Lecture 19 — Truncation Methods and Infinite-Mean Laws of Large Numbers
Truncation arguments
- Weak Law of Large Numbers with infinite mean
- The St. Petersburg paradox
- Choosing normalizing sequences $b_n$ so that $\frac{S_b}{b_n}\to 1$ in probability for heavy-tailed distributions.
1. General Truncation Inequality
Let ${X_{n,k}}_{1 \le k \le n}$ be independent for each $n$, and define truncated variables
\[\bar X_{n,k} = X_{n,k} \mathbf{1}\{\vert X_{n,k}\vert \le b_n\}.\]Let
\[S_n = \sum_{k=1}^n X_{n,k}, \qquad a_n = \sum_{k=1}^n \mathbb{E}(\bar X_{n,k}).\]A standard inequality (via Markov + variance control on truncated part) gives:
\[P\!\left( \frac{S_n - a_n}{b_n} > \varepsilon \right) \le \underbrace{\sum_{k=1}^n P(\vert X_{n,k}\vert > b_n)}_{(I)} + \underbrace{\frac{1}{\varepsilon^2 b_n^2}\sum_{k=1}^n \mathbb{E}(X_{n,k}^2 \mathbf{1}\{\vert X_{n,k}\vert \le b_n\})}_{(II)}.\]Conclusion.
If
- (I) → 0, and
- (II) → 0 as $n\to\infty$,
then
\[\frac{S_n - a_n}{b_n} \xrightarrow{P} 0.\]2. Specialization to i.i.d. Variables
If ${X_k}$ are iid with partial sums $S_n = \sum_{k=1}^n X_k$ then:
\[(I)=n\,P(\vert X\vert > b_n), \qquad (II)=\frac{n\,\mathbb{E}(X^2 \mathbf{1}\{\vert X\vert \le b_n\})}{\varepsilon^2 b_n^2},\]and
\[a_n = n \,\mathbb{E}(X \mathbf{1}\{\vert X\vert \le b_n\}).\]If both vanish, then
\[\frac{S_n - a_n}{b_n} \xrightarrow{P} 0.\]The problem becomes choosing a normalizing sequence $b_n$ that makes both terms go to zero.
3. The St. Petersburg Paradox
Durrett Example 2.2.7.
\[P(X = 2^k) = 2^{-k},\quad k\ge 1.\]Then
\[E[X] = \sum_{k=1}^\infty 2^k 2^{-k} = \infty.\]Since the mean is infinite, the usual LLN does not apply.
In fact,
How to prove it? Truncation.
Fix $M > 0$, and define truncated variables:
\[X^{(M)} = X \wedge M.\]Then:
- $X^{(M)}$ are iid,
- $\mathbb{E}[X^{(M)}] < M < \infty$,
- by the standard WLLN:
Since $S_n \ge \sum_{k=1}^n X_k^{(M)}$,
\[P\!\left( \frac{S_n}{n} > \mathbb{E}[X^{(M)}] - \varepsilon \right) \to 1.\]Now let $M\to\infty$.
By MCT:
Thus for each fixed $M$, with probability → 1, the sample average exceeds a constant that can be made arbitrarily large. Hence,
\[\frac{S_n}{n} \xrightarrow{P} \infty.\]4. A More Refined Normalization:
Claim
\(\frac{S_n}{n\log_2 n} \xrightarrow{P} 1.\)
Your notes derive this using the truncation conditions (I), (II).
Let
\[b_n = n\log_2 n.\]Compute the truncated mean:
\[a_n = n\, \mathbb{E}(X\,;\,X \le n\log n) = n \sum_{2^k \le n\log n} 2^k 2^{-k} = n (\log n + \log\log n).\]Then $a_n \sim n\log n = b_n$.
So the normalization is correct.
Check (I)
\[nP(X > b_n) = n \sum_{2^k > n\log n} 2^{-k} \le \frac{2n}{n\log n} = \frac{2}{\log n}\to 0.\]Check (II)
\[\frac{n\, \mathbb{E}(X^2; X\le b_n)}{b_n^2} \approx \frac{n\cdot 2b_n}{b_n^2} = \frac{2}{\log n} \to 0.\]Thus both truncation conditions vanish, implying:
\[\frac{S_n}{b_n} = \frac{S_n}{n \log_2 n} \xrightarrow{P} 1.\]This is a Weak Law of Large Numbers with infinite mean, but with a different normalizing sequence.
5. General Criterion Using $\mu(s)$
Define the truncated first moment:
\[\mu(s)= E[X\,; X\le s].\]- $\mu(s)$ is increasing.
- $\lim_{s\to\infty} \mu(s) = \infty$ iff $E[X]=\infty$.
A useful fact derived in the notes:
\[\frac{\mu(s)}{s} = \frac{E[X/s\,;\,X\le s]}{1} \longrightarrow 0.\]This uses the DCT because $X/s \le 1$ on ${X\le s}$ and $X/s\to 0$.
The normalizing sequence $b_n$
Define $b_n$ by
\[\frac{\mu(b_n)}{b_n} = \frac{1}{n}.\]This guarantees that the truncation method will succeed and often that:
\[\frac{S_n}{b_n} \xrightarrow{P} 1.\]Jump points
If $P(X=S_0)>0$, there may be jumps in $\mu$. The notes mention this as a warning: the minimizing $b_n$ may occur at the point mass.
6. A Feller Example
The notes conclude with another heavy-tailed example (attributed to William Feller):
\(P(X = 2^k - 1) = \frac{1}{2^k k(k+1)},\qquad k\ge 1,\) \(P(X=-1) = 1 - P(X\ge 1).\)
Using the telescoping identity
\[\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1},\]one can show $E[X]=0$ but the variance and higher moments diverge.
A similar truncation calculation yields:
The message: “Truncation leads to the right normalization.”
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