Is it true that $\sum_{i=1}^{\infty}\left(M_{i-2} M_{i}-M_{i-1} M_{i-2}\right)<\infty, \text { a.s? }$

martingalesuniform-convergence

Prove or disprove. Suppose that $$\left(M_{n}\right)_{n}$$ is a martingale with $$M_{n} \geqslant-10 \quad \forall n$$, a.s.

Is it true that
$$\sum_{i=1}^{\infty}\left(M_{i-2} M_{i}-M_{i-1} M_{i-2}\right)<\infty, \text { a.s? }$$

This is what I have done so far

I thought of applying the lemma of discrete time stochastic integral which states:

Let $$\left(X_{n}\right)_{n \in \mathbb{N}_{0}}$$ be a supermartingale w.r.t. a filtration $$\left(\mathcal{F}_{n}\right)_{n \in \mathbb{N}_{0}}$$. Let $$\left(C_{n}\right)_{n \in \mathbb{N}}$$ be a predictable process w.r.t. $$\left(\mathcal{F}_{n}\right)_{n \in \mathbb{N}_{0}}$$ which is non-negative and bounded, i.e., $$C_{n} \leq K$$ uniformly in $$n \in \mathbb{N}$$. Then the discrete time stochastic integral
$$Y_{n}=(C \circ X)_{n}:=\sum_{k=1}^{n} C_{k}\left(X_{k}-X_{k-1}\right)$$
is a supermartingale.

Thus in my case:
$$\sum_{i=1}^{\infty}\left(M_{i-2} M_{i}-M_{i-1} M_{i-2}\right)= \sum_{i=1}^{\infty} M_{i-2}\left(M_{i}-M_{i-1}\right)\leq \sum_{i=1}^{\infty}\left|M_{i-2}\right|\left(M_{i}-M_{i-1}\right) = Y_n$$

Hence $$Y_n$$ would be a super martingale hence bounded by $$E[Y_0]$$. Since $$Y_n$$ is a positive martingale and bounded, then by the monotone convergence theorem, it converges a.s.

I am not sure if this is right and moreover, I am not sure how I should go about showing $$|M_{i-2}| \leq K$$ uniformily in n.

Attempt # 2

First I am going to prove than $$M_n$$ is bounded in $$L^1$$:

Since it is given that $$M_{n} \geqslant-10 \quad \forall n \text {, a.s. }$$, then $$M_{n}^{-} \leq -10^{-}$$so $$E M_{n}^{-}$$is bounded. By martinagle property, we know that $$M_{n}=E M_{0}$$ for all $$n$$ we also know that $$E M_{n}^{+}-E M_{n}^{-}=E M_{n}$$ is bounded. This makes $$E\left|M_{n}\right|=E M_{n}^{+}+E M_{n}^{-}$$ bounded. Hence $$M_n$$ is bounded in $$L^1$$.

Hence, I can use Austin's theorem: Let $$\left(M_{n}\right)_{n \in \mathbb{N}_{0}}$$ be a martingale which is bounded in $$L^{1}$$. Then
$$\sum_{k=1}^{\infty}\left(M_{k}-M_{k-1}\right)^{2}<\infty \text { a.s. }$$
Hence this proves that $$M_n$$ is bounded in $$L^2$$.

Now let $$Y_{n}=\sum_{i=1}^{n} M_{i-2} (M_{i} – M_{i-1})$$ Then $$Y_n$$ is a martingale and moreover $$Y_n$$ is bounded in $$L^2$$ since $$M_n$$ is bounded in $$L^2$$. Therefore, $$Y_n$$ converges to $$Y_{\infty}$$ a.s as by the lemma of $$L^2$$ results which state: Let $$\left(X_{n}\right)_{n \in \mathbb{N}_{0}}$$ be a martingale bounded in $$L^{2}$$. Then $$X_{n} \rightarrow X_{\infty}$$ a.s. and in $$L^{2}$$ for some random variable $$X_{\infty}$$. Moreover, $$\mathbb{E}\left[X_{\infty}^{2}\right]<\infty$$.

For any process $$X=(X_n)_{n \ge 0}$$, write $$G_n(X):=\sum_{i=2}^{n}\left(X_{i-2}\cdot (X_{i}-X_{i-1})\right)\,. \tag{1}$$ If $$\, \widetilde{M}_n:=M_n+10 \, \,$$ for all $$n$$, then $$G_n(\widetilde{M})-G_n(M)=10 \sum_{i=2}^{n} \left(\widetilde{M}_{i}-\widetilde{M}_{i-1} \right)=10(\widetilde{M}_n-\widetilde{M}_1)$$ which converges almost surely, since non-negative martngales converge by the Martingale convergence theorem. Therefore, to prove that $$G_n(M)$$ tends to a finite limit a.s., it suffices to prove that $$\lim_n G_n(\widetilde{M})<\infty\, \,$$ a.s.

Thus it will be sufficient to prove the following

Claim Given a Martingale $$(M_n)$$ with respect to $$({\mathcal F}_n)$$ such that $$M_n \ge 0$$ for all $$n$$, we have that $$\lim_n G_n( M )<\infty \, \,$$ a.s.

The proof below is not short, but the idea is simple: We truncate $$M$$ at a height $$L$$ that it is unlikely to reach, correct the truncation to be a bounded Martingale $$Y^T$$, prove that $$G_n(Y^T)$$ converges almost surely, and then handle the errors due to the correction.

Proof of claim: We first recall the construction from Almost sure convergence of martingale increment , which is included here for convenience:

Given $$L>E(M_0)$$, let $$\tau=\inf\{n \ge 0 :M_n \ge L \}$$ where by convention, $$\, \inf \, \emptyset=\infty$$. By optional stopping, $$E(M_0)=E(M_{\tau \wedge n}) \ge P(\tau \le n)L \,,$$ so letting $$n \to \infty$$ gives $$P(\tau <\infty) \le E(M_0)/L \,. \tag{2}$$ Write $$M^{\tau}_n:=M_{\tau \wedge n}$$ and $$\delta_0:=E[M_0-(M_0 \wedge L)]$$, and for $$n \ge 1$$, $$\delta_n:=E[M^{\tau}_n-(M^{\tau}_n\wedge L)|{\mathcal F}_{n-1}] \ge 0\,.$$ Note that $$\delta_n \le E[M^{\tau}_n |{\mathcal F}_{n-1}] \le L .$$ Define a martingale $$\{Y_n\}$$ by $$Y_0=M_0\wedge L$$ and $$Y_n=Y_{n-1}+(M^{\tau}_n \wedge L)-M^{\tau}_{n-1}+\delta_n$$ Induction on $$n$$ shows that
$$Y^{\tau}_n:=Y_{\tau \wedge n}=(M^{\tau}_n \wedge L)+\sum_{k=0}^{\tau\wedge n} \delta_k \,. \tag{3}$$ Clearly $$Y^{\tau}_n \ge 0$$ and $$|Y^{\tau}_n-Y^{\tau}_{n-1}| \le 2L$$ for all $$n \ge 1$$. Let $$T=\tau \wedge \inf\{ n\ge 0 : Y^{\tau}_n \ge L\}\,,$$ so that $$0 \le Y^T_n=Y_{T \wedge n} \le 3L \tag{4}$$ for all $$n$$. From $$(3)$$ we infer that $$\Delta_n:=\sum_{k=0}^n \delta_k \quad \text{satisfy} \quad \Delta_{T\wedge n} \le 3L \quad \text{for all } \; n\,. \tag{5}$$

Since $$E(M_0) \ge E(Y_0) =E(Y_{T\wedge n}) \ge L \cdot P(T \le n <\tau)$$ we deduce that $$P(T \le n) \le P(\tau \le n)+ P(T \le n <\tau) \le 2E(M_0)/L\,,$$ so $$P(T<\infty) \le 2E(M_0)/L \,. \tag{*}$$

The two processes $$(Y^T_n)$$ and $$\bigl(G_n(Y^T)\bigr)$$ are both martingales with respect to $$({\mathcal F}_n)$$. Moreover, by orthogonality of Martingale increments and $$(4)$$, $$E\Bigl[ G_n^2(Y^T)\Bigr] =\sum_{k=2}^n E \Bigl[ \bigl(Y^T_{ k-2 }\bigr)^2 \bigl(Y^T_{k}-Y_{k-1}^T \bigr)^2 \Bigr] \le 9L^2\cdot \sum_{k=2}^n E\Bigl[ \bigl(Y^T_{k}-Y_{k-1}^T \bigr)^2 \Bigr] \le 9L^2 E(Y^2_{T \wedge n}) \,.$$

By applying $$(4)$$ again we deduce that $$E\Bigl[ G_n^2(Y^T)\Bigr] \le 81L^4$$, so $$\lim_n G_n(Y^T) < \infty \quad \text{a.s.} \,. \tag{6}$$ On the event $$\{T=\infty\}$$, we have $$M_n=Y_n-\Delta_n$$ for all $$n$$, so $$M_{k-2}(M_k-M_{k-1})=(Y_{k-2}-\Delta_{k-2})\cdot(Y_k-Y_{k-1}-\delta_k) \,,$$ whence $$G_N(M)=G_n(Y)-\sum_{k=2}^n (Y_{k-2}-\Delta_{k-2})\delta_k -\sum_{k=2}^n \Delta_{k-2} \cdot(Y_k-Y_{k-1})\,. \tag{7}$$

On the event $$\{T=\infty\}$$, we have for $$m that $$|\sum_{k=m}^n (Y_{k-2}-\Delta_{k-2})\delta_k | \le 3L \sum_{k=m}^n \delta_k$$ which tends to $$0$$ as $$m,n \to \infty$$ by $$(5)$$. Since Cauchy sequences converge, $$\lim_n \sum_{k=m}^n (Y_{k-2}-\Delta_{k-2})\delta_k <\infty \,. \tag{8}$$

Observe that $$\sum_{k=2}^n \Delta_{(k\wedge T)-2} \cdot(Y_k^T-Y_{k-1}^T)$$ is an $$L^2$$ bounded martingale, so it converges a.s. Therefore, $$\sum_{k=2}^n \Delta_{k-2} \cdot(Y_k-Y_{k-1})$$ converges a.s. on the event $$\{T=\infty\}$$,

Combining the last observation with $$(6), (7), (8)$$, we conclude that

$$P(\lim_n G_n(M) <\infty) \le P(T<\infty) \le 2E(M_0)/L$$ by $$(*)$$. Since $$L$$ can be chosen arbitrarily large, this proves the claim.