
# 9: Laws of Large Numbers

[ "article:topic", "Laws of Large Numbers" ]

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Let $$X_1, X_2, X_3, \ldots$$ be a sequence of independent and identically distributed random variables. A common abbreviation for independent and identically distributed'' is i.i.d.. What this can mean is that we are taking repeated independent samples from some distribution, for example when doing a poll or an experiment in quantum mechanics (when doing a poll, it might be better to have the samples not be independent -- after all, does it really make sense to call up the same person twice? But if the population is large the effect of having the samples be independent is negligible, and can make the analysis of the results a bit simpler).

Assume that $$\expec X_1$$, the mean of $$X_1$$ (therefore of every $$X_n)$$, is defined and finite, and denote it by $$\mu$$. In a real-life situation, we might not know what the mean is, and wish to estimate it. So we look at the sum of the first $$n$$ samples,

$S_n = \sum_{k=1}^n X_k,$

and use it to form the empirical average,

$\overline{X}_n = \frac1n S_n = \frac{1}{n} \sum_{k=1}^n X_k.$

The natural question is whether we can expect the empirical average to be close to the true mean $$\mu=\expec X_1$$ as $$n\to\infty$$, and in what sense of close''. A theorem or statement to that effect, making some assumptions, is called a law of large numbers. This can be generalized to a large extent to cases when the $$X_n$$'s are not identically distributed, or not independent, or both, etc. Thus, the Law of Large Numbers'' is not a single theorem in the normal sense of the word but rather a class of theorems, or even a principle'' or meta-theorem'' that you might hear a probabilist refer to in a somewhat vague, metaphorical way.

### Weak laws of large numbers

Theorem: Weak Law of Large Numbers

If $$\expec|X_1|<\infty$$ then

$\frac{S_n}{n} \xrightarrow[n\to\infty]{\prob} \mu.$

Proof in the case of finite variance

In the most general case, proving this requires some work -- we shall deduce it from its stronger cousin, the Strong Law of Large Numbers. However, if we assume that $$\sigma^2 = \var(X_1)<\infty$$, the proof is extremely easy! It suffices to note that in this case $$\var(S_n) = n \sigma^2$$ (this is true even when the $$X_n$$'s are not independent but only uncorrelated), or $$\var(S_n/n) = \sigma^2/n$$.

From Chebyshev's inequality we then have that for any $$\epsilon>0$$,

\prob(|n^{-1} S_n - \mu|>\epsilon) \le \frac{\sigma^2}{n \epsilon^2} \xrightarrow[n\to\infty]{} 0.

### Strong laws of large numbers

Our goal in this section will be to prove:

Theorem: Strong Law of Large Numbers

As in the case of the weak law, it turns out that this is easier to prove when making more restrictive assumptions about the distribution of $$X$$, and specifically about the existence of moments. So we will prove it several times, successively weakening our assumptions until we get to the most general (but least easy to prove) result.

Proof

[Proof in the case of a finite fourth moment] To prove that $$S_n/n\to \mu$$, we want to prove that $$\prob(|S_n/n-\mu|>\epsilon\textrm{ i.o.})=0$$. Looking at the bound \eqref{eq:variancebound}, we see that if the bound $$\sigma^2/n\epsilon^2$$ were to form the $$n$$-th general term of a convergent series, then by the Borel-Cantelli lemma this would be enough to get the desired consequence. This is not the case, but if we assume that $$X_1$$ has a finite fourth moment, then we could do a similar trick that will give a convergent series. In that case, using Markov's inequality we can get that

\label{eq:previousbound}
\prob(|S_n/n-\mu|>\epsilon) = \prob\left(\left( \sum_{k=1}^n (X_k-\mu) \right)^4 > n^4 \epsilon^4\right)
\le \frac{\expec(S_n-n\mu)^4}{n^4 \epsilon^4}.

Denote $$\hat{X_k}=X_k-\mu$$, and $$T_n = S_n-n\mu= \sum_{k=1}^n \hat{X_k}$$. To bound $$\expec(T_n^4)$$, note that we can write

\begin{eqnarray*} T_n^4 &=& \sum_{k=1}^n \hat{X_k}^4 + \binom{4}{1} \sum_{1 \le i\neq j \le n} \hat{X_i}^3 \hat{X_j}
+ \frac12\binom{4}{2}\sum_{1 \le i\neq j \le n} \hat{X_i}^2 \hat{X_j}^2
\\ & & + \frac12\binom{4}{2,1,1} \sum_{1\le i,j,k \le n\textrm{ distinct}} \hat{X_i}^2 \hat{X_j} \hat{X_k}
+ \frac{1}{4!}\binom{4}{1,1,1,1} \sum_{1 \le i,j,k,\ell \le n\textrm{ distinct}} \hat{X_i} \hat{X_j} \hat{X_k} \hat{X_\ell}
\end{eqnarray*}

(where $$\binom{4}{2,1,1}=4!/2!1!1!$$, $$\binom{4}{1,1,1,1}=4!/1!1!1!1!$$ are multinomial coefficients).
Now take the expectations on both sides, and use the fact that $$\expec(\hat{X_k}) = 0$$ and that the $$\hat{X_k}'s are independent, to get $\expec(T_n^4) = n m_4 + 3n(n-1) m_2^2,$ where we denote \(m_2 = \expec(\hat{X}_k^2) = \var(X_k)$$, $$m_4=\expec(\hat{X}_k^4)<\infty$$. This gives us the bound we wanted! In fact, all that matters is that $$\expec(T_n^4)\le C n^2$$ for some constant $$C>0$$. Combining this with \eqref{eq:previousbound}, we get that

$\prob(|S_n/n-\mu|>\epsilon) \le \frac{C}{n^2 \epsilon^4}.$

In particular, we have that $$\sum_{n=1}^\infty \prob(|S_n/n-\mu|>\epsilon) <\infty$$, so by the Borel-Cantelli lemma, the probability of the event

$A_\epsilon := \left\{|S_n/n-\mu|>\epsilon \textrm{ i.o.}\right\}$

is $$0$$. This is true for all $$\epsilon>0$$, therefore the probability of

$\{S_n/n\nrightarrow \mu\} \subset \bigcup_{\epsilon>0} A_\epsilon = \bigcup_{n=1}^\infty A_{1/n}$

is also $$0$$, since it is contained in a countable union of events of probability 0.
Proof in the case of finite variance

We are slowly refining our techniques to weaken the assumptions required to prove the SLLN. The following nice proof manages to harness the Chebyshev variance bound \eqref{eq:variancebound} after all to deduce the theorem in the case of r.v.'s with finite variance. Observe that while the series of terms on the right-hand side of \eqref{eq:variancebound} diverges, if we restrict it to a subsequence $$n_k = k^2$$, it will become a convergent series: $$\sum_{k=1}^\infty \sigma^2/k^2\epsilon^2<\infty$$. This implies, again by the Borel-Cantelli lemma, that

$\prob(|S_{n_k}/{n_k} - \mu| > \epsilon \textrm{ i.o.}) = 0$

for all $$\epsilon>0$$ (the i.o.'' here refers to the running index'' $$k$$, not $$n$$). This implies as before that

$\prob\left(\frac{S_{n_k}}{n_k}\xrightarrow[k\to\infty]{\textrm{a.s.}} \mu\right) = 1.$

So, while we have not shown almost sure convergence of the empirical averages to the true mean for all values of $$n$$, at least we have done so for the subsequence $$n_k=k^2$$. But this subsequence is relatively dense. In particular, if we show that in between elements of the subsequence $$n_k=k^2$$, the empirical average cannot fluctuate too wildly, then the theorem would follow. This will again follow from a combination of variance-based (i.e., Chebyshev) bounds and the Borel-Cantelli lemma. Fix some $$k$$, and take an integer $$n$$ satisfying $$n_k\le n < n_{k+1}$$. How likely is the $$n$$-th empirical average to deviate significantly from the $$n_k$$-th empirical average? Using the notation $$T_n=S_n-n\mu$$ as before, we have

\begin{eqnarray*}
\prob\left( \left| \frac{T_n}{n}-\frac{T_{n_k}}{n_k} \right| > \epsilon \right) &=& \prob\left( \left|
\frac{n_k T_n - n T_{n_k}}{n\cdot n_k} \right|>\epsilon\right)
\\ \textrm{(by triangle ineq.)}  &\le&
\prob\left( \left| \frac{T_n -T_{n_k}}{n_k} \right| > \epsilon/2 \right)+ \prob\left( \left|T_n\right| \frac{n-n_k}{n\cdot n_k}
> \epsilon/2 \right)
\\ \textrm{(by Chebyshev's ineq.)} &\le&
\frac{4 \textrm{Var}(T_n-T_{n_k})}{\epsilon^2 n_k^2} + \frac{4\textrm{Var}(T_n)(n-n_k)^2}{n^2 n_k^2}
\\ &\le&
\frac{10 k \sigma^2}{\epsilon^2 k^4} + \frac{20 k^4 \sigma^2}{k^8} < \frac{20 \sigma^2}{k^3},
\end{eqnarray*}

where we have denoted $$\sigma^2=\var(X_1)$$.

The estimate that we obtained is valid for a single $$n$$. We are actually interested in the maximal fluctuation of the empirical averages $$S_n/n$$ from $$S_{n_k}/n_k$$, when $$n$$ ranges in the interval $$[n_k,n_{k+1})$$. By a simple subadditivity argument, usually referred to as a union bound (i.e., bounding the probability of a union of events by the sum of the probabilities, which is the most naive bound you can imagine), we get easily that

\begin{eqnarray*}
\prob\left(\max_{n_k \le n < n_{k+1}} \left| \frac{T_n}{n} - \frac{T_{n_k}}{n_k} \right| > \epsilon \right)
&=&
\prob\left(\bigcup_{n_k \le n < n_{k+1}} \left\{\left| \frac{T_n}{n} - \frac{T_{n_k}}{n_k} \right| > \epsilon \right\}\right)
\\ &\le& \sum_{n=n_k}^{n_{k+1}-1}
\prob\left(\left| \frac{T_n}{n} - \frac{T_{n_k}}{n_k} \right| > \epsilon \right) < \frac{20 \sigma^2\cdot 2k}{k^3}
= \frac{40\sigma^2}{k^2}.
\end{eqnarray*}

Once again, we have obtained as our bound the general term of a convergent series! Denoting

$A_{k,\epsilon} = \left\{ \max_{n_k \le n < n_{k+1}} \left| \frac{T_n}{n} - \frac{T_{n_k}}{n_k} \right| > \epsilon \right\},$

by the Borel-Cantelli lemma this implies that for any $$\epsilon>0$$, $$\prob(A_{k,\epsilon}\textrm{ i.o.})=0$$ (with the i.o.'' qualifier again referring to the running index $$k$$). What about the chance that this will happen for some $$\epsilon>0$$? After all, there are so many $$\epsilon's out there... But no, we also get that $\prob\left(\bigcup_{\epsilon>0} \left\{ A_{k,\epsilon}\textrm{ i.o.} \right\} \right) = \prob\left( \bigcup_{m=1}^\infty \left\{ A_{k,1/m}\textrm{ i.o. (w.r.t.\ \(k$$)}\right\} \right) = 0,$

by subadditivity for countable unions and the fact that $$A_{k,\epsilon} \subset A_{k,\epsilon'}$$ if $$\epsilon>\epsilon'$$.

Finally, combining our two main results, namely that the two events

\begin{eqnarray*}
E_1 &=& \left\{ \frac{S_{n_k}}{n_k} \xrightarrow[n\to\infty]{} \mu \right\},
\\
E_2 &=& \left( \bigcup_{m=1}^\infty \left\{
\max_{n_k\le n< n_{k+1}} \left|\frac{T_n}{n}-\frac{T_{n_k}}{n_k} \right| > \frac{1}{m}
\textrm{ for inifinitely many $$k's} \right\} \right)^c \end{eqnarray*} both have probability 1, we get that their intersection \(E_1\cap E_2$$ also has probability 1. But we have the event inclusion

$E_1 \cap E_2 \subset \left\{ \frac{S_n}{n} \xrightarrow[n\to\infty]{} \mu \right\}$

(if the conditions in both the events $$E_1, E_2$$ occur, then $$S_n/n$$ must converge to $$\mu$$), so this latter event also has probability 1.

The above proof was a little involved, but is based on a few simple ideas: 1. Use Chebyshev bounds together with the Borel-Cantelli lemma. 2. Break up the event $$\{S_n/n\to \mu\}$$ in a clever way into events which can be managed using this technique, namely (in this case) the weaker convergence along the subsequence $$n_k = k^2$$, and separately the control of the fluctuations of the empirical averages in each range $$[n_k,n_{k+1})$$ between successive elements of the subsequence.

An extra benefit of the above proof is that it doesn't use the full power of the assumption that the $$X_n$$'s are independent. In fact, everything is based on variance computations of sums of the $$X_k$$'s, so the proof works equally well for \emph{uncorrelated} random variables!

Theorem 4.1: You see me

[SLLN for uncorrelated r.v.'s with finite variance] If $$X_1,X_2,\ldots$$ are a sequence of uncorrelated and identically distributed r.v.'s with finite variance, then

$\frac{1}{n}\sum_{k=1}^n X_k \xrightarrow[n\to\infty]{\textrm{a.s.}} \expec X_1.$

Finally, we turn to the proof of the full SLLN for an i.i.d.\ sequence (Theorem~\ref{thm-slln}), assuming only a finite first moment. Etemadi's 1981 proof presented in [Dur2010] is based on a similar idea of proving convergence first for a subsequence, and introduces another useful technique, that of truncation.

Proof

[Proof of Theorem~\ref{thm-slln}]
First, observe that it is enough to prove the theorem in the case where $$X_1\ge 0$$, since in the general case we may decompose each $$X_n$$ into its positive and negative parts, and this gives two i.i.d.\ sequences of nonnegative r.v.'s. The validity of the SLLN for each of the two sequences implies the result for their difference.

Second, for each $$n\ge 1$$ let $$Y_n = X_n \ind_{\{X_n\le n\}}$$ ($$X_n$$ truncated at $$n$$''), and denote $$T_n=Y_1+Y_2+\ldots +Y_n$$.
Since $$\expec X_1<\infty$$, by a homework exercise we know that $$\sum_{n=1}^\infty \prob(X_n> n)<\infty$$,

which by the Borel-Cantelli lemma implies that $$\prob(X_n>n\textrm{ i.o.}) = \prob(X_n\neq Y_n\textrm{ i.o.}) = 0$$. It follows that the event

$\left\{ \sup_{n\to\infty} |S_n-T_n|=\infty \right\}$

has probability 0, and therefore the even smaller event

$\left\{ \limsup_{n\to\infty} \left|\frac{S_n}{n}-\frac{T_n}{n} \right|>0 \right\}= \left\{ \limsup_{n\to\infty} \left|\left(\frac{S_n}{n}-\mu\right)-\frac{T_n-\expec(T_n)}{n} \right|>0 \right\},$

has probability 0 (the equivalence between these last two events follows from the observation that $$\expec(Y_n)\uparrow \mu$$ by the monotone/dominated convergence theorems, and therefore also $$\expec(T_n)/n\to\mu$$ as $$n\to\infty$$ since $$(n^{-1} \expec(T_n))_{n=1}^\infty$$ is the sequence of arithmetic averages of the $$\expec(Y_n)$$'s).

So we have shown that to prove the theorem, it is enough to prove that

$\prob\left(\frac{T_n-\expec T_n}{n}\to0\right) = 1.$

Now, to prove this, as before we establish a.s.\ convergence first along a subsequence $$n_k$$, this time taking $$n_k = \lfloor \alpha^k \rfloor$$ where $$\alpha>1$$. (Here, $$\lfloor x\rfloor$$ denotes the floor'' function, namely the largest integer $$\le x$$). This is done by again combining the Borel-Cantelli lemma with a Chebyshev variance bound, except that showing that the sum of the bounds gives a convergent series requires more work than before. We have

\begin{eqnarray*}
\prob\left(\left|\frac{T_{n_k}}{n_k} - \frac{\expec(T_{n_k})}{n_k}\right| > \epsilon\right)
&\le&
\frac{\var(T_{n_k})}{\epsilon^2 n_k^2}
= \frac{1}{\epsilon^2 n_k^2} \sum_{m=1}^{n_k} \var(Y_m).
\end{eqnarray*}

Therefore

\begin{eqnarray}
\sum_{k=1}^\infty
\prob\left(\left|\frac{T_{n_k}}{n_k} - \frac{\expec(T_{n_k})}{n_k}\right| > \epsilon\right)
&\le&
\sum_{k=1}^\infty  \frac{1}{\epsilon^2 n_k^2} \sum_{m=1}^{n_k} \var(Y_m)
= \frac{1}{\epsilon^2} \sum_{m=1}^\infty \var(Y_m) \sum_{n_k\ge m} \frac{1}{n_k^2}
\nonumber \\
&\le&
\frac{1}{\epsilon^2} \sum_{m=1}^\infty \var(Y_m) \frac{1}{m^2}(1+\alpha^{-2}+\alpha^{-4}+\alpha^{-6}+\ldots)
\nonumber \\
&=&
\frac{1}{\epsilon^2(1-\alpha^{-2})} \sum_{m=1}^\infty  \frac{\var(Y_m)}{m^2}
\label{eq:infinitesum}
\end{eqnarray}

Lemma

We have

$\sum_{m=1}^\infty \frac{\var(Y_m)}{m^2} \le 4 \expec X_1 < \infty.$

Proof

Using the formula $$\expec(Z)=\int_0^\infty \prob(Z>x)\,dx$$ that's valid for any nonnegative r.v.\ $$Z$$, we have that
\begin{eqnarray*}
\sum_{m=1}^\infty \frac{\var(Y_m)}{m^2} &\le& \sum_{m=1}^\infty \frac{1}{m^2} \expec(Y_m^2)
=\sum_{m=1}^\infty \frac{1}{m^2}\expec\left(X_m^2 \ind_{\{X_m\le m\}}\right)
\\ &=& \sum_{m=1}^\infty \frac{1}{m^2} \int_0^\infty \prob\left(X_m^2 \ind_{\{X_m\le m\}}>t\right)dt
\\ &=& \sum_{m=1}^\infty \frac{1}{m^2} \int_0^\infty \prob\left(\sqrt{t} \le X_m \le m \right)dt
\\ &=& \sum_{m=1}^\infty \frac{1}{m^2} \int_0^{m^2} \prob\left(\sqrt{t} \le X_1 \le m \right)dt
\\ &\le& \sum_{m=1}^\infty \frac{1}{m^2} \int_0^{m^2} \prob\left(X_1 \ge \sqrt{t} \right)dt
\\ &=& \int_0^\infty \left(\sum_{m\ge \sqrt{t}} \frac{1}{m^2}\right) \prob\left(X_1\ge \sqrt{t}\right)dt
\\ &\le& \int_0^\infty \frac{2}{\sqrt{t}} \prob\left(X_1\ge \sqrt{t}\right)dt = 4 \int_0^\infty \prob(X_1\ge u)du = 4\expec(X_1).
\qedhere
\end{eqnarray*}
\end{proof}

It follows that the infinite sum on the left-hand side of \eqref{eq:infinitesum} converges, and therefore by the Borel-Cantelli lemma, we have that

$\prob\left(\left|\frac{T_{n_k}}{n_k} - \frac{\expec(T_{n_k})}{n_k}\right| > \epsilon\textrm{ infinitely often (w.r.t.\ $$k$$)}\right) =0.$

Since this is true for all $$\epsilon>0$$, using the standard trick we get that

$\prob\left(\frac{T_{n_k}}{n_k} - \frac{\expec(T_{n_k})}{n_k}\xrightarrow[k\to\infty]{}0\right) = \prob\left(\frac{T_{n_k}}{n_k} \xrightarrow[k\to\infty]{}\mu\right) = 1.$

The last step is now to show that convergence along this subsequence forces $$(T_n-\expec T_n)/n$$ to behave well between successive $$n_k$$'s. Observe that if $$n_k \le n < n_{k+1}$$ then

$\frac{T_{n_k}}{n_{k+1}} \le \frac{T_n}{n} \le \frac{T_{n_{k+1}}}{n_k}.$

Since $$n_k=\lfloor \alpha^k\rfloor$$, in the limit this gives that almost surely the bounds

$\frac{1}{\alpha} \mu \le \liminf_{n\to\infty} \frac{T_n}{n} \le \limsup_{n\to\infty} \frac{T_n}{n} \le \alpha \mu$

hold. Since $$\alpha>1$$ was arbitrary, the intersection of these events for $$\alpha=1+1/d$$, $$d=1,2,\ldots$$, implies that

$\prob\left( \frac{T_n}{n} \to\mu \right) = 1,$

which finishes the proof of Theorem~\ref{thm-slln}.