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? } $

Question

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, $\mathbb{E}\left[X_{\infty}^{2}\right]<\infty$.

Answer 1

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 $m<n$ 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.