$X$ supermartingale, $\mathsf E_0[X_2] = X_0$. Prove $X$ martingale

Question

$$X_n \ge \mathsf E_n[X_{n+1}] \therefore X_1 \ge \mathsf E_1[X_2] \therefore X_0 \ge \mathsf E_0[X_1] \ge \mathsf E_0[\mathsf E_1[X_2]] \stackrel{(LIE)}= \mathsf E_0[X_2]$$

Given: $\mathsf E_0[X_2] = X_0$

$$\therefore X_0 \ge \mathsf E_0[X_1] \ge X_0 \therefore X_0 = \mathsf E_0[X_1]$$

Sample space $\Omega = \{HH,HT,TH,TT\}$ so all I need to show now is $X_1 = \mathsf E_1[X_2]$ but here I'm stuck. Thanks in advance.

Answer 1

I doubt the question itself is incorrect.

Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $\mathbb{F}=\{\mathcal{F}_{n}\mid n=0,1,2,\ldots\}$ be a filtration. For sufficiently compliacted probability space $(\Omega,\mathcal{F},P)$ and filtration $\mathbb{F}$, we can construct a $X=\{X_{n},n=0,1,2,\ldots\}$ such that $X$ is a supermartingale but not a martingale. Define $\tilde{X}_{0}=E\left[X_{2}\mid\mathcal{F}_{0}\right]$ and $\tilde{X}_{1}=E\left[X_{2}\mid\mathcal{F}_{1}\right]$, $\tilde{X}_{n}=X_{n}$ for $n\geq2$. It is easy to check that $\tilde{X}=\{\tilde{X}_{n},n=0,1,2,\dots\}$ is a supermartingale with $E\left[\tilde{X}_{2}\mid\mathcal{F}_{0}\right]=\tilde{X}_{0}$. However, there is no reason that $\tilde{X}$ is a martingale.

It is because the condition $E\left[\tilde{X}_{2}\mid\mathcal{F}_{0}\right]=\tilde{X}_{0}$ only imposes restriction on terms $\tilde{X}_{0}$, $\tilde{X}_{1}$, and $\tilde{X}_{2}$ but not $\tilde{X}_{n}$ for $n>2$.

Answer 2

$E(X_2|X_1) \leq X_1$ and $E(X_1|X_0) \leq X_0$. Hence $EX_2 \leq EX_1\leq EX_0$. But it is given that $EX_2=EX_0$. Hence we get $EX_2 = EX_1= EX_0$. But then we can go back to the first set of inequalities and conclude that $X_1-E(X_2|X_1)$ and $X_)-E(X_1|X_0)$ are non-negative random variables with expectation $0$. This implies that $E(X_2|X_1)=X_1$ and $E(X_1|X_0)=X_0$. This means $(X_0,X_1,X_2)$ is martingale.

We cannot say anything about $(X_0,X_1,X_2,...)$.