n\choose q} p^q (1-p)^{n-q} = \sum_{q=m}^n {n\choose q} [p^{k-q}] (1-p)^{n-q} \\ = \sum_{q=m}^n {n\choose q} (-1)^{k-q} {n-q\choose k-q}.$$

Here we assume that $m\ge 1$ so that $k\ge 1$ also. (The integral is singular at zero when $m=0.$) We have for the RHS putting $px=t$ the integral

$$m {n\choose m} \int_0^1 p^{m-1} x^{m-1} (1-px)^{n-m} \; p \; dx \\ = p^m m {n\choose m} \int_0^1 x^{m-1} (1-px)^{n-m} \; dx.$$

Extracting coefficients we find

$$m {n\choose m} \int_0^1 x^{m-1} [p^{k-m}] (1-px)^{n-m} \; dx \\ = m {n\choose m} \int_0^1 x^{m-1} (-1)^{k-m} {n-m\choose k-m} x^{k-m} \; dx \\ = m {n\choose m} (-1)^{k-m} {n-m\choose k-m} \int_0^1 x^{k-1} \; dx \\ = \frac{m}{k} {n\choose m} (-1)^{k-m} {n-m\choose k-m}.$$

The task is therefore to simplify the sum form of $[p^k].$ We may write

$$\sum_{q=0}^n {n\choose q} (-1)^{k-q} {n-q\choose k-q} - \sum_{q=0}^{m-1} {n\choose q} (-1)^{k-q} {n-q\choose k-q}.$$

We get for the first piece

$$(-1)^k \sum_{q=0}^n {n\choose q} (-1)^q [z^{k-q}] (1+z)^{n-q} \\ = (-1)^k [z^k] (1+z)^n \sum_{q=0}^n {n\choose q} (-1)^q z^q (1+z)^{-q} \\ = (-1)^k [z^k] (1+z)^n \left(1-\frac{z}{1+z}\right)^n = (-1)^k [z^k] 1 = 0.$$

The second piece is

$$(-1)^k \sum_{q=0}^n {n\choose q} (-1)^q [z^{k-q}] (1+z)^{n-q} [w^{m-1-q}] \frac{1}{1-w} \\ = (-1)^k [z^k] (1+z)^n \sum_{q=0}^n {n\choose q} (-1)^q z^q (1+z)^{-q} [w^{m-1}] \frac{w^q}{1-w} \\ = (-1)^k [z^k] (1+z)^n [w^{m-1}] \frac{1}{1-w} \sum_{q=0}^n {n\choose q} (-1)^q z^q (1+z)^{-q} w^q \\ = (-1)^k [z^k] (1+z)^n [w^{m-1}] \frac{1}{1-w} \left(1-\frac{wz}{1+z}\right)^n \\ = (-1)^k [z^k] [w^{m-1}] \frac{1}{1-w} \left(1+z(1-w)\right)^n \\ = (-1)^k [w^{m-1}] \frac{1}{1-w} {n\choose k} (1-w)^k \\ = (-1)^k [w^{m-1}] {n\choose k} (1-w)^{k-1} = (-1)^{m-1+k} {n\choose k} {k-1\choose m-1}.$$

The second piece was being subtracted from the first, which was zero, so that we get

$$(-1)^{k-m} {n\choose k} {k-1\choose m-1} = \frac{m}{k} (-1)^{k-m} {n\choose k} {k\choose m} \\ = \frac{m}{k} (-1)^{k-m} \frac{n!}{(n-k)! \times m! \times (k-m)!} = \frac{m}{k} (-1)^{k-m} {n\choose m} {n-m\choose k-m}.$$

This is the claim and hence concludes the argument.