Let $(X_n)_{n=1}^{\infty}$ be a sequence of pairwise independent, identically distributed random variables with finite mean.

Let $Y_n = X_n\mathbf1\{{|X_n| \leq n\}}$, then $$ \sum_{n=1}^{\infty} P(Y_n \neq X_n) = \sum_{n=1}^{\infty} P(|X_n| >n) = \sum_{n=1}^{\infty} P(|X_1|>n) \leq \mathbb{E}[|X_1|] < \infty. $$

Hence from Borel-Cantelli lemma $\lim_{n\rightarrow \infty}\frac{X_1 + \dots + X_n}{n} = \mathbb{E}[X_1]~a.s$ if and only if $\lim_{n\rightarrow \infty}\frac{Y_1 + \dots + Y_n]}{n} = \mathbb{E}[X_1]~a.s.$

Now I have to show that $\lim_{n\rightarrow \infty}\frac{\mathsf {Var}[Y_1 + \dots + Y_n]}{n^2} = 0$, unfortunately I don't know how.

I would really appreciate any hints or tips.

  • $\begingroup$ Suppose that the $X_i$ have a finite second moment. Since they are pairwise independent $V(\sum_{i=1}^n X_i)=\sum_{i=1}^n V(X_i) = nV(X_1)$ and the conclusion follows. Why are you interested in $V(\sum_{i=1}^n Y_i)$ ? $\endgroup$ Aug 21, 2019 at 15:07
  • $\begingroup$ The interest of this kind of truncation methods is precisely to bypass these kind of assumptions ! Nice (but tough) exercise, where does it come from ? $\endgroup$
    – Olivier
    Aug 21, 2019 at 15:23
  • 2
    $\begingroup$ By the way, I don't get your "iff" statement on the Var. Could you copy the full and precise statement of your problem please ? $\endgroup$
    – Olivier
    Aug 21, 2019 at 15:28
  • $\begingroup$ Could you give some elements on my question please ? $\endgroup$
    – Olivier
    Aug 22, 2019 at 12:46
  • 1
    $\begingroup$ From Borel-Cantelli lemma and Toeplitz lemma we have that $ \lim_{n\rightarrow \infty}\frac{X_1 + \dots + X_n}{n} = \mathbb{E}[X_1] a.s $ iff $ \lim_{n\rightarrow \infty}\frac{Y_1 + \dots + Y_n]}{n} \mathbb{E} = [X_1] $ a.s. For that reason I assumed also that $\lim_{n\rightarrow \infty}\frac{\mathsf {Var}[X_1 + \dots + X_n]}{n^2} = 0$ iff $\lim_{n\rightarrow \infty}\frac{\mathsf {Var}[Y_1 + \dots + Y_n]}{n^2} = 0.$ But now I'm not sure is it true. $\endgroup$ Aug 22, 2019 at 13:34

3 Answers 3


I come after the battle, but here it is : yet another answer,

\begin{align*} Var(Y_1+\ldots+Y_n) & =\sum_{i=1}^n Var(Y_i) \\ &=\sum_{i=1}^n Var(X_i 1_{|X_i|\le i}) \\ &=\sum_{i=1}^n Var(X_1 1_{|X_1|\le i}) \\ &\le \sum_{i=1}^n \mathbb E[X_1^2 1_{|X_1|\le i}] \\ & = \mathbb E[X_1^2 \sum_{i=1}^n 1_{|X_1|\le i}] \\ & \le n \mathbb E[X_1^2 1_{|X_1| \le n}] \end{align*}

Therefore : $$ \frac{Var(Y_1+\ldots+Y_n)}{n^2} \le \mathbb E\Big[\frac{X_1^2}{n} 1_{|X_1| \le n}\Big] $$

Now we have the bound

$$\frac{X_1^2}{n} 1_{|X_1| \le n} \le |X_1|$$

the left hand side converges to $0$ a.s., while the right hand side is an integrable bound (assuming $X_1$ has a finite expectation) that allows to apply Dominated Convergence Theorem to the effect that :

$$\mathbb E\Bigg[ \frac{X_1^2}{n} 1_{|X_1| \le n}\Bigg] =o(1)$$

  • $\begingroup$ Why does $\frac{X_1^2}{n} 1_{|X_1| \le n}$ converge a.s to $0$ ? $\endgroup$ Aug 22, 2019 at 14:16

For this proof we need $E|X|<\infty$ (it seems that this is the case).

Notice that $$E(Y_k^2)\leq \int_0^k 2tP(|X|>t)dt. $$ Hence, by Fubini's theorem

\begin{align} \sum^\infty_{k=1}\frac{E(Y_k^2)}{k^2} &\leq \int_0^\infty \left(\sum^\infty_{k=1}\frac{1_{(t\leq k)}}{k^2}\right)2tP(|X|>t)dt\\ & \int_0^\infty\left( \sum^\infty_{k=t}\frac{1}{k^2}\right)2tP(|X|>t)dt\\ & \leq \int_0^\infty \left( \sum^\infty_{k=t}\frac{2}{k(k+1)}\right)2tP(|X|>t)dt\\ & = \int_0^\infty \left( \sum^\infty_{k=t}\frac{2}{k} - \frac{2}{k+1}\right)2tP(|X|>t)dt\\ & \leq \int_0^\infty \frac{2}{t}2tP(|X|>t)dt = 4E|X|<\infty \end{align}

So, we proved that $$\sum^\infty_{k=1}\frac{E(Y_k^2)}{k^2} $$ converges. Then, by Kronecker's lemma, $$\lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^n_{k=1}E(Y_k^2) =0.$$

Additional thoughts: After reading the comments, I think this may help you.

You can use Kolmogorov's maximal inequality (or a Martingale in $L_2$ convergence Theorem) to prove that if $$ \lim_{n\rightarrow \infty}\sum^n_{k=1}\frac{\text{Var}(Y_k^2)}{k^2}$$ converges, then $$\lim_{n\rightarrow \infty}\sum^n_{k=1}\frac{Y_k -EY_k}{k} $$ converges a.s.. Hence, by Kronecker's lemma, $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=1}(Y_k -EY_k)=0. $$ Using Dominated Convergence Theorem, it's easy to show $$E Y_k \rightarrow EX. $$ Therefore, by Cesàro's Lemma, $$\frac{1}{n}\sum_{k=1}^n Y_k = EX,\ a.s. .$$

Now, you can use your Borel-Cantelli Argument to prove that the above expression implies $$\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=1}X_k=EX,\ a.s.. $$

This is basically the proof for the Strong Law of Large Numbers in the case i.i.d. with $E|X|<\infty.$


Since the $Y_i$ are independent, it is enough to show that $\frac{1}{n^2}\sum_{k=1}^n{Var(Y_k)} \rightarrow 0$.

Now, $\mathbb{E}[Y_k^2]=\int_0^{k^2}{P(Y_k^2 > a)\,da}=\int_0^k{2uP(k > |X| > u)\,du} \leq \int_0^k{2uP(X > u)\,du}$.

Denote as $F(u)=\int_u^{\infty}{P(|X|>a)\,da}=\mathbb{E}[|X|1(|X| > u)]$, then $uP(X>u)=F(u)-\frac{d}{du}(uF(u))$.

As a consequence, $\mathbb{E}[Y_k^2] \leq 2\int_0^k{F(u)\,du}-kF(k) \leq 2\int_0^k{F(u)\,du}$.

Note that $F$ is non-increasing and goes to $0$, so $\int_0^k{F(u)\,du}=o(k)$, hence $Var(Y_k) =o(k)$, which ends the proof.

  • $\begingroup$ Very nice proof ! +1 $\endgroup$
    – Olivier
    Aug 21, 2019 at 21:04

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