Prove that $\int \binom{n}{k}p^k(1−p)^{n−k}\,dp$ from $0$ to $1$ is $\frac{1}{n+1}$

Question

Prove that$$ \int_0^1 \binom{n}{k} p^{k}( 1-p ) ^{n-k}dp=\frac{1}{n+1}. $$

I met this question in "The Art and Craft of Problem Solving". The whole question is from Bay Area Math Meet 2000, and by using an alternative method to interpret that question, we can get the result.

The book then asked us to use 2 different methods to solve it,

1. repeat integration by part. This is easy, I solve it.
2. Using manipulation of the binomial series. I can not solve the question using this method. How can I use the binomial series to solve it?

Answer 1

Using this, the left-hand side is $$\binom{n}{k}\operatorname{B}(k+1,\,n-k+1)=\frac{\Gamma(n+1)}{\color{blue}{\Gamma(k+1)}\color{green}{\Gamma(n-k+1)}}\frac{\color{blue}{\Gamma(k+1)}\color{green}{\Gamma(n-k+1)}}{\Gamma(n+2)}=\frac{1}{n+1},$$where the blue factors cancel, as do the green ones.

I'm not sure how we can "use the binomial series", if by that you mean expanding $(1-p)^{n-k}$. (Edit: @MarkusScheuer showed how.) But there's another way to use it. Summing over $k$ from $0$ to $n$ inclusive, these integrals give $\int_0^1(p+1-p)^ndp=\int_0^1dp=1$, so we only need show why the integrals are $k$-independent. Indeed, they are values, averaged over a $U(0,\,1)$ distribution of $p$, for the probabilities of a $B(n,\,p)$-distributed variable taking each possible value of $k$. We can argue they are equiprobable when we are that ignorant of the distribution of $p$.

Answer 2

It isn't either of the two methods, but you can use the beta function here: $$\int_0^1 p^k(1-p)^{n-k}dp=B(k+1,n+1-k)=\frac{k!(n-k)!}{(n+1)!}$$ and so: $$\int_0^1 \begin{pmatrix}n\\k\end{pmatrix}p^k(1-p)^{n-k}dp=\frac{n!}{k!(n-k)!}\frac{k!(n-k)!}{(n+1)!}=\frac{1}{n+1}$$

Answer 3

In order to apply the binomial series we introduce a parameter $z$ so that $z^k$ can be used to select the term with $\binom{n}{k}$. We also use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series.

We obtain \begin{align*} \color{blue}{\int_{0}^1}&\color{blue}{\binom{n}{k}p^k(1-p)^{n-k}\,dp}\\ &=[z^k]\int_{0}^1\binom{n}{k}\left(pz\right)^k(1-p)^{n-k}\,dp\\ &=[z^k]\int_{0}^1\left(pz+(1-p)\right)^n\,dp\\ &=[z^k]\int_{0}^{1}\left(1+(z-1)p\right)^n\,dp\\ &=[z^k]\frac{1}{n+1}\cdot\left.\frac{(1+(z-1)p)^{n+1}}{z-1}\right|_{p=0}^1\\ &=[z^k]\frac{1}{n+1}\cdot\frac{z^{n+1}-1}{z-1}\\ &=[z^k]\frac{1}{n+1}\left(1+z+\cdots+z^k+\cdots+z^n\right)\\ &\,\,\color{blue}{=\frac{1}{n+1}} \end{align*} and the claim follows.

Answer 4

Another possible approach.

Let's indicate the integrand function as $$ f(p;n,k) = \left( \matrix{ n \cr k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k} $$ the derivative wrt $p$ is $$ \eqalign{ & {d \over {dp}}f(p;n,k) = \cr & = \left( \matrix{ n \cr k \cr} \right)\left( {kp^{\,k - 1} \left( {1 - p} \right)^{\,n - k} - \left( {n - k} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k - 1} } \right) = \cr & = k\left( \matrix{ n \cr k \cr} \right)p^{\,k - 1} \left( {1 - p} \right)^{\,n - k} - \left( {n - k} \right)\left( \matrix{ n \cr n - k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k - 1} = \cr & = n\left( \matrix{ n - 1 \cr k - 1 \cr} \right)p^{\,k - 1} \left( {1 - p} \right)^{\,n - k} - n\left( \matrix{ n - 1 \cr n - k - 1 \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k - 1} = \cr & = n\left( {f(p;n - 1,k - 1) - f(p;n - 1,k)} \right) \cr} $$

Then the integral becomes $$ \eqalign{ & S(n,k) = \int_{\,0}^{\,1} {\left( \matrix{ n \cr k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k} dp} = \cr & = \int_{\,0}^{\,1} {f(p;n,k)dp} = \cr & = \left. {\left( {p\,f(p;n,k)} \right)\,} \right|_{\,p = 0}^{\;1} - \int_{\,0}^{\,1} {p\,df(p;n,k)} = \cr & = \left[ {k = n} \right] - n\int_{\,0}^{\,1} {p\,\left( {f(p;n - 1,k - 1) - f(p;n - 1,k)} \right)dp} = \cr & = \left[ {k = n} \right] - n\int_{\,0}^{\,1} {p\,\left( {\left( \matrix{ n - 1 \cr k - 1 \cr} \right)p^{\,k - 1} \left( {1 - p} \right)^{\,n - k} - \left( \matrix{ n - 1 \cr k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - 1 - k} } \right)dp} = \cr & = \left[ {k = n} \right] - n\left( \matrix{ n - 1 \cr k - 1 \cr} \right)\int_{\,0}^{\,1} {p^{\,k} \left( {1 - p} \right)^{\,n - k} dp} + n\left( \matrix{ n - 1 \cr k \cr} \right)\int_{\,0}^{\,1} {\,p^{\,k + 1} \left( {1 - p} \right)^{\,n - 1 - k} dp} = \cr & = \left[ {k = n} \right] - {{n\left( \matrix{ n - 1 \cr k - 1 \cr} \right)} \over {\left( \matrix{ n \cr k \cr} \right)}}\int_{\,0}^{\,1} {\left( \matrix{ n \cr k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k} dp} + {{n\left( \matrix{ n - 1 \cr k \cr} \right)} \over {\left( \matrix{ n \cr k + 1 \cr} \right)}}\int_{\,0}^{\,1} {\,\left( \matrix{ n \cr k + 1 \cr} \right)p^{\,k + 1} \left( {1 - p} \right)^{\,n - 1 - k} dp} = \cr & = \left[ {k = n} \right] - k\int_{\,0}^{\,1} {\left( \matrix{ n \cr k \cr} \right)p^{\,k} \left( {1 - p} \right)^{\,n - k} dp} + \left( {k + 1} \right)\int_{\,0}^{\,1} {\,\left( \matrix{ n \cr k + 1 \cr} \right)p^{\,k + 1} \left( {1 - p} \right)^{\,n - 1 - k} dp} = \cr & = \left[ {k = n} \right] - kS(n,k) + \left( {k + 1} \right)S(n,k + 1) \cr} $$ that is: $$ S(n,k) = S(n,k + 1) + {{\left[ {k = n} \right]} \over {k + 1}} $$ or better $$ \eqalign{ & S(n,n - k) = S(n,n - k - 1) + {{\left[ {0 = n - k} \right]} \over {n + 1}} \cr & S(n,j) = S(n,j - 1) + {{\left[ {0 = j} \right]} \over {n + 1}} \cr} $$ where $[P]$ denotes the Iverson bracket

And that demonstrates the thesis.

Answer 5

By integration by parts:

$$\int_0^1 \binom{n}{k} p^{k}( 1-p ) ^{n-k}dp=\int_0^1 \binom{n}{k+1} p^{k+1}( 1-p ) ^{n-(k+1)}dp :=X$$ Then, $$1 =\int_0^1 1 dp = \int_0^1 (p+ (1-p))^n dp= \sum_{k=0}^n \int_0^1 \binom{n}{k} p^{k}( 1-p ) ^{n-k} dp=\sum_{k=0}^n X= (n+1)X$$ Hence, $$X= \int_0^1 \binom{n}{k} p^{k}( 1-p ) ^{n-k}dp = \frac{1}{n+1}$$