Show that $M_{n}=\left(\cfrac{N}{N-1}\right)^n X_n(N-X_n)$ is a martingale

Suppose we have a random variable $(X_n)$ such that :

• $X_0=k\in\{1,..,N\}$

• The conditional distribution of $X_{n+1}$ conditioned to $\mathcal F_n=\sigma(X_1,..,X_n)$ follows the binomial distribution with parameters $(N,X_n/N)$

Let $M_{n}=\left(\cfrac{N}{N-1}\right)^n X_n(N-X_n)$. Prove that $M_n$ is a martingale with respect to $\mathcal F_n$.

Attempt: I already proved that $X_n$ is a martingale.

$E(M_{n+1} \, | \, F_n)=E\left( \left(\cfrac{N}{N-1}\right)^{n+1}X_{n+1}(N-X_{n+1}) \, | \, F_n \right)$

I noticed that $X_{n}(N-X_{n})=N*Var(X_{n+1})$

And we know that $E[X_{n+1}\, | \, F_n]=X_n$

But I got stuck there, did I start this wrong? Am I missing a property that I can use?

Solution 1:

Since $X_{n+1} \, | \, X_n = x \sim B(N, x/N)$, we have: $$\begin{align*} E(X_{n+1} (N - X_{n+1}) \, | \, X_n = x) &= N E(X_{n+1} \, | \, X_n=x) - E(X_{n+1}^2 \, | X_n = x) \\ &= Nx - \left( {x}\left( 1- \frac{x}{N} \right) + x^2 \right) \\ &= (N-1)x\left( 1 - \frac{x}{N} \right) \end{align*}$$ Where in the second equality we used $E(X^2) = Var(X) + E(X)^2$ and in the last equality we expanded and collected terms. Since $M_n$ is a Markov process, it follows that $$E(X_{n+1} (N - X_{n+1}) \, | \, \mathcal{F}_n) = E(X_{n+1} (N - X_{n+1}) \, | \, X_n) = (N-1)X_n \left( 1 - \frac{X_n}{N} \right)$$

Thus,

$$\begin{align*} E(M_{n+1} \, | \, \mathcal{F}_n) = \left( \frac{N}{N-1} \right)^{n+1} (N-1) X_n \left( 1 - \frac{X_n}{N} \right) = \left( \frac{N}{N-1} \right)^{n} X_n \left( N - {X_n} \right) = M_n \end{align*}$$ Thus $M_n$ is an $\mathcal{F}_n$-martingale.