Let $h_0 = 0, h_n = 1, h_i = \sum_{j=1}^nh_jp_{ij}$ when $i = 1$ to $n-1$ and $p_{ij} = \binom{n}{j} (\frac{i}{n})^j(1-\frac{i}{n})^{n-j}$, how could I prove $h_i = \frac{i}{n}$ for all $i = 0$ to $n$.
I tried to find the relation between $h_i$ and $h_{i+1}$ and then use induction method, but the relation is too complicated. Could anyone help me with this problem? Thank you very much!
I can prove the identity, so is, asuming $h_i=i/n$ the equality is satisfied.
$$S_n=\sum_{j=1}^n\frac{j}{n}p_{ij} =\sum_{j=1}^n\frac{j}{n} \binom{n}{j} (\frac{i}{n})^j(1-\frac{i}{n})^{n-j}=$$
$$=\frac{1}{n}\sum_{j=1}^nj \binom{n}{j} (\frac{i}{n})^j(1-\frac{i}{n})^{n-j}$$
We now consider this identity:
$$j \dbinom n j = n \dbinom {n - 1} {j - 1}$$
$$S_n=\frac{1}{n}\sum_{j=1}^nn \dbinom {n - 1} {j - 1}(\frac{i}{n})^j(1-\frac{i}{n})^{n-j}=\frac{n}{n}\frac{i}{n}\sum_{j=1}^n\dbinom {n - 1} {j - 1}(\frac{i}{n})^{j-1}(1-\frac{i}{n})^{n-j}=$$
$$=\frac{i}{n}\sum_{j=0}^{n-1}\dbinom {n-1} {j}(\frac{i}{n})^j(1-\frac{i}{n})^{n-j-1}=\frac{i}{n}\left(\frac{i}{n}+1-\frac{i}{n}\right)^{n-1}=\frac{i}{n}$$
This proves the identity for $n>0$. It's trivialy satisfied for $n=0$.