Expectation of two successive martingales

Question

Consider a martingale $(M_n)_{n \geq 0}$ adapted to the filtration $\{\mathcal{F}_n\}_{n \geq 0}$. I want to prove that: \begin{align} \mathbb{E}[M_nM_{n+1}] = \mathbb{E}[M_n^2]. \end{align} Therefore, \begin{align} \mathbb{E}[M_nM_{n+1}] &= \mathbb{E}[M_nM_{n+1} \mid \mathcal{F}_0] \\ &= \mathbb{E}\big[ \mathbb{E}[M_{n+1} \mid \mathcal{F}_n] \cdot M_{n+1} \mid \mathcal{F}_0 \big] \\ &\text{$= \mathbb{E}[ M_{n+1} \cdot M_{n+1} \mid \mathcal{F}_0 ]\qquad$ since $\mathcal{F}_0 \subseteq \mathcal{F}_n^*$} \\ &= \mathbb{E}[ M_{n+1} \cdot M_{n+1}] \\ &= \mathbb{E}[ M_{n+1}^2]. \end{align}

*If $\mathcal{A} \subseteq \mathcal{B}$, then \begin{align} \mathbb{E}\big[ \mathbb{E}[X \mid \mathcal{A} ] \mid \mathcal{B} \big] = \mathbb{E}\big[ \mathbb{E}[X \mid \mathcal{B} ] \mid \mathcal{A} \big] = \mathbb{E}[X \mid \mathcal{A} ]. \end{align} What happened? Why is this solution not correct?

Answer 1

The identity

$\mathbb{E}(\mathbb{E}(M_{n+1} \mid \mathcal{F}_n) \cdot M_{n+1} \mid \mathcal{F}_0) = \mathbb{E}(M_{n+1} M_{n+1} \mid \mathcal{F}_0)$

does not hold true. You cannot apply $(\ast)$ to justify this identity because the term at the left-hand side is not of the form

$$\mathbb{E}(\mathbb{E}(X \mid \mathcal{A}) \mid \mathcal{B})$$

but of the (more general) form

$$\mathbb{E}(\mathbb{E}(X \mid \mathcal{A}) \cdot Y \mid \mathcal{B}).$$

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Hint: Write

$$\mathbb{E}(M_n \cdot M_{n+1}) = \mathbb{E}(\mathbb{E}(M_{n+1} \cdot M_n \mid \mathcal{F}_n))$$

and use the pull out property.

Answer 2

I believe it is your third equality. The property (*) that you are trying to use may no longer be valid as you are multiplying $\mathbb{E}[M_{n+1} | F_n]$ by $M_{n+1}$. Have you checked the proof to see if it could be modified to account for the addition of such a term?