Wright-Fisher Model: Prove that $P_X(V_N<V_0)=x/N$?

Question

Let $X_n$ be the wright-Fisher model that is $X_n$ is a Markov chain with transition probability $P(i,j)$~$Binomial(N,i/N)$. Show that $X_N$ is a martingale and prove that exit distribution $h(x)=P_X(V_N<V_0)=x/N$?

I have proved that $X_n$ is martingale.

Since $X_n$~$B(N, X_{n-1}/N$), so $E(X_n)=NX_{n-1}/N=X_{n-1}$

But I am not sure how to prove $P_X(V_N<V_0)=x/N$?

I need some help or suggestion to prove this.

Any help will be appreciated. Thanks.

Answer 1

I guess $V_j$ is the hitting time of $j$.

To finish, use the optional stopping theorem.

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From optional stopping, we have $$E [ X_{V_N \wedge V_0} ]= E[X_0].$$ But lefthand side is equal to $$ N P(V_N<V_0)+ 0 P(V_0 < V_N)=N P(V_N<V_0).$$ Since you assumed $X_0=x$, the result follows.