This question regards Theorem 1.8.6 on Durrett page 52 which states as

(The strong law of large numbers) Let $X_{1}, X_{2},\cdots$ be i.i.d random variables with $E|X_{i}|<\infty$. Let $E(X_{i})=\mu$ and $S_{n}=X_{1}+\cdots+X_{n}$. Then $S_{n}/n\longrightarrow \mu$ a.s.

The proof it requires at least two lemmas and Kolmogorov's One Series Theorem, I have read through them without problems.

However, I am thinking about a little transformation of this law. What if we set $E(X_{i})=0$ and there is a bounded sequence of non-random constants $c_{n}$ and we let $S_{n}:=c_{1}X_{1}+\cdots+c_{n}X_{n}$, will $S_{n}/n$ converges to $0$ almost surely? That is:

Let $X_{1},X_{2},\cdots$ be i.i.d integrable random variables with $E(X_{i})=0$. If $c_{n}$ is a bounded sequence of non-random constants, and we set $S_{n}:=c_{1}X_{1}+\cdots+c_{n}X_{n}$, show that $S_{n}/n\longrightarrow 0$ a.s.

I've been thinking about using the similar proof of the strong law, since the strong law is just let $c_{n}=1$ for all $n$, and $\mu\neq 0$.

So firstly I tried to directly make $Y_{n}:=c_{n}X_{n}$ for each $n$, and argue just for $Y_{n}$. The good thing here is that $E(Y_{n})=0$ so we don't change the limit in the almost sure convergence, however, since $c_{n}$ are different, even though $Y_{n}$'s are still independent, they are not identically distributed any longer.

How could I make all those different $c_{n}$ to be one thing? (so that in this way they are i.i.d again). Or perhaps I am heading on a wrong direction?

Thank you in advance for any discussion, hint, or solution!

Edit 1:

Since $c_{n}$ is bounded, $c_{n}\leq M$ for all $n$. Thus, if replace $Y_{n}:=MX_{n}$, and write $Z_{n}:=Y_{1}+\cdots+Y_{n}$ then they are still i.i.d, and then I can definitely show that $$\dfrac{Z_{n}}{n}\longrightarrow\mu\ \text{a.s.}$$

Now, note that $S_{n}:=c_{1}X_{1}+\cdots+c_{n}X_{n}\leq Z_{n}$, so this problem can be reduced to if if $Z_{n}/n\longrightarrow 0$ a.s. and $Z_{n}/n\geq S_{n}/n$, then $S_{n}/n\longrightarrow 0$ a.s.

I don't really know if this is true... If it is, how could I prove it?

Edit 2:

Okay I figured it out. The point here is that even though $c_{k}X_{k}$ is not i.i.d, you can still use i.i.d when you have $P(|c_{k}X_{k}|>n)$, since you can directly divided by $|c_{k}|$. Since it is bounded, everything will be fine.

For details, please see my answer of my own post.

Edit 3:

Since I noticed that some users voted me and favorite this post during me writing the proof in my answer, I make an edit here to let the system alert you that there is an edit so that you could see my answer. Thank you for your vote and favorite :) Enjoy my proof!

  • 1
    $\begingroup$ Can you check Kolmogorov's three series theorem? It is theorem 2.5.8 on Durrett PTE 5th edition. $\endgroup$ – E-A Oct 16 '19 at 19:40
  • $\begingroup$ @E-A Yes I know this theorem, but I am afraid that it may be hard to be applied. $\endgroup$ – JacobsonRadical Oct 16 '19 at 20:07
  • $\begingroup$ @E-A but indeed it is the only theorem which does not require identical distribution, but then it will be hard for me to verify those three conditions in the theorem if they are not identically distributed. $\endgroup$ – JacobsonRadical Oct 16 '19 at 20:12

Below is what I wanted to prove

Let $(X_{k})$ be integrable i.i.d random variables such that $E[X_{k}]=0$. Show that if $c_{n}$ is a bounded sequence of non-random constants, then $$n^{-1}\sum_{k=1}^{n}c_{k}X_{k}\longrightarrow 0\ \text{a.s.}$$

To prove this, we need two lemmas and Kolmogorov's One Series Theorem, which are stated below:

Lemma 1: (Kronecker's Lemma) Let $\{x_{n}\}$ and $\{a_{n}\}$ be two sequences of real numbers such that $a_{n}>0$ and $a_{n}\nearrow\infty$. Show that if $\sum_{n=1}^{\infty}x_{n}/a_{n}$ converges, then $$a_{n}^{-1}\sum_{m=1}^{n}x_{m}\longrightarrow 0.$$

Lemma 2: Show that $\sum_{k=1}^{\infty}k^{-2}Var(X_{k}\mathbb{1}_{(|X_{k}|\leq k)})\leq 4E|X_{1}|.$

Kolmogorov's One Series Theorem: Suppose $X_{1},X_{2},\cdots$ are independent and have $EX_{n}=0$. If $$\sum_{n=1}^{\infty}Var(X_{n})<\infty,$$ then we have $\sum_{n=1}^{\infty}Var(X_{n})$ converges a.s.

I will firstly use these three things to prove my statement, and then I will prove the first two lemmas. For the one series theorem, the proof of it is everywhere.

Proof of Statement:

Set $S_{n}:=c_{1}X_{1}+\cdots+c_{n}X_{n}$, $Y_{k}:=(c_{k}X_{k})\mathbb{1}_{(|c_{k}X_{k}|\leq k)}$, and $T_{n}:=Y_{1}+\cdots+Y_{n}.$

It then suffices to show that $T_{n}/n\longrightarrow 0$. Indeed, since $c_{k}$ is a bounded sequence, there exists $N$ such that $|c_{k}|\leq N$ for all $k$, and thus if $\omega\in\Omega$ satisfied $|X_{k}|>k/|c_{k}|$, it must satisfied $|X_{k}|>k/N$, and thus by monotonicity, we have $$\sum_{k=1}^{\infty}P(|c_{k}X_{k}|>k)\leq\sum_{k=1}^{\infty}P(|X_{k}|>k/N)\leq\int_{0}^{\infty}P(|X_{1}|>t)dt=E|X_{1}|<\infty,$$ which implies that $$P(c_{k}X_{k}\neq Y_{k}\ \text{i.o.})=0,$$ and thus $$|S_{n}(\omega)-T_{n}(\omega)|\leq R(\omega)<\infty\ \text{a.s. for all}\ n.$$

Secondly, glance at the proof of Lemma 2, we know that the inequality can be modified as $$\sum_{k=1}^{\infty}Var(Y_{k}/k^{2})\leq 4|c_{k}|E|X_{1}|<\infty,$$ where it is finite since $c_{k}$ is bounded and $X_{1}$ is integrable.

Now, set $Z_{k}:=Y_{k}-EY_{k}$, so that $E(Z_{k})=0$ and thus $Var(Z_{k})=Var(Y_{k})$. By Lemma 2, we have $$\sum_{k=1}^{\infty}Var(Z_{k})/k^{2}=\sum_{k=1}^{\infty}Var(Y_{k})/k^{2}\leq 4|c_{k}|E|X_{1}|<\infty,$$ and thus by Kolmogorov's One Series Theorem, we conclude that $\sum_{k=1}^{\infty}Z_{k}/k$ converges a.s., so use Lemma 1, we have $$n^{-1}\sum_{k=1}^{n}(Y_{k}-EY_{k})\longrightarrow 0,$$ and thus $$\dfrac{T_{n}}{n}-n^{-1}\sum_{k=1}^{n}EY_{k}\longrightarrow 0\ \text{a.s.}$$

Now, since $|c_{k}|$ is bounded, we can use dominated convergence theorem to get $EY_{k}\longrightarrow 0\ \text{as}\ k\longrightarrow\infty,$ and thus it follows that $$n^{-1}\sum_{k=1}^{n}EY_{k}\longrightarrow 0,$$ and hence $$T_{n}/n\longrightarrow 0.$$

Proof of Lemma 1:

By the setting of this problem, I believe it means that $a_{n}>0$ for $n\geq 1$, so let us set $a_{0}:=0$. For $m\geq 1$, define $b_{m}:=\sum_{k=1}^{m}x_{k}/a_{x}$, and define $b_{0}=0$. Then, $x_{m}=a_{m}(b_{m}-b_{m-1}).$ Therefore, we have (don't forget $a_{0}=b_{0}=0$): \begin{align*} a_{n}^{-1}\sum_{m=1}^{n}x_{m}&=a_{n}^{-1}\sum_{m=1}^{n}a_{m}(b_{m}-b_{m-1})\\ &=a_{n}^{-1}\Big(\sum_{m=1}^{n}a_{m}b_{m}-\sum_{m=1}^{n}a_{m}b_{m-1}\Big)\\ &=a_{n}^{-1}\Big(a_{n}b_{n}+\sum_{m=2}^{n}a_{m-1}b_{m-1}-\sum_{m=1}^{n}a_{m}b_{m-1}\Big)\\ &=b_{n}-\sum_{m=1}^{n}\dfrac{(a_{m}-a_{m-1})}{a_{n}}b_{m-1}. \end{align*}

By hypothesis, $b_{n}\longrightarrow b_{\infty}<\infty$ as $n\longrightarrow\infty$. It then remains for us to show that $$\sum_{m=1}^{n}\dfrac{(a_{m}-a_{m-1})}{a_{n}}b_{m-1}\longrightarrow b_{\infty}.$$

Let $\epsilon>0$, set $B:=\sup|b_{n}|$, and choose an $M$ such that for all $m\geq M$, we have $|b_{m}-b_{\infty}|<\epsilon/2$. Also, pick an $N$ such that $a_{M}/a_{n}<\epsilon/4B$ for all $n\geq N$.

Now, for $n\geq N$, recalling $a_{m}- a_{m-1}\geq 0$, and noting $\sum_{m=1}^{n}\dfrac{(a_{m}-a_{m-1})}{a_{n}}=1,$ we have: \begin{align*} \Big|\sum_{m=1}^{n}\dfrac{(a_{m}-a_{m-1})}{a_{n}}b_{m-1}-b_{\infty}\Big|&\leq\sum_{m=1}^{n}\dfrac{(a_{m}-a_{m-1})}{a_{n}}|b_{m-1}-b_{\infty}|\\ &\leq \dfrac{a_{M}}{a_{n}}\cdot 2B+\dfrac{a_{n}-a_{M}}{a_{n}}\cdot\dfrac{\epsilon}{2}\\ &<\epsilon. \end{align*}

The result follows immediately by taking $\epsilon\searrow 0$.

Proof of Lemma 2:

Set $Y_{k}:=X_{k}\mathbb{1}_{(|X_{k}|\leq k)}$.

Firstly, we observe that $$Var(Y_{k})\leq E[Y_{k}^{2}]=\int_{0}^{\infty}2yP(|Y_{k}|>y)dy\leq\int_{0}^{k}2yP(|X_{1}|>y)dy,$$ then since everything is $\geq 0$ and the sum is just an integral with respect to counting measure on $\{1,2\cdots\}$, we can use Fubini's Theorem to yield: \begin{align*} \sum_{k=1}^{\infty}\dfrac{E(Y_{k}^{2})}{k^{2}}&\leq\sum_{k=1}^{\infty}k^{-2}\int_{0}^{\infty}\mathbb{1}_{(y<k)}2yP(|X_{1}|>y)dy\\ &=\int_{0}^{\infty}\Big(\sum_{k=1}^{\infty}k^{-2}\mathbb{1}_{(y<k)}\Big)2yP(|X_{1}|>y)dy \end{align*}

Now, since $E|X_{1}|=\int_{0}^{\infty}P(|X_{1}|>y)dy$, the result follows immediately from the lemma below.

Lemma 2.1: If $y\geq 0$, then $2y\sum_{k>y}k^{-2}\leq 4$.

Proof of Lemma 2.1:

Note that if $m\geq 2$, we then have $$\sum_{k\geq m}k^{-2}\leq\int_{m-1}^{\infty}x^{-2}dx=(m-1)^{-1}.$$

Now, when $y\geq 1$, the sum starts at $k=[y]+1\geq 2$. Also, note that $y/[y]\leq 2$ for $y\geq 1$, since the worst case is $y$ being really close to $2$. Thus, we have $$2y\sum_{k>y}k^{-2}\leq 2y/[y]\leq 4.$$

Finally, for $0\leq y<1$, we have $$2y\sum_{k>y}k^{-2}\leq 2\Big(1+\sum_{k=2}^{\infty}k^{-2}\Big)\leq 4.$$

The whole proof now is complete. Please let me know if there is any problem in my proof. Enjoy :)

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