# How to evaluate this integral? (relating to binomial)

I saw some result that some article used, (without proving) that stated:$$\int_0^1 p^k (1-p)^{n-k} \mathrm{d}p = \frac{k!(n-k)!}{(n+1)!}$$

But I was wondering, how would you integrate it? How did this integral come about? Is it something to do with the binomial distribution?

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Beta function –  Steven Taschuk Mar 19 '12 at 23:36
see too Weisstein's derivation –  Raymond Manzoni Mar 19 '12 at 23:43
Ahh okay, thanks for the heads up! :) –  Heijden Mar 20 '12 at 16:10

Not a natural derivation, but there is slightly different approach toward it.

Let's consider the quantity

$$I(n,k) = \int_{0}^{1} \binom{n}{k} p^k (1-p)^{n-k} \; dp.$$

Then by integration by parts, as in two former answers, we have

$$I(n, k+1) = I(n, k).$$

Let $I$ denote this common value. Thus

$$1 = \int_{0}^{1} 1 \; dp = \int_{0}^{1} \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \; dp = \sum_{k=0}^{n} I = (n+1)I$$

and the result follows.

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Thanks for your help sos440, nice to see a diff. way also! –  Heijden Mar 20 '12 at 16:11

This can be proven using repeated integration by parts: $$\begin{eqnarray} \int_0^1 p^k(1-p)^{n-k} &=& \frac{1^{k+1}(1-1)^{n-k}}{k+1}-\frac{0^{k+1}(1-0)^{n-k}}{k+1}+\frac{n-k}{k+1}\int_0^1 p^{k+1}(1-p)^{n-k-1}\\ &=& \frac{n-k}{k+1}\int_0^1 p^{k+1}(1-p)^{n-k-1}\\ &=& \frac{(n-k)(n-k-1)}{(k+1)(k+2)}\int_0^1 p^{k+2}(1-p)^{n-k-2}\\ &\vdots&\\ &=& \frac{k!(n-k)!}{n!}\int_0^1 p^{n}=\frac{k!(n-k)!}{n!}\frac{1}{n+1}=\frac{k!(n-k)!}{(n+1)!}\\ \end{eqnarray}$$

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Thanks for your help Alex :) –  Heijden Mar 20 '12 at 16:11

There are probably several ways. An easy one is by induction on $k$.

If $k=0$, then $$\int_0^1(1-p)^n\,dp=\left.-\frac{(1-p)^{n+1}}{n+1}\right|_0^1=\frac1{n+1}=\frac{0!(n-0)!}{(n+1)!}.$$ Now assume that the formula holds for some $k$. Then, integrating by parts, $$\begin{eqnarray} \int_0^1p^{k+1}(1-p)^{n-(k+1)}\,dp&=&\int_0^1p^{k+1}(1-p)^{(n-1)-k}\,dp\\ &=&\left.-\frac{p^{k+1}(1-p)^{n-k}}{n-k}\right|_0^1+\int_0^1\frac{(k+1)p^k(1-p)^{n-k}}{n-k}\,dp\\ &=&\frac{(k+1)}{n-k} \frac{k!(n-k)!}{(n+1)!}\\ &=&\frac{(k+1)!(n-(k+1))!}{(n+1)!}. \end{eqnarray}$$ The induction principle then guarantees that the formula holds for all $k$.

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Thanks martin, didnt realise I could do it by induction! –  Heijden Mar 20 '12 at 16:12

You can also proof it "by story". Let "random number" mean a number picked from $[0,1]$ with uniform probability. Then the formula below can be interpreted as follows.

$$\int_0^1 p^k (1-p)^{n-k} \mathrm{d}p = \frac{k!(n-k)!}{(n+1)!}$$

The left-hand side is the probability of taking random $p$ and then drawing a sequence of $n$ numbers from which some $k$ numbers are smaller than $p$ and some $n-k$ are larger.

To understand the right-hand side, consider $n+1$ random numbers sorted, so that first $k$ are the smallest and last $n-k$ are the largest (with $(k+1)$-th being our $p$ from left-hand side interpretation); however, there are $(n+1)!$ permutations total, with $k!(n-k)!$ having the desired property (in a sorted sequence we disregard the order of first $k$ and last $n-k$), thus the right-hand side fraction denotes the same probability.

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Hmm interesting way, cheers! –  Heijden Mar 20 '12 at 16:11