Almost sure convergence of martingale increment

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 $$ \sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{4}<\infty \quad \text { a.s. ? } $$

This was my attempt which I'm unsure of:

Proved by contradiction: Suppose $\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{4} \geq \infty$. Since $\left(M_{n}\right)_{n}$ is a martingale, we know that $ E\left[M_{n}-M_{0} \mid F_{0}\right]=0$. Thus:

$$ \sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{4} \geq \sum_{i=1}^{\infty}\left|M_{i}-M_{i-1}\right| \geq \infty $$

This would imply that: $E\left[\left|M_{n}\right|\right]>\infty$ which is a contraction as $M_n$ is a martingale. Therefore: $\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{4}<\infty \quad \text { a.s.}$

Attempt # 2

The increments of a martingale are orthogonal hence:

$$ \mathbb{E}\left[\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{4}\right]=\mathbb{E}\left[\left(\sum_{i=1}^{\infty} M_{i}-M_{i-1}\right)^{4}\right] $$

Since $M_n$ is a martingale then $M_n$ is finite, hence the increments are finite. Thus by Doob's martingale convergence theorem, $M_n$ converges a.s to $M_{\infty}$

Sorry if I went the wrong way with this proof. Any help or tips will be highly appreciated.

Answer 1

By considering $M_n+10$ instead of $M_n$, we may and shall assume that $\{M_n\}$ is a Martingale with respect to $\{{\mathcal F}_n\}$ that satisfies $M_n \ge 0$ for all $n$. We will show that $$ \sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{2}<\infty \quad \text { a.s. } \tag{1} $$ which implies that the same holds when the exponent 2 is replaced by 4.

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 $$P(\tau <\infty) \le E(M_0)/L \,. \tag{2}$$ Next, 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 $$\sum_{k=0}^{T\wedge n} \delta_k \le 3L \,. \tag{5}$$

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 $P(T<\infty) \le 2E(M_0)/L$. By orthogonality of Martingale increments $$9L^2 \ge E(Y_{T \wedge n}^2)=\sum_{k=1}^n E[(Y_{T\wedge k}-Y_{T\wedge (k-1)})^2] \,. \tag{6}$$ Thus on the event $\{T=\infty\}$, we have $$ (M_{i}-M_{i-1} )^{2} = (Y_i-Y_{i-1}-\delta_i)^2 \le 2(Y_i-Y_{i-1})^2 +2\delta_i^2 \le 2(Y_i-Y_{i-1})^2 +2L\delta_i \,$$ so
$$\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{2} \le 2\sum_{i=1}^{\infty} [(Y_i-Y_{i-1})^2 +L\delta_i] <\infty \; \text{a.s.}$$ by $(5)$ and $(6)$. Thus $$P\Bigl(\sum_{i=1}^{\infty}\left(M_{i}-M_{i-1}\right)^{2} =\infty\Bigr) \le P(T<\infty) \le 2E(M_0)/L\,.$$ Taking $L \to \infty$ shows this probability is $0$.