Let $M_n$ be a square integrable martingale with $M_0 = 0$ and let $M_n^2 = M_0^2 + X_n + A_n$ be the Doob decomposition of $M_n^2$
i.e. $A_n = \sum_{i=1}^nE(M_i^2 – M_{i-1}^2|\mathcal{F}_{i-1}) = \sum_{i=1}^nE([M_i – M_{i-1}]^2|\mathcal{F}_{i-1})$
and $X_n = $ a martingale. It is easily shown that $E(M_n^2) = E(A_n)$
I want to show:
(1) $A_\infty = \text{lim}_{n \rightarrow \infty}A_n$ is bounded, then $\text{lim}_{n \rightarrow \infty}M_n$ exists a.s.
(2) $\{A_\infty < \infty\} \subseteq \{\text{lim}_{n \rightarrow \infty}M_n \quad \text{exists}\}$ and
(3) If $M_n$ has uniformly bounded increments, $\{A_\infty < \infty\} = \{\text{lim}_{n \rightarrow \infty}M_n \quad \text{exists}\}$
I can do (1) by saying that $$\text{sup}_{n \in \mathbb{N}}E[(M_n^\pm)^2] \leq \text{sup}_{n \in \mathbb{N}}E[M_n^2] = \text{sup}_{n \in \mathbb{N}}E(A_n) = E(A_\infty) < \infty$$
which implies that $\text{lim}_{n \rightarrow \infty} M_n^\pm$ exist as the square roots of $(M_n^\pm)^2$ by the martingale convergence theorem.
I do not know where to start for (2) and (3) here. I know that if I can show $\text{sup}_{n \in \mathbb{N}}E(M_n^+) < \infty$ then the limit exists by the martingale convergence theorem, but I cannot see how $A_n$ being a convergent sequence (as it is bounded and increasing due to $M_n^2$ being a submartingale) implies this. Is there a way to do this with stopping rules?
Any insight/hints/solutions would be greatly appreciated. Thanks!
Best Answer
Define
$$T_k := \inf\{n \in \mathbb{N}; |A_n| \geq k\}.$$
By the optional stopping theorem, $(M_{n \wedge T_k})_{n \in \mathbb{N}}$ is a martingale and its compensator is bounded by $k$. It follows from (1) that $\lim_{n \to \infty} M_{n \wedge T_k}$ exists almost surely. In particular, $\lim_{n \to \infty} M_n$ exists on $\{T_k = \infty\}$. Since
$$\{A_{\infty}<\infty\} = \bigcup_{k \in \mathbb{N}} \{T_k=\infty\}$$
this proves (2).
For (3) define
$$\tau_k := \inf\{n \in \mathbb{N}; |M_n| \geq k\}.$$ Then
$$\mathbb{E}(A_{n \wedge \tau_k}) =\mathbb{E}(M_{n \wedge \tau_k}^2) = \mathbb{E}\bigg[ \big((M_{n \wedge \tau_k} - M_{(n \wedge \tau_k)-1}) + \underbrace{M_{(n \wedge \tau_k)-1}}_{\leq k} \big)^2 \bigg] \leq (k+c)^2$$
where
$$c := \sup_{n \in \mathbb{N}} |M_n-M_{n-1}|.$$
Applying Fatou's lemma gives
$$\mathbb{E}(A_{\infty \wedge \tau_k}) \leq (k+c)^2.$$
In particular,
$$\mathbb{E}(A_{\infty} 1_{\{\tau_k=\infty\}})< \infty,$$
and so $A_{\infty}<\infty$ a.s. on $\{\tau_k = \infty\}$. As $$\left\{\lim_{n \to \infty} M_n \, \text{exists}\right\} \subseteq \bigcup_{k \in \mathbb{N}} \{\tau_k = \infty\},$$ this proves $$\left\{\lim_{n \to \infty} M_n \, \text{exists}\right\} \subseteq \{A_{\infty} < \infty\}.$$ The converse is clear from (2).