Integral equation involving Binomial distribution

Question

I am trying to find the form of a function $u^{(n)}(p)$ which satisfies $\forall k \in [0,n] \int_0^1 dp\, u^{(n)}(p) \binom{n}{k} p^k(1-p)^{n-k} = \frac{1}{n+1}$.

This is a private case of a more general problem I am trying to solve, which is to find $u_m^{(n)}(p)$ with $(m\le n)$ such that $\forall k \in [0,m-1] \int_0^1 dp\, u_m^{(n)}(p) \binom{n}{k} p^k(1-p)^{n-k} = \frac{1}{m+1}$ and $\int_0^1 dp\,u_m^{(n)}(p)(1-I_{1-p}(n-m+1,m))=\int_0^1 dp\,u_m^{(n)}(p)I_{p}(m,n-m+1)=\frac{1}{m+1}$ with $I$ the regularized incomplete beta function. Using $I_p(n,1)=p^n$ it is easy to show that this is the same as the first problem when $m=n$.

EDIT: $u^{(n)}(p)$ should be a distribution function, so $u^{(n)}(p)\ge 0$ for $p \in [0,1]$

Answer 1

Hint: summing over $k$, and using the fact that finite summations may always be taken inside the integral, we can derive a much simpler necessary condition on $u(p)$:

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

Answer 2

Apparently, the first part was rather simple to solve. Let $u^{(n)}(p)=1=\text{const}$ then

$\int_0^1 dp\, \binom{n}{k} p^k(1-p)^{n-k} = \binom{n}{k} \int_0^1 dp\, p^k(1-p)^{n-k}=\binom{n}{k}B(k+1,n-k+1)=\frac{n!}{k!(n-k)!}\,\frac{(k+1-1)!(n-k+1-1)!}{(k+1+n-k+1-1)!}=\frac{n!}{k!(n-k)!}\,\frac{k!(n-k)!}{(n+1)!}=\frac{1}{n+1}$.