# 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?

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.

• Hmm interesting way, cheers! – Heijden Mar 20 '12 at 16:11
• $p^k(1-p)^{n-k}$ is the probability you described. But what about the integral? Integral makes us talk not about single $p$ but about whole bunch of them. – Yola Jan 22 '18 at 13:55
• @Yola $p^k(1-p)^{n-k}$ is the probability of drawing a sequence of $n$ numbers, from which some $k$ numbers are smaller than $p$ and some $n−k$ are larger, given a particular, fixed value of $p$. And then, we sum all these probabilities for any possible value of $p$ using the integral. Does that explain your question? – dtldarek Jan 22 '18 at 14:52
• That's exactly that i thought, but i just can't coneive this part with integration. Probably i should just get used to it, and this will become clearer for me later. Thanks! +1 – Yola Jan 22 '18 at 15:14
• @Yola Then I suggest this: imagine the same problem, but say that $p \in \{0, 1/2\}$ (each happening with probability $1/2$). Work out the formula on the left. Then consider $p \in \{0/4, 1/4, 2/4, 3/4\}$ (all with probabilities $1/4$). Then consider $p \in \{0/8, 1/8, \ldots, 7/8\}$, and so on. When you spot the common theme, do $p \in \{0/2^m, 1/2^m, \ldots, (2^m-1)/2^m\}$. Going with $m$ to infinity is exactly the step that makes that sum an integral (here such a simplified integration is possible, because we are integrating a polynomial, which is a very well-behaved function on [0,1]). – dtldarek Jan 22 '18 at 16:49

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

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.

• Thanks for your help sos440, nice to see a diff. way also! – 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 $k$.

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