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

Question

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

I have to prove that it is martingale

MY ATTEMPT :

$X_i$ 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. I am having trouble with the martingale property

$E[Y_{i+1}|\mathcal F_i]= \frac{1}{2n-i-1}(2E[X_{i+1}|\mathcal F_i]-i-1)= ?$

I am stuck here, but I cannot apply the formula for the mean with the conditioning on the way and I don't think $X_{i+1}$ is independent from $\mathcal F_i$ since the money I have up to the $(i+1)$-th extraction depends on how much money I have up to the $i$-th extraction. Even if I am wrong and there is independence, using the formula for the mean of a hypergeometric distribution yields $0$:

$E[Y_{i+1}|\mathcal F_i]= \frac{1}{2n-i-1}(2E[X_{i+1}|\mathcal F_i]-i-1)= \frac{1}{2n-i-1}(2E[X_{i+1}]-i-1)= \frac{1}{2n-i-1}(2E[X_{i+1}]-i-1)=\frac{1}{2n-i-1}(2(i+1)n/2n -i-1)=0$

What I am missing and how do I proceed?

Answer 1

What you only need to do is calculate $E(X_{i+1}|\sigma(X_{1},...,X_{i}))$. Now notice that if $X_{i}$ is known, then you can calculate $E(X_{i+1}|X_{1},...,X_{i})=E(X_{i+1}|X_{i})$. This is because the information $X_{i}$ specifies the number of white balls and number of black balls that are left before the $i+1$-th draw. So, conditioning on that, you can easily find the expected value of $X_{i+1}$.

So,

Suppose $X_{i}=k$ (here you could additionally assume that $X_{1}=k_{1},...X_{i-1}=k_{i-1}$ as you know $\sigma(X_{1},..,X_{i})$ but those informations will not matter to us in the calculation) . Then, there are $n-k$ white balls left and $n-i+k$ black balls left.

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

Hence $E(X_{i+1}|X_{i}=k)=\dfrac{(k+1)(n-k)+k(n-i+k)}{2n-i}$

Now just verify that

$\displaystyle E(Y_{i+1}|\sigma(X_{1},...,X_{i}))=\frac{2E(X_{i+1}|X_{i}=k)-(i+1)}{2n-i-1}=\frac{2k-i}{2n-i}=\frac{2X_{i}-i}{2n-i}=Y_{i}$

To elaborate,

\begin{align}E(Y_{i+1}|X_{i}=k)&=\dfrac{\frac{2(k+1)(n-k)+k(n-i+k)}{2n-i}-(i+1)}{2n-i-1}\\\\ &=\frac{2kn-2k^{2}+2n-2k+2kn-2ki-2k^{2}-2ni-2n-i^{2}-i}{(2n-i)(2n-i-1)}\\\\ &=\frac{2k(2n-i-1)-i(2n-i-1)}{(2n-i)(2n-i-1)}\\\\ &=\frac{(2n-i-1)(2k-i)}{(2n-i)(2n-i-1)}\\\\ &=\frac{2k-i}{2n-i}\\\\ &=\frac{2X_{i}-i}{2n-i}\end{align}