2024.Q2 – Heavy-Tailed Sums, Truncation, and Almost Sure Convergence
2024 Probability Prelim Exam (PDF)
Problem Statement (verbatim) :contentReference[oaicite:0]{index=0}
Let $X, X_1, X_2, \ldots$ be i.i.d. sequence of positive random variables. Let $0 < \beta < 1$. Assume
\(P(X > x) \le x^{-\beta}, \quad x > 1.\)
Let ${a_n}_{n=1,2,\dots}$ be a sequence of positive real numbers that satisfies
\(\sum_{n=1}^\infty a_n^\beta < \infty.\)
Prove the following:
(a) $\displaystyle \sum_{n=1}^\infty P(a_n X > 1) < \infty.$
(b)(i) $\displaystyle \sum_{n=1}^\infty E\big(a_n X \cdot \mathbf{1}_{{a_n X < 1}}\big) < \infty.$
(b)(ii) $\displaystyle \sum_{n=1}^\infty E\big(a_n^2 X^2 \cdot \mathbf{1}_{{a_n X < 1}}\big) < \infty.$
(c)
(i) $\displaystyle \sum_{n=1}^\infty a_n X_n < \infty,$ a.s.
(ii) Assume also that ${a_n}_{n=1,2,\dots}$ is non-decreasing. Prove:
\(a_n \cdot \sum_{k=1}^n X_k \longrightarrow 0, \quad \text{a.s.}\)
Part (a)
Claim.
\(\sum_{n=1}^\infty P(a_n X > 1) < \infty.\)
Proof.
For each $n$, \(P(a_n X > 1) = P\!\left(X > \frac{1}{a_n}\right).\)
We consider two cases.
-
Case 1: $a_n < 1$. Then $\frac{1}{a_n} > 1$, so we can use the tail bound: \(P\!\left(X > \frac{1}{a_n}\right) \le \left(\frac{1}{a_n}\right)^{-\beta} = a_n^\beta.\)
-
Case 2: $a_n \ge 1$. Then $\frac{1}{a_n} \le 1$, so we just use the trivial bound \(P\!\left(X > \frac{1}{a_n}\right) \le 1 \le a_n^\beta\) because $a_n^\beta \ge 1$ when $a_n \ge 1$.
In both cases, \(P(a_n X > 1) \le a_n^\beta.\) Therefore \(\sum_{n=1}^\infty P(a_n X > 1) \le \sum_{n=1}^\infty a_n^\beta < \infty.\)
Conclusion.
The series $\sum_n P(a_n X > 1)$ converges, controlled by the given summability of $\sum a_n^\beta$.
Key Takeaways
- It is often useful to manipulate inside the probability first, and then apply given tail bounds.
- Splitting into cases $a_n < 1$ and $a_n \ge 1$ lets you combine the nontrivial tail bound with the trivial inequality $P(\cdot)\le 1$.
- The condition $\sum a_n^\beta < \infty$ not only gives summability, it also implies $a_n \to 0$.
Part (b)(i)
Claim.
\(\sum_{n=1}^\infty E\big[a_n X \,\mathbf{1}_{\{a_n X < 1\}}\big] < \infty.\)
Proof.
Let $Y_n = a_n X$. Then \(E\big[Y_n \mathbf{1}_{\{Y_n<1\}}\big] = \int_0^1 P(Y_n > t)\,dt = \int_0^1 P\!\left(X > \frac{t}{a_n}\right) dt.\)
We split the integral at $t = a_n$:
-
Region $0 \le t \le a_n$:
Here $\frac{t}{a_n} \le 1$, so we only know \(P\!\left(X > \frac{t}{a_n}\right) \le 1.\) Hence \(\int_0^{a_n} P\!\left(X > \frac{t}{a_n}\right) dt \le \int_0^{a_n} 1\,dt = a_n.\) -
Region $a_n \le t \le 1$:
Here $\frac{t}{a_n} \ge 1$, so we can use the tail bound: \(P\!\left(X > \frac{t}{a_n}\right) \le \left(\frac{t}{a_n}\right)^{-\beta} = a_n^\beta t^{-\beta}.\) Thus \(\int_{a_n}^1 P\!\left(X > \frac{t}{a_n}\right) dt \le a_n^\beta \int_{a_n}^1 t^{-\beta} dt.\) For $0<\beta<1$, \(\int_{a_n}^1 t^{-\beta} dt = \frac{1 - a_n^{1-\beta}}{1-\beta} \le \frac{1}{1-\beta}.\) Hence \(\int_{a_n}^1 P\!\left(X > \frac{t}{a_n}\right) dt \le \frac{a_n^\beta}{1-\beta}.\)
Combining the two regions, \(E[a_n X \mathbf{1}_{\{a_nX<1\}}] \le a_n + \frac{a_n^\beta}{1-\beta}.\)
Since $\sum a_n^\beta < \infty$, we get $a_n \to 0$. So for large $n$, $a_n \le a_n^\beta$, which implies \(\sum_{n=1}^\infty a_n < \infty \quad\text{and}\quad \sum_{n=1}^\infty a_n^\beta < \infty.\) Therefore $$ \sum_{n=1}^\infty E[a_n X \mathbf{1}{{a_nX<1}}] \le \sum{n=1}^\infty a_n
- \frac{1}{1-\beta} \sum_{n=1}^\infty a_n^\beta < \infty. $$
Conclusion.
The series of truncated expectations $\sum_n E[a_nX\,1_{{a_nX<1}}]$ converges.
Key Takeaways
- Use the tail-integral formula
\(E[Y1_{\{Y<1\}}] = \int_0^1 P(Y>t)\,dt\) to turn expectations into integrals of tail probabilities. - Split the integral at the point where the given tail bound becomes valid.
- The given summability $\sum a_n^\beta$ can be used twice: once directly, and once to deduce $a_n \to 0$, which helps to control $\sum a_n$.
Part (b)(ii)
Claim.
\(\sum_{n=1}^\infty E\big[a_n^2 X^2 \,\mathbf{1}_{\{a_n X < 1\}}\big] < \infty.\)
Proof.
Let $Z_n = a_n^2 X^2$. By the same tail-integral idea, \(E[Z_n \mathbf{1}_{\{Z_n<1\}}] = \int_0^1 P(Z_n > t)\,dt = \int_0^1 P\!\left(X > \frac{\sqrt{t}}{a_n}\right) dt.\)
Again, split at $t=a_n^2$:
-
Region $0 \le t \le a_n^2$:
Here $\frac{\sqrt{t}}{a_n} \le 1$, so \(P\!\left(X > \frac{\sqrt{t}}{a_n}\right) \le 1 \quad\Rightarrow\quad \int_0^{a_n^2} P\!\left(X > \frac{\sqrt{t}}{a_n}\right) dt \le a_n^2.\) -
Region $a_n^2 \le t \le 1$:
Here $\frac{\sqrt{t}}{a_n} \ge 1$, so \(P\!\left(X > \frac{\sqrt{t}}{a_n}\right) \le \left(\frac{\sqrt{t}}{a_n}\right)^{-\beta} = a_n^\beta t^{-\beta/2}.\) Thus \(\int_{a_n^2}^1 P\!\left(X > \frac{\sqrt{t}}{a_n}\right) dt \le a_n^\beta \int_{a_n^2}^1 t^{-\beta/2} dt.\) Since $0<\beta<2$, the integral is finite and bounded uniformly in $n$: \(\int_{a_n^2}^1 t^{-\beta/2} dt = \frac{1 - (a_n^2)^{1-\beta/2}}{1-\beta/2} \le \frac{1}{1-\beta/2}.\)
Therefore, \(E[a_n^2 X^2 \mathbf{1}_{\{a_nX<1\}}] \le a_n^2 + \frac{a_n^\beta}{1-\beta/2}.\)
As before, from $\sum a_n^\beta<\infty$ we get $a_n\to 0$, hence eventually $a_n^2 \le a_n^\beta$. So both $\sum a_n^2$ and $\sum a_n^\beta$ converge, and \(\sum_{n=1}^\infty E[a_n^2 X^2 \mathbf{1}_{\{a_nX<1\}}] < \infty.\)
Conclusion.
The series of truncated second moments $\sum_n E[a_n^2 X^2 1_{{a_nX<1}}]$ is finite.
Key Takeaways
- The same integral-splitting trick works for higher powers.
- The exponent $\beta < 2$ guarantees that $\int t^{-\beta/2}\,dt$ is finite near $t=0$.
- Heavy-tail bounds combined with shrinking weights can make even second moments summable.
Part (c)(i)
Claim.
\(\sum_{n=1}^\infty a_n X_n < \infty \quad \text{a.s.}\)
Proof.
We want to apply Kolmogorov’s Three-Series Theorem to the independent sequence ${a_n X_n}$. Define \(Y_n = a_n X_n.\)
The theorem says $\sum Y_n$ converges a.s. if and only if the following three series are all finite:
-
$\displaystyle \sum_{n=1}^\infty P( Y_n > 1) < \infty.$ -
$\displaystyle \sum_{n=1}^\infty E\big(Y_n \mathbf{1}_{{ Y_n \le 1}}\big) < \infty.$ -
$\displaystyle \sum_{n=1}^\infty E\big(Y_n^2 \mathbf{1}_{{ Y_n \le 1}}\big) < \infty.$
| Here $Y_n \ge 0$ (since $X_n\ge 0$), so $ | Y_n | = Y_n$, and $ | Y_n | \le1$ is the same as $Y_n<1$. |
- Condition (1) is precisely $\sum P(a_n X_n>1) < \infty$, which holds by part (a).
- Condition (2) is $\sum E[a_n X_n \mathbf{1}_{{a_nX_n\le 1}}] < \infty$, which we proved in part (b)(i).
- Condition (3) is $\sum E[a_n^2 X_n^2 \mathbf{1}_{{a_nX_n\le 1}}] < \infty$, which we proved in part (b)(ii).
Thus all three conditions hold. By Kolmogorov’s Three-Series Theorem, \(\sum_{n=1}^\infty a_n X_n \quad\text{converges almost surely.}\)
Conclusion.
The weighted series $\sum a_n X_n$ converges a.s. by verifying the three-series conditions using parts (a) and (b).
Key Takeaways
- Kolmogorov’s Three-Series Theorem is a standard tool for a.s. convergence of sums of independent (possibly heavy-tailed) random variables.
- Parts (a), (b)(i), and (b)(ii) were all aimed at setting up its three conditions.
- Truncation at level 1 is conventional in the theorem, but any fixed threshold would work up to constants.
Part (c)(ii)
Problem statement: “Assume also that ${a_n}_{n=1,2,\dots}$ is non-decreasing. Prove:
\(a_n \sum_{k=1}^n X_k \to 0 \quad \text{a.s.}\)”In the standard form of the lemma we will use, one actually assumes the weights are non-increasing and tend to $0$. This is how the official solution proceeds. So we implicitly use that condition (and you correctly noticed this issue in your reflection). :contentReference[oaicite:1]{index=1}
Claim.
Suppose ${a_n}$ is non-increasing with $a_n \to 0$, and $\sum_{n=1}^\infty a_n X_n$ converges a.s. Then \(a_n \sum_{k=1}^n X_k \longrightarrow 0 \quad \text{a.s.}\)
Proof. (Kronecker’s lemma)
Let $b_n = a_n$, and define \(S_n = \sum_{k=1}^n X_k.\)
We know from part (c)(i) that \(\sum_{n=1}^\infty b_n X_n = \sum_{n=1}^\infty a_n X_n\) converges a.s.
A standard version of Kronecker’s Lemma states:
If ${b_n}$ is a non-increasing sequence of positive numbers with $b_n \to 0$, and $\sum b_n x_n$ converges, then \(b_n \sum_{k=1}^n x_k \to 0.\)
Apply this lemma with $x_k = X_k$ and $b_n = a_n$. Since the weighted sum converges a.s. and $a_n \downarrow 0$, we obtain \(a_n S_n \to 0 \quad \text{a.s.}\)
Conclusion.
Under the usual monotone-decreasing assumption on the weights, convergence of $\sum a_n X_n$ implies that the scaled partial sums $a_n \sum_{k=1}^n X_k$ vanish a.s.
Key Takeaways
- Kronecker’s Lemma allows you to go from
“$\sum a_n X_n$ converges” to
“$a_n \sum_{k=1}^n X_k \to 0$”
when ${a_n}$ is decreasing and $a_n \to 0$. - In problems like this, if you see both:
- a convergent weighted sum $\sum a_n X_n$, and
- a question about scaled partial sums $a_n S_n$,
you should immediately think of Kronecker’s Lemma.
- It is important to check the monotonicity direction: Kronecker’s Lemma is usually stated for non-increasing weights that tend to zero.
Global Key Takeaways for Question 2
- Heavy-tail bounds like $P(X>x)\le x^{-\beta}$ plus a summability condition on weights ${a_n}$ are a classic setup for:
- controlling tail probabilities $\sum P(a_nX>1)$,
- controlling truncated expectations via tail integrals,
- applying Kolmogorov’s Three-Series Theorem.
- Integral representation of expectations (tail integrals) is extremely powerful for bounding $E[g(X)]$ when we are given information about tail probabilities $P(X>x)$.
- Once you show that the weighted sum $\sum a_n X_n$ converges a.s., Kronecker’s Lemma is the right tool to deduce that the scaled partial sums vanish.
- This problem is an archetype of how to handle:
- heavy-tailed variables,
- truncation + summability,
- three-series theorem,
- and Kronecker’s lemma in one coherent package.
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