The following question has me stumped: Let $X_n$ be a square integrable martingale with $E((X_n)^2)\leq n$ for all $n$. Prove that $X_n/n$ tends to $0$ almost surely. (this is in a sense a law of large numbers, generalizing the case where $X_n$ is a sum of $n$ iid zero mean random variables.)

Any ideas?


2 Answers 2


The claim in question is a corollary of a standard SLLN for martingale difference sequences (MDS).


The statement of SLLN for MDS is as follows. Let $N_t$ be a martingale difference sequence (MDS) such that $\sum\limits_{t=1}^{\infty} \frac{E[N_t^2]}{t^2} < \infty$, then

$$ \frac{1}{n} \sum_{t=1}^n N_t \rightarrow 0 \;\;a.s. $$

(In this case, the martingale difference sequence $N_t$ is given by differencing the martingale $X_t$: $N_t = X_t - X_{t-1}$. Then summation by parts gives \begin{align*} \sum_{t=1}^n \frac{E[N_t^2]}{t^2} &= \sum_{t=1}^n \frac{E[X_t^2] - E[X_{t-1}^2]}{t^2} \\ &= \frac{E[X_n^2]}{n^2} - \sum_{t = 1}^{n} E[X_{t-1}^2] \left( \frac{1}{t^2} - \frac{1}{(t-1)^2} \right). \end{align*}

The assumption that $E[X_{t}^2] = O(t)$ implies that $$ E[X_{t-1}^2] ( \frac{1}{(t-1)^2} - \frac{1}{t^2} ) = O(\frac{1}{t^2}). $$ Therefore $\sum\limits_{t=1}^{\infty} \frac{E[N_t^2]}{t^2} < \infty$. )

In turn, the SLLN for MDS can be shown via two arguments. Both are standard devices for results of this type, one via the martingale convergence theorem and another via Kolmogorov's martingale maximal inequality.

Via Martingale Convergence Theorem

(The previous answer is a variation of this argument.)

If $\sum\limits_{t=1}^{\infty} \frac{E[N_t^2]}{t^2} < \infty$, the martingale $Y_n = \sum\limits_{t = 1}^n \frac{N_t}{t}$, $n \geq 1$, is bounded in $L^2$, therefore converges almost surely (and in $L^2$). Therefore, by Kronecker's lemma, $$ \frac{1}{n}\sum_{t = 1}^n N_t \stackrel{a.s.}{\rightarrow} 0 $$ as $n \rightarrow \infty$.

Via Maximal Inequality

Consider again the $L^2$-martingale $Y_n = \sum\limits_{t = 1}^n \frac{X_t}{t}$, $n \geq 1$. Let $\sigma^2_t = \frac{E[ X_t^2 ]}{t^2}$.

By the maximal inequality, for all $n > 0$ and for all $\epsilon > 0$, $$ P( \sup_{m \geq n} | S_m - S_n | \geq \epsilon ) \leq \frac{K}{\epsilon^2} \sum_{t \geq n} \sigma^2_t $$ for some constant $K$ independent of $n$. Therefore $$ P( \inf_n \sup_{m \geq n} | S_m - S_n | \geq \epsilon ) = 0 $$ for all $\epsilon > 0$. In other words, the sequence $S_n$, $n \geq 1$, is Cauchy, therefore converges, with probability $1$. Again by Kronecker's lemma, $$ \frac{1}{n}\sum_{t = 1}^n N_t $$ converges to zero as $n \rightarrow \infty$ with probability $1$.

  • 1
    $\begingroup$ I just looked up my intro probability script because I could not believe it: The strong law of large numbers was proved in 2 pages using 4th moments. You proved a more general statement in 3 lines and I can not find an issue with it. wtf $\endgroup$
    – Felix B.
    Jul 28, 2021 at 20:11
  • $\begingroup$ ah I did not notice the change from 1/t to 1/n at first. I guess there is some of the difficulty hidden in kronecker's lemma $\endgroup$
    – Felix B.
    Jul 28, 2021 at 20:21
  • $\begingroup$ The martingale theorems (martingale convergence theorem and maximal inequality) are kind of big hammers. (Correct me if otherwise, but I don't believe the argument you refer to exploits the martingale structure.) $\endgroup$
    – Michael
    Aug 7, 2021 at 2:44
  • $\begingroup$ @FelixB. No, not "handwaved." It's the Martingale Convergence Theorem---as stated. $\endgroup$
    – Michael
    Aug 9, 2021 at 12:19
  • $\begingroup$ hm, no I made a mistake we do not have monotonicity since we do not care about $\sum_{t=1}^n \frac{N_t^2}{t^2}$ but rather $Y_n^2=\left(\sum_{t=1}^n\frac{N_t}{t}\right)^2$ and I don't think the statement "bounded in $L^2$ implies almost sure convergence" is true: math.stackexchange.com/a/138054/445105 $\endgroup$
    – Felix B.
    Aug 11, 2021 at 15:03

We assume $X_0=0$ without loss of generality. Define $$Y_n:=\sum_{i=1}^n\frac{X_i-X_{i-1}}i, n\geqslant 1, \quad Y_0:=0.$$ Then $\left(Y_n\right)_{n\geqslant 1}$ is a martingale (for the same filtration as $\left(X_n\right)_{n\geqslant 1}$) and using the fact that $\left(X_i-X_{i-1}\right)_{i\geqslant 1}$ is a martingale differences sequence, we have
\begin{align} \mathbb E\left[Y_n^2\right]=\sum_{i=1}^n\frac 1{i^2}\left(\mathbb E\left[X_i^2\right]-\mathbb E\left[X_{i-1}^2\right]\right). \end{align} Now, using Abel's transformation and the assumption on $\mathbb E\left[X_i^2\right]$, we derive boundedness of the sequence $\left(\mathbb E\left[Y_n^2\right]\right)_{n\geqslant 1}$. Using the martingale convergence theorem, we get that the sequence $\left(Y_n\right)_{n\geqslant 1}$ converges almost surely to some random variable $Y$.

Now, we have (accounting $X_0=0$) \begin{align} \frac{X_n}n&=\frac 1n\sum_{l=1}^n\left(X_l-X_{l-1}\right)\\ &=\frac 1n\sum_{l=1}^n\frac{X_l-X_{l-1}}l\cdot l\\ &=\frac 1n\sum_{l=1}^n\left(Y_l-Y_{l-1}\right)\cdot l\\ &=\frac 1n\sum_{k=1}^nY_k\cdot k-\frac 1n\sum_{k=0}^{n-1}Y_k\cdot (k+1)\\ &=\frac 1n\sum_{k=1}^nY_k\cdot k-\frac 1n\sum_{k=1}^{n-1}Y_k\cdot (k+1)\\ &=Y_n-\frac 1n\sum_{k=1}^nY_k, \end{align} from which it follows that $X_n/n\to 0$ almost surely.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .