Let $X_1,X_2,\dots$ be a sequence of independent random variables, such that the series $$\sum_{n=1}^\infty\frac{\operatorname{Var}(X_n)}{n^2}$$ converges. Show that as $n\to\infty$, $$\frac{1}{n}\sum_{k=1}^n(X_k-\mathbb E[X_k])$$ converges almost surely to $0$.

There is a quick solution via the martingale convergence theorem: we have that $$Y_n=\sum_{k=1}^n\frac{X_k-\mathbb E[X_k]}{k}$$ is martingale, and $\sup_n\mathbb E[|Y_n|]$ is finite, so $Y_n$ converges almost surely to some random variable $Y$, and we can finish with Kronecker's lemma.

I'm interested though in any approaches avoiding the heavy machinery of the martingale convergence theorem. I feel like defining the $Y_n$'s as I did above could be fruitful. For example, the Kolmogorov inequality gives the bound $$\mathbb P\left(\max_{1\leq i\leq n}|Y_i|>\varepsilon\right)\leq\frac{1}{\varepsilon^2}\mathbb E\left[Y_n^2\right]=\frac{1}{\varepsilon^2}\cdot\sum_{k=1}^n\frac{\operatorname{Var}(X_k)}{k^2}.$$ But I'm unsure what to make from all this.

  • $\begingroup$ With independence, it looks like the strong law of large numbers. $\endgroup$ Commented Dec 28, 2020 at 22:16
  • $\begingroup$ @herbsteinberg : The variables are not necessarily identically distributed. $\endgroup$
    – Michael
    Commented Dec 28, 2020 at 22:17
  • $\begingroup$ @Michael I believe the LLN proof could be used with the variance condition as given. $\endgroup$ Commented Dec 28, 2020 at 22:27
  • $\begingroup$ Whatever method is used, I suspect it would require a lot of work. $\endgroup$
    – Michael
    Commented Dec 28, 2020 at 22:37
  • $\begingroup$ Perhaps look at Section 8.2 (pages 3-4) here: www2.stat.duke.edu/courses/Fall17/sta711/lec/wk-08.pdf. They use Lemma 1, Levy's result about $\to_p \iff \to_\text{a.s.}$ for sums of independent r.v.s, Lemma 2 (Kronecker's Lemma) to prove Theorem 3 (Kolmogorov's Convergence Criterion) and Theorem 4 (Kronecker's $L_2$ SLLN), a generalization of your desired result (generalizing by considering $1/b_n^2$ and $1/b_n$ instead of $1/n^2$ and $1/n$, where the $b_n$ are positive monotone increasing) $\endgroup$
    – D.R.
    Commented Dec 28, 2020 at 22:43

2 Answers 2


A proof, which you kind of started, can be derived from Kolmogorov inequality,

For $n < m$, Kolmogorov inequality applied to the martingale $Y_k-Y_n$, $n \leq k \leq m$, gives that, for all $\epsilon > 0$ $$ P(\max_{n \leq k \leq m} |Y_k - Y_n| > \epsilon) \leq \frac{1}{\epsilon^2} \sum_{k = n}^m \frac{Var(X_k)}{k^2}. $$

So, for all $\epsilon > 0$

$$ P(\inf_l \max_{n,m \geq l} |Y_m - Y_n| > \epsilon) \leq \lim_{l \rightarrow \infty} P(\max_{n,m \geq l} |Y_m - Y_n| > \epsilon) = 0. $$

Now take $\epsilon_p \rightarrow 0$. Define $$ \Omega_p = \{ \inf_l \max_{n,m \geq l} |Y_m - Y_n| \leq \epsilon_p \}, $$ and $\Omega' = \bigcap_p \Omega_p$. Then $P(\Omega') = 1$ and, for all $\omega \in \Omega'$, $Y_n(\omega)$ is a Cauchy sequence, therefore converges. Kronecker's Lemma then finishes the proof, as before.


This does not really "avoid the martingale machinery", though. This argument substitutes martingale convergence by Kolmogorov inequality, which is a maximal inequality for martingales.

It does not seem easy to get away from using the martingale property in one way or another.

The alternative argument proposed by previous answer uses the fact that a series of independent summands that converges in probability must also converge almost surely. The standard proof of this fact is a stopping time argument, that also uses the stronger independence property.

  • $\begingroup$ Ah that's nice and more similar to the kind of thing I was trying. I just felt that using the martingale convergence theorem was nuking an ant -- the Kolmogorov inequality is much easier to prove. $\endgroup$
    – jlammy
    Commented Dec 29, 2020 at 15:47

It is well known that a series of independent random variables converges almost surely iff it converges in distribution iff it converges in probability. If you are willing to assume this result the rest is very easy. To show that $Y_n$ converges in probability it is enough to prove convergence in $L^{2}$. But convergence in $L^{2}$ follows from the following basic fact from FA: if $(x_n)$ is s sequence in a Banach space such that $\sum \|x_n\| <\infty$ then $\sum x_n$ converges in the norm.

  • $\begingroup$ If $\{X_n\}$ is a sequence of independent random variables such that $X_n=1$ with probability $\frac{1}{n}$ and $X_n=0$ otherwise, then $\{X_n\}$ converges to $0$ in probability but doesn't converge almost surely to $0$. $\endgroup$
    – user801306
    Commented Dec 28, 2020 at 23:56
  • $\begingroup$ @MatthewPilling There is a difference between sequence of independent random variables and series of independent random variables. $\endgroup$ Commented Dec 29, 2020 at 0:13
  • $\begingroup$ My bad! I misread you first sentence. $\endgroup$
    – user801306
    Commented Dec 29, 2020 at 0:20

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