Consider successive generations of a population of genes of fixed population size M. Each gene can be just one of two alleles, type a or A. Let $Y_n \in S = \{0,1,\dots,M\}$ denote the number of type A alleles in generation n under the standard Wright-Fisher model. You may assume that $\{Y_n : n = 0,1,2,\dots\}$ is a Markov chain with finite sample space S and stationary transition probabilities
$$p_{ij} = P(Y_{i + 1} = j|Y_n = i) = {M \choose j} \left( \frac{i}{M} \right) ^j \left( 1 - \frac{i}{M} \right) ^ {M - j}$$
I have shown $E(|Y_n|) < \infty$, and need help showing $E(Y_{n+1} | Y_n, \dots, Y_0) = Y_n$.
I have attempted:
$$E(Y_{n+1} | Y_n, \dots, Y_0) = \sum_{k=1}^M k{M \choose k} \left( \frac{Y_n}{M} \right) ^k \left( 1 - \frac{Y_n}{M} \right) ^ {M - k}$$
and messing around with it but haven't come out with the result.
I have an exam (tomorrow actually) and am just doing some last minute revision. Any help will be appreciated. Thanks.
By definition of the Wright-Fisher (WF) model (your $p_{ij}$ equation), we have $(Y_{n+1}|Y_n,...,Y_0)=(Y_{n+1}|Y_n) \sim Bin\Big(M, \frac{Y_n}{M}\Big)$.
Since $(Y_{n+1}|Y_n,...,Y_0)$ is a binomial, it's expectation is the product of its parameters, $E[Y_{n+1}|Y_n,...,Y_0] = M \frac{Y_n}{M} = Y_n$.
Indeed, if you plug the expression for the expectation into Mathematica $$ E[Y_{n+1}|Y_n] = \sum_{k=0}^{M}{k {M \choose k} \Big(\frac{Y_n}{M}\Big)^k \Big(1-\frac{Y_n}{M}\Big)^{M-k}}, $$ you get $$ Y_n \Big(\frac{M}{M - Y_n}\Big)^M \Big(1 - \frac{Y_n}{M}\Big)^M = Y_n. $$