Prove $Z_i = \frac{2n-i}{2n-i-1}Y_i^2 - \frac1{2n-i-1} (1 \le i \le 2n-2) $ is a martingale

Question

$Y_i = \frac{2X_i-i}{2n-i}, (1\le i \le 2n-1)$ $Z_i = \frac{2n-i}{2n-i-1}Y_i^2 - \frac1{2n-i-1} (1 \le i \le 2n-2) $

I have to prove that $Y_i$ and $Z_i$ are martingales and calculate the mean and variance of $X_i$

MY ATTEMPT :

Here it was proven that $Y_i$ is a martingale Prove $Y_i = \frac{2X_i-i}{2n-i}, (1\le i \le 2n-1)$ is a martingale

As in that question:

$X_i$ is the money I have after the i-th extraction which is equivalent to the number of white balls that I have up to the i-th extraction. I notice that $0 \le X_i \le i$

Looks like $X_i$ follows a hypergeometric distribution $X_i\sim $Hypergeometric$(2n,n,i)$ , $0\le i\le n$

Assuming the natural filtration $\mathcal F_i=\sigma(X_1,...,X_i)$ I have already proven adaptability and finite mean of $Z_i$. For the martingale property

$E[Z_{i+1}|\mathcal F_i]= \frac{2n-i-1}{2n-i-2}E[Y_{i+1}^2|\mathcal F_i] - \frac1{2n-i-2} $

To compute $E[Y_{i+1}^2|\mathcal F_i]$ I am unsure if the same simplified formula holds as in the non-conditioning case:

$E[Y_{i+1}^2|\mathcal F_i]=E[Y_{i+1}|\mathcal F_i]^2 +Var[Y_{i+1}|\mathcal F_i]\tag{*} ? $

If this is true, how do I compute that variance? I am not arriving to the answer. The problem did warn that it was a complicated calculation, though.

Answer 1

It's the exact same question when it comes to showing that $Z_{i}$ is a martingale. If you know $X_{i}=k$, then $E(X_{i+1}^{2}|X_{i}=k)= (k+1)^{2}P(X_{i+1}=k+1|X_{i}=k)+k^{2}P(X_{i+1}=k|X_{i}=k)$. And I already showed you what $P(X_{i+1}=k+1|X_{i}=k)$ etc are.

To quote the other question directly

$P(X_{i+1}=k+1|X_{i}=k)=\frac{n-k}{2n-i}$ and $P(X_{i+1}=k|X_{i}=k)=\frac{n-i+k}{2n-i}$

So I'll not even bother to do the verification that $Z_{i}$ is a martingale and you should verify it yourself.

Now, as to calculating the mean and variance, observe that since $Y_{i}$ is a martingale $E(Y_{i+1}|\mathcal{F}_{i})=Y_{i}$ . Hence $E(Y_{i+1})=E(E(Y_{i+1}|\mathcal{F}_{i}))=E(Y_{i})$

Thus, the expectations remains constant.

Thus $\displaystyle\frac{2E(X_{i})-i}{2n-i}=E(Y_{i})=E(Y_{1})=\frac{2E(X_{1})-1}{2n-1}$ .

Now calculate $E(X_{1})=1\cdot\frac{n}{2n}+0\cdot\frac{n}{2n}=\frac{1}{2}$ which is just your expected profit after first draw and use the above relation to find $E(X_{i})$.

Hence, you now know $E(Y_{i})$ . Now as $Z_{i}$ is a martingale, $E(Z_{i})=E(Z_{1})$. From this, you can find a relation between $E(Y_{1}^{2})$ and $E(Y_{i}^{2})$ and use that to find $E(Y_{i}^{2})$ in terms of $E(Y_{1}^{2})$ which is easily calculable. Now you expand $Y_{i}^{2}$ out and use everything that you have calculated to find $E(X_{i}^{2})$ in terms of $E(X_{1}^{2})$ and $E(X_{1})$ and then find the variance by using $Var(X_{i})=E(X_{i}^{2})-(E(X_{i}))^{2}$.

Expect to do some long and tedious but elementary calculations though. It is often the case that in such questions, it essentially becomes a middle school level tedious algebra question where you have some equations given and you have to find a certain value by eliminating variables.