4
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ wouldn't that just be a binomial distribution? and what is the mean of a binomial distribution? (Note I am asking - i could be wrong) $\endgroup$
    – PeterR
    May 26 '14 at 17:58
  • $\begingroup$ @PeterR I got so absorbed in the question (this was the final part) that I completely overlooked something so simple since there were a whole bunch of symbols involved. Yeah, it's binomial. Making the expectation $M \frac{Y_n}{M} = Y_n$. Perfect, thanks! $\endgroup$
    – user117682
    May 26 '14 at 18:01
  • $\begingroup$ Been there, done that. Glad it helped. $\endgroup$
    – PeterR
    May 26 '14 at 19:26
5
$\begingroup$

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 $$ Y_n \Big(\frac{M}{M - Y_n}\Big)^M \Big(1 - \frac{Y_n}{M}\Big)^M = Y_n. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy