Lecture 24 — Formal SLLN Proof via Truncation and Block Arguments
This lecture continues the formal proof of the Strong Law of Large Numbers using truncation, “good” sequences, and block decomposition (the Chung–Erdős / Chandrasekharan–style argument your professor sketches).
We want to prove the SLLN in the general case:
Theorem (SLLN).
If ${X_k}_{k\ge 1}$ are iid with
\(E\vert X\vert < \infty, \qquad E[X]=\mu,\)
then
\(\frac{S_n}{n} \xrightarrow{a.s.} \mu, \qquad S_n=\sum_{k=1}^n X_k.\)
This lecture gives a classical, more delicate proof (following Durrett), using truncation and carefully selected subsequences $T_{2^n}$.
1. Reduction and Notation
WLOG, assume $X\ge 0$ a.s.
(General case: apply to $X^+$ and $X^-$ separately.)
Define:
\[X_k = X_k^+ - X_k^-, \qquad X_k^+ = \max(X_k,0), \qquad E[X^+] < \infty.\]Let
\[\delta_n^+ = \sum_{k=1}^n X_k^+, \qquad \delta_n^- = \sum_{k=1}^n X_k^-, \qquad \delta_n = \delta_n^+ - \delta_n^-.\]Then,
\[\frac{\delta_n}{n} = \frac{\delta_n^+}{n} - \frac{\delta_n^-}{n}.\]If we can show both terms converge almost surely, we are done.
2. Step 1 — Truncation Removes Only Finitely Many Terms
Define truncated variables:
\[Y_k = X_k\,\mathbf{1}_{\{\vert X_k\vert \le k\}}.\]Then:
\[P(X_k \ne Y_k) = P(\vert X_k\vert > k) = P(\vert X\vert > k).\]Since $E\vert X\vert = \int_0^\infty P(\vert X\vert >t)\,dt < \infty$,
\[\sum_{k=1}^\infty P(\vert X\vert >k) < \infty.\]Thus by Borel–Cantelli I,
\[P(X_k \ne Y_k \text{ i.o.}) = 0.\]Hence
\[X_k = Y_k \quad\text{for all large } k,\ a.s.\]So it suffices to prove the SLLN for the truncated variables $Y_k$.
3. Interlude — “Good” Sequences ${a_n}$
Your notes (bottom of page 1) define:
A sequence $a_n \uparrow \infty$ is good if
\(\sum_{n=m}^\infty a_n^{-2} \le C\, a_m^{-2},\qquad m\ge1.\)
Examples:
- $a_n = n^p$ for any $0<p<2$,
- more generally: if $a_n / n^{1/p}$ is non-decreasing, then $a_n$ is good.
For this proof, we will use the good sequence:
\[a_n = n.\]4. Step 2 — Second-Moment Control
Let
\[Y_n = X\,\mathbf{1}_{\{\vert X\vert \le a_n\}},\qquad a_n = n.\]Assume as above:
\[\sum_{n=1}^\infty P(\vert X\vert >a_n) < \infty.\]Then we want to show:
\[\sum_{n=1}^\infty \frac{E[Y_n^2]}{a_n^2} < \infty.\]Proof sketch (page 2):
Decompose according to the annuli $a_{m-1} < \vert X\vert \le a_m$:
\[\sum_{n=1}^\infty \frac{E[Y_n^2]}{a_n^2} = \sum_{n=1}^\infty a_n^{-2} \sum_{m=1}^n E(X^2; a_{m-1}<\vert X\vert \le a_m).\]Swap summation order:
\[\sum_{m=1}^\infty E(X^2; a_{m-1}<\vert X\vert \le a_m) \sum_{n=m}^\infty a_n^{-2}.\]Since ${a_n}$ is good,
\[\sum_{n=m}^\infty a_n^{-2} \le C a_m^{-2}.\]Thus:
\[\sum_{n=1}^\infty \frac{E[Y_n^2]}{a_n^2} \le C \sum_{m=1}^\infty E\!\left(\frac{X^2}{a_m^2}; a_{m-1}<\vert X\vert \le a_m\right) \le C \sum_{m=1}^\infty m\, P(\vert X\vert >a_m) <\infty.\](The last inequality uses the summability assumption.)
5. Step 3 — Block Averages
Define partial sums of truncated variables:
\[T_n = \sum_{k=1}^n Y_k.\]We will show for a geometric subsequence $n=\alpha^m$:
\[\frac{T_{\alpha^m} - E[T_{\alpha^m}]}{\alpha^m} \xrightarrow{a.s.} 0.\]Your notes use $\alpha=2$.
Write $2^m$ as the block endpoint and define the block deviations:
Then:
\[\frac{T_{2^{m+1}} - E[T_{2^{m+1}}]}{2^{m+1}} = \sum_{j=1}^{m} U_j.\]So it suffices to show:
\[\sum_{j=1}^\infty E[U_j^2] < \infty.\]Compute $E[U_m^2]$
From page 2 of your notes:
\[E[U_m^2] \le 4 \sum_{k=2^m+1}^{2^{m+1}} \frac{\operatorname{Var}(Y_k)}{k^2} \le 4 \sum_{k=1}^\infty \frac{E[Y_k^2]}{k^2}.\]But Step 2 showed the latter sum is finite.
Thus $\sum_m E[U_m^2]<\infty$.
By the Kolmogorov convergence criterion (or BC I applied to $U_m^2$):
\[U_m \to 0 \quad a.s.\]Therefore:
\[\frac{T_{2^m} - E[T_{2^m}]}{2^m} \xrightarrow{a.s.} 0.\]6. Step 4 — Identify the Limit
We already have:
\(E[Y_k] = E(X; \vert X\vert \le k) \xrightarrow{k\to\infty} E[X] = \mu\) by Dominated Convergence.
Thus:
\[\frac{E[T_{2^m}]}{2^m} = \frac{1}{2^m} \sum_{k=1}^{2^m} E[Y_k] \xrightarrow{m\to\infty} \mu.\]Combine with the previous step:
\[\frac{T_{2^m}}{2^m} = \frac{E[T_{2^m}]}{2^m} + \frac{T_{2^m}-E[T_{2^m}]}{2^m} \xrightarrow{a.s.} \mu.\]7. Step 5 — Extend From the Subsequence to All $n$
For $2^m \le n \le 2^{m+1}$,
\[\frac{T_{2^m}}{2^{m+1}} \le \frac{T_n}{n} \le \frac{T_{2^{m+1}}}{2^m}.\]Since both endpoints converge almost surely to $\mu$, the middle term must as well.
Thus:
\[\boxed{ \frac{T_n}{n} \xrightarrow{a.s.} \mu. }\]Finally, since $X_k=Y_k$ eventually almost surely,
\[\frac{S_n}{n} = \frac{T_n}{n} + o(1) \xrightarrow{a.s.} \mu.\]This completes the proof of the SLLN.
Cheat-Sheet Summary — Lecture 24
-
Use truncation $Y_k = X_k \mathbf{1}_{{\vert X_k\vert \le k}}$.
BC I shows truncation changes only finitely many terms. -
Introduce good sequences ${a_n}$ to guarantee
\(\sum E[Y_n^2]/a_n^2 < \infty.\) -
For the geometric subsequence $n=2^m$, define block deviations
\(U_m = \sum_{k=2^m+1}^{2^{m+1}} (Y_k - E[Y_k]) / 2^m.\) -
Show $\sum E[U_m^2] < \infty$ so $U_m\to 0$ a.s.
-
Conclude
\(\frac{T_{2^m}}{2^m} \to \mu.\) -
Squeeze all intermediate $n$, giving
\(\frac{S_n}{n} \to \mu \quad \text{a.s.}\)
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