Truncation Estimates for Integrable IID Variables

Why Truncation Is Useful

The strong law for IID variables with finite variance applies directly when the common distribution has a finite second moment. For an integrable variable, that moment may be infinite, so the variance argument cannot be applied to the original sequence. Truncation replaces each variable by a bounded one, while two summability estimates control the changes: large values occur only finitely often almost surely, and the variances of the truncated variables, with the appropriate weights, have a finite sum.

Throughout, $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, so $\mathbb{P}(\Omega)=1$. Let $X_1,X_2,\ldots$ be real-valued IID random variables, meaning they are independent and have the same distribution. Assume $\mathbb{E}|X_1|<\infty$. No finite second moment is assumed. For each integer $n\ge1$, define the truncation at level $n$ by

Z_n=X_n\chi_{\{|X_n|\le n\}}.

The retained event uses $\le n$; the discarded tail event uses the strict inequality $|X_n|>n$. In particular, a value with $|X_n|=n$ is retained. Each $Z_n$ is measurable and bounded in absolute value by $n$, so it has a finite variance.

The Tail-Probability Estimate

Proposition: Under the assumptions above, $$ \sum_{n=1}^{\infty}\mathbb{P}(|X_n|>n) =\sum_{n=1}^{\infty}\mathbb{P}(|X_1|>n) \le \mathbb{E}|X_1|<\infty. $$

Proof. The equality follows because all $X_n$ have the same distribution. Set $Y=|X_1|$. For each fixed value $y\ge0$, the number of positive integers $n$ satisfying $n<y$ is at most $y$. This uses the strict tail condition: if $y$ is an integer, the index $n=y$ is not counted. Thus, pointwise, $\sum_{n=1}^{\infty}\chi_{\{Y>n\}}\le Y$. Apply Tonelli's theorem to the nonnegative measurable functions $\chi_{\{Y>n\}}$: the sum of their expectations equals the expectation of their sum, and so is at most $\mathbb{E}Y$. This proves the inequality and finiteness. $\square$

Summability also has an almost-sure consequence. Let $A_n=\{|X_n|>n\}$. For every $N\ge1$, countable subadditivity gives $\mathbb{P}(\bigcup_{n\ge N}A_n)\le\sum_{n\ge N}\mathbb{P}(A_n)$. The right side tends to zero as $N$ tends to infinity because the series converges. Since $\{A_n\text{ infinitely often}\}=\bigcap_{N\ge1}\bigcup_{n\ge N}A_n$, continuity from above of the probability measure gives $\mathbb{P}(A_n\text{ infinitely often})=0$. Consequently $X_n=Z_n$ for all but finitely many $n$, almost surely. This conclusion concerns the original and truncated variables at their corresponding indices.

Worked Example: A Discrete Distribution and Its Truncations

Suppose $X_n$ have the common distribution $\mathbb{P}(X_n=2)=1/4$, $\mathbb{P}(X_n=-1)=1/4$, and $\mathbb{P}(X_n=0)=1/2$. Its first absolute moment is $\mathbb{E}|X_1|=2(1/4)+1(1/4)=3/4$. For $n=1$, the strict event $|X_1|>1$ occurs only at $X_1=2$, so its probability is $1/4$. For $n\ge2$, all three possible values satisfy $|X_1|\le n$, so the tail probability is zero. Therefore $\sum_n\mathbb{P}(|X_n|>n)=1/4\le3/4$.

At level $n=1$, $Z_1=-1$ with probability $1/4$ and is zero otherwise. Hence $\mathbb{E}Z_1=-1/4$, $\mathbb{E}Z_1^2=1/4$, and $\operatorname{Var}(Z_1)=1/4-(-1/4)^2=3/16$. For every $n\ge2$, $Z_n$ has the same distribution as $X_1$: its mean is $2(1/4)-1(1/4)=1/4$, its second moment is $4(1/4)+1(1/4)=5/4$, and its variance is $5/4-(1/4)^2=19/16$. Thus

\sum_{n=1}^{\infty}\frac{\operatorname{Var}(Z_n)}{n^2} =\frac{3}{16}+\frac{19}{16}\sum_{n=2}^{\infty}\frac1{n^2} \le \frac{3}{16}+\frac{19}{16}=\frac{11}{8}.

For the inequality, since $x^{-2}$ is decreasing, $1/n^2\le\int_{n-1}^{n}x^{-2}\,dx$ for each $n\ge2$; summing gives $\sum_{n=2}^{\infty}1/n^2\le\int_1^\infty x^{-2}\,dx=1$. This also verifies directly that the displayed variance series is finite.

The Variance-Summability Estimate

Proposition: For $Z_n=X_n\chi_{\{|X_n|\le n\}}$ as above, $$ \sum_{n=1}^{\infty}\frac{\operatorname{Var}(Z_n)}{n^2} \le 2\mathbb{E}|X_1|<\infty. $$

Proof. Since $Z_n$ is square-integrable, $\operatorname{Var}(Z_n)=\mathbb{E}[Z_n^2]-(\mathbb{E}Z_n)^2\le\mathbb{E}[Z_n^2]$. The variables have the same underlying distribution before truncation, so Tonelli's theorem for nonnegative functions gives

\sum_{n=1}^{\infty}\frac{\mathbb{E}[Z_n^2]}{n^2} =\mathbb{E}\left[Y^2\sum_{n=1}^{\infty}\frac{\chi_{\{Y\le n\}}}{n^2}\right], \qquad Y=|X_1|.

It remains to bound the expression inside the expectation for every $y\ge0$. If $0\le y\le1$, then $\chi_{\{y\le n\}}=1$ for every $n\ge1$. The decreasing-function comparison gives $\sum_{n=1}^{\infty}n^{-2}\le1+\int_1^\infty x^{-2}\,dx=2$. Therefore $y^2\sum_n\chi_{\{y\le n\}}/n^2\le2y^2\le2y$.

If $y>1$, let $k=\lceil y\rceil$, the least integer greater than or equal to $y$. The condition $y\le n$ is then equivalent to $n\ge k$. Another decreasing-function comparison yields

\sum_{n=k}^{\infty}\frac1{n^2} \le\frac1{k^2}+\int_k^\infty\frac{dx}{x^2} =\frac1{k^2}+\frac1k \le\frac2k \le\frac2y.

Here $1/k^2\le1/k$ because $k\ge1$, and $1/k\le1/y$ because $k\ge y$. Multiplying by $y^2$ gives a bound of $2y$. The same bound holds in both cases, including $y=0$. Taking expectations in the Tonelli identity therefore gives $\sum_n\mathbb{E}[Z_n^2]/n^2\le2\mathbb{E}Y$. The variance inequality proves the proposition. $\square$

Worked Example: Finite First Moment but Infinite Variance

Let the common distribution of the IID variables have density $f(x)=2x^{-3}$ for $x\ge1$, and density zero elsewhere, with respect to Lebesgue measure on $\mathbb{R}$. It is a probability density because $\int_1^\infty2x^{-3}\,dx=1$. Its first moment is $\mathbb{E}|X_1|=\int_1^\infty2x^{-2}\,dx=2$, but its second moment is $\int_1^\infty2x^{-1}\,dx=\infty$.

For every integer $n\ge1$, the strict tail probability is $\mathbb{P}(|X_1|>n)=\int_n^\infty2x^{-3}\,dx=1/n^2$. Thus the tail-probability proposition bounds their sum by $2$. For $n=1$, $Z_n=0$ almost surely, since the distribution has no mass at the endpoint $x=1$. For $n\ge2$,

\mathbb{E}[Z_n^2]=\int_1^n x^2(2x^{-3})\,dx=2\ln n.

Consequently the weighted second-moment series is $\sum_{n=2}^{\infty}2\ln(n)/n^2$, and it is finite by the variance proposition's proof, which bounds the full series by $2\mathbb{E}|X_1|=4$. Since $\operatorname{Var}(Z_n)\le\mathbb{E}[Z_n^2]$, the weighted variance series is finite as well. This example verifies that the estimate does not require a finite second moment of $X_1$.

How These Bounds Enter the Strong-Law Argument

The truncated variables are independent: each $Z_n$ is a measurable function of the corresponding independent variable $X_n$. They are generally not identically distributed, because the truncation level changes with $n$. Their centered versions $Z_n-\mathbb{E}Z_n$ are independent, have mean zero, and have the same variances as $Z_n$. The variance proposition supplies a summable bound on those variances after division by $n^2$, the form used with the earlier material on Kolmogorov's maximal inequality.

The truncation does not permanently alter the average expectation. For a real-valued $X_1$, $X_1\chi_{\{|X_1|\le n\}}\to X_1$ pointwise, and $|X_1\chi_{\{|X_1|\le n\}}|\le|X_1|$. Since $|X_1|$ is integrable, the Dominated Convergence Theorem applies and gives $\mathbb{E}[X_1\chi_{\{|X_1|\le n\}}]\to\mathbb{E}X_1$. Identical distributions give the corresponding expectation for each $Z_n$. Together with the almost-sure finite number of discarded terms, these estimates are ingredients in extending the strong law beyond finite variance; they are not, by themselves, the full strong law.

Common Pitfall: It is not valid to bound $\mathbb{E}[X_1^2\chi_{\{|X_1|\le n\}}]$ by $\mathbb{E}X_1^2$ and then conclude that its weighted series is finite: under the present assumptions $\mathbb{E}X_1^2$ may be infinite. The valid estimate sums the truncated second moments first and uses the pointwise bound $y^2\sum_{n:y\le n}n^{-2}\le2y$, which is integrable by the finite-first-moment assumption.

Worked Example: Why the First-Moment Assumption Matters

Consider the density $f(x)=x^{-2}$ for $x\ge1$, zero elsewhere. Its integral is $\int_1^\infty x^{-2}\,dx=1$, but its first moment is $\int_1^\infty x^{-1}\,dx=\infty$. For its IID sequence, $\mathbb{P}(|X_n|>n)=\int_n^\infty x^{-2}\,dx=1/n$. The tail-probability series diverges: for each integer $j\ge0$, the block $2^j\le n<2^{j+1}$ contains $2^j$ terms, each at least $1/2^{j+1}$, so the block sum is at least $1/2$. Infinitely many such blocks force divergence. Thus finite first moment is not a dispensable condition for the tail-summability estimate.

Check Your Understanding

Use the strict tail event $|X_n|>n$ and the inclusive truncation event $|X_n|\le n$ when working through these questions.

For a fixed $y\ge0$, explain why the number of positive integers $n$ with $n<y$ is at most $y$, including when $y$ is an integer.
In the variance estimate, why does the case $0\le y\le1$ use the full series $\sum_{n\ge1}n^{-2}$?
For the discrete distribution in the first worked example, recompute $\operatorname{Var}(Z_1)$ from its two possible values.
Give the Dominated Convergence hypotheses that justify $\mathbb{E}Z_n\to\mathbb{E}X_1$.
For the density $x^{-2}$ on $[1,\infty)$, verify the lower bound on each dyadic block of the tail-probability series.