2025-Q4 – Truncation, Tail Control, and a Strong Law with Variable Normalization
2025 Probability Prelim Exam (PDF) (link pending)
Problem 4 (verbatim)
Let $X$ be a random variable that satisfies
\(\sum_{k=1}^{\infty} P(|X| \ge a_k) < \infty,\)
where ${a_k}_{k\ge1}$ is a non negative and non decreasing sequence of real numbers that satisfies
- $a_k \to \infty$ as $k\to\infty$.
- There exists $C<\infty$ so that for each $n\ge1$, \(\sum_{k=n}^{\infty} a_k^{-2} \le C n a_n^{-2}.\)
Let ${X,X_k}_{k\ge1}$ be an i.i.d. sequence of random variables, and denote
\(Y_k = X_k \cdot 1_{\{|X_k|\le a_k\}},\qquad k\ge1.\)
Prove the following.
(a)(i)
\(\frac{\sum_{k=1}^n X_k \cdot 1_{\{|X_k|>a_k\}}}{a_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0.\)
(a)(ii)
If
\(\frac{\sum_{k=1}^n (Y_k - E Y_k)}{a_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0,\)
then
\(\frac{\sum_{k=1}^n X_k - E Y_k}{a_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0.\)
(b)(i)
Let $a_0=0$. Prove \(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) = \sum_{k=1}^{\infty} E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right) \left(\sum_{n=k}^{\infty} a_n^{-2}\right).\)
(b)(ii)
Use the result of part (i) to prove that \(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) \le C \sum_{k=1}^{\infty} k\, P(a_{k-1}<|X|\le a_k).\)
(c)(i)
Show that \(\sum_{k=1}^{\infty} \frac{Y_k - E(Y_k)}{a_k} \quad\text{converges a.s.}\)
Hint:
\(\sum_{k=1}^{\infty} k\,P(a_{k-1}<|X|\le a_k)
= \sum_{k=0}^{\infty} P(|X|>a_k)\)
by summation by parts.
(c)(ii)
\(\frac{\sum_{k=1}^{n} X_k - E(Y_k)}{a_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0.\)
Solutions in Claim–Proof–Conclusion Format
(a)(i) Tail part is negligible
Claim
\(\frac{1}{a_n}\sum_{k=1}^n X_k 1_{\{|X_k|>a_k\}} \xrightarrow[n\to\infty]{\text{a.s.}} 0.\)
Proof
Define the events \(A_k = \{|X_k|\ge a_k\}.\)
Since $X_k$ has the same distribution as $X$, \(P(A_k) = P(|X_k|\ge a_k) = P(|X|\ge a_k).\)
The assumption gives \(\sum_{k=1}^\infty P(A_k) = \sum_{k=1}^\infty P(|X|\ge a_k) < \infty.\)
By the first Borel Cantelli lemma,
\(P(A_k\ \text{i.o.}) = 0.\)
So with probability one there exists a random index $K(\omega)$ such that for all $k\ge K(\omega)$,
\(|X_k(\omega)| \le a_k.\)
On that event, \(\sum_{k=1}^n X_k(\omega) 1_{\{|X_k(\omega)|>a_k\}} = \sum_{k=1}^{K(\omega)-1} X_k(\omega)1_{\{|X_k(\omega)|>a_k\}}\) for all $n\ge K(\omega)$. Hence the numerator stabilizes to a finite constant $C(\omega)$, while $a_n\to\infty$. Thus \(\frac{\sum_{k=1}^n X_k 1_{\{|X_k|>a_k\}}}{a_n} \to \frac{C(\omega)}{\infty}=0 \quad\text{a.s.}\)
Conclusion
The contribution from the large deviations $|X_k|>a_k$ is asymptotically negligible after normalization by $a_n$.
Key Takeaways
- Recognize the tail indicator sum as a Borel Cantelli setup.
- Once only finitely many “bad” events occur, the normalized partial sum must vanish since $a_n\to\infty$.
(a)(ii) Passing from truncated to full sum
Claim
If
\(\frac{\sum_{k=1}^n (Y_k - EY_k)}{a_n} \to 0 \quad\text{a.s.},\)
then
\(\frac{\sum_{k=1}^n X_k - EY_k}{a_n} \to 0\quad\text{a.s.}\)
Proof
Decompose $X_k$ into truncated and tail parts: \(X_k = X_k 1_{\{|X_k|\le a_k\}} + X_k 1_{\{|X_k|>a_k\}} = Y_k + X_k 1_{\{|X_k|>a_k\}}.\)
Then $$ \sum_{k=1}^n X_k - E Y_k = \sum_{k=1}^n (Y_k -EY_k)
- \sum_{k=1}^n X_k 1_{{|X_k|>a_k}}. $$
Divide by $a_n$: $$ \frac{\sum_{k=1}^n X_k - E Y_k}{a_n} = \frac{\sum_{k=1}^n (Y_k -EY_k)}{a_n}
- \frac{\sum_{k=1}^n X_k 1_{{|X_k|>a_k}}}{a_n}. $$
By the assumption, the first term tends to zero almost surely.
By part (a)(i), the second term also tends to zero almost surely.
Therefore the sum tends to zero almost surely.
Conclusion
As soon as the truncated sum behaves well, the full sum with the same normalization also behaves well.
Key Takeaways
- Break the full sum into truncated part plus tail part.
- Use part (a)(i) to control the tails.
- Linearity of expectation and algebraic decompositions are your main tools.
(b)(i) Rearranging the double sum
Claim
With $a_0=0$, \(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) = \sum_{k=1}^{\infty} E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right) \left(\sum_{n=k}^{\infty} a_n^{-2}\right).\)
Proof
For each $n$, \(Y_n^2 = X_n^2 1_{\{|X_n|\le a_n\}}.\)
Since $X_n\stackrel{d}{=} X$, \(E(Y_n^2) = E(X^2 1_{\{|X|\le a_n\}}).\)
Partition ${|X|\le a_n}$ into disjoint “annuli”: \(\{|X|\le a_n\} = \bigcup_{k=1}^{n} \{a_{k-1}<|X|\le a_k\}.\)
Therefore \(E(Y_n^2) = \sum_{k=1}^{n} E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right).\)
Now sum in $n$ with weight $a_n^{-2}$: \(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) = \sum_{n=1}^{\infty} a_n^{-2} \sum_{k=1}^{n} E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right).\)
All terms are non negative, so Tonelli’s theorem allows us to swap sums: \(= \sum_{k=1}^{\infty} E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right) \sum_{n=k}^{\infty} a_n^{-2}.\)
Conclusion
We have expressed the global series as a sum over “shells” in $|X|$, each weighted by a tail sum of $a_n^{-2}$.
Key Takeaways
-
Use the decomposition ${ X \le a_n} = \bigcup_{k\le n}{a_{k-1}< X \le a_k}$. - Non negative terms justify interchanging sums via Tonelli.
(b)(ii) Bounding by tail probabilities
Claim
\(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) \le C \sum_{k=1}^{\infty} k\, P(a_{k-1}<|X|\le a_k).\)
Proof
From part (b)(i), \(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) = \sum_{k=1}^{\infty} E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right) \left(\sum_{n=k}^{\infty} a_n^{-2}\right).\)
By the condition on $(a_n)$, \(\sum_{n=k}^{\infty} a_n^{-2} \le C k a_k^{-2}.\)
Also on ${a_{k-1}<|X|\le a_k}$ we have $|X|\le a_k$, so \(X^2 1_{\{a_{k-1}<|X|\le a_k\}} \le a_k^2 1_{\{a_{k-1}<|X|\le a_k\}}.\)
Hence \(E\!\left(X^2 1_{\{a_{k-1}<|X|\le a_k\}}\right) \le a_k^2 P(a_{k-1}<|X|\le a_k).\)
Combine: \(\sum_{n=1}^{\infty} a_n^{-2} E(Y_n^2) \le \sum_{k=1}^{\infty} a_k^2 P(a_{k-1}<|X|\le a_k)\, C k a_k^{-2} = C\sum_{k=1}^{\infty} k\, P(a_{k-1}<|X|\le a_k).\)
Conclusion
The series of second moments of $Y_n/a_n$ is bounded by a weighted tail sum involving $P(a_{k-1}<|X|\le a_k)$.
Key Takeaways
- Use the growth condition on $(a_n)$ to control $\sum_{n=k}^{\infty} a_n^{-2}$.
-
Bound $X^2$ by $a_k^2$ on the event where $ X \le a_k$.
(c)(i) Almost sure convergence of the normalized series
Claim
The series
\(\sum_{k=1}^{\infty} \frac{Y_k - E(Y_k)}{a_k}\)
converges almost surely.
Proof
Let \(Z_k := \frac{Y_k - E(Y_k)}{a_k},\qquad k\ge1.\)
The $Y_k$ are independent (functions of independent $X_k$), hence the $Z_k$ are independent and centered: \(E(Z_k)=0.\)
Compute their variances: \(\mathrm{Var}(Z_k) = a_k^{-2}\,\mathrm{Var}(Y_k) \le a_k^{-2} E(Y_k^2).\)
From part (b)(ii), \(\sum_{k=1}^{\infty} a_k^{-2} E(Y_k^2) \le C \sum_{k=1}^{\infty} k P(a_{k-1}<|X|\le a_k).\)
By the hint and the assumption $\sum_{k=1}^{\infty}P(|X|\ge a_k)<\infty$, one checks \(\sum_{k=1}^{\infty} k P(a_{k-1}<|X|\le a_k) = \sum_{k=0}^{\infty} P(|X|>a_k) < \infty.\)
Thus \(\sum_{k=1}^{\infty} \mathrm{Var}(Z_k) \le \sum_{k=1}^{\infty} a_k^{-2} E(Y_k^2) < \infty.\)
By the standard theorem for series of independent centered random variables
(if $\sum \mathrm{Var}(Z_k)<\infty$, then $\sum Z_k$ converges a.s.),
we get that
\(\sum_{k=1}^{\infty} Z_k
= \sum_{k=1}^{\infty} \frac{Y_k - E(Y_k)}{a_k}\)
converges almost surely.
Conclusion
The series of normalized centered truncations converges almost surely.
Key Takeaways
- Use the bound on $\sum a_k^{-2}E(Y_k^2)$ from (b)(ii).
- The “variance test” for independence is a powerful tool for series of random variables.
(c)(ii) Final strong law type result
Claim
\(\frac{\sum_{k=1}^{n} X_k - E(Y_k)}{a_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0.\)
Proof
From part (c)(i), the series
\(\sum_{k=1}^{\infty} \frac{Y_k - E(Y_k)}{a_k}\)
converges a.s.
Let
\(T_n := \sum_{k=1}^{n} \frac{Y_k - E(Y_k)}{a_k},
\quad\text{so } T_n\to T \text{ a.s. for some finite }T.\)
Note that \(Y_k - E(Y_k) = a_k (T_k - T_{k-1}),\quad T_0=0.\)
Let \(S_n := \sum_{k=1}^n (Y_k - E(Y_k)).\)
Then \(S_n = \sum_{k=1}^n a_k (T_k - T_{k-1}).\)
Apply discrete summation by parts: \(\sum_{k=1}^n a_k (T_k - T_{k-1}) = a_n T_n - \sum_{k=0}^{n-1} T_k (a_{k+1}-a_k).\)
Divide by $a_n$: \(\frac{S_n}{a_n} = T_n - \frac{1}{a_n}\sum_{k=0}^{n-1} T_k (a_{k+1}-a_k).\)
Since $T_n\to T$ a.s., the sequence $(T_k)$ is bounded a.s. by some random $M(\omega)$.
The weights
\(w_{k,n} := \frac{a_{k+1}-a_k}{a_n},\quad k=0,\dots,n-1\)
are non negative and sum to $1$:
\(\sum_{k=0}^{n-1} w_{k,n}
= \frac{a_n - a_0}{a_n} = 1.\)
Hence the second term is a weighted average: \(\frac{1}{a_n}\sum_{k=0}^{n-1} T_k (a_{k+1}-a_k) = \sum_{k=0}^{n-1} T_k w_{k,n}.\)
Because $T_k\to T$ and $(w_{k,n})$ form a probability distribution on ${0,\dots,n-1}$, this weighted average also tends to $T$. Therefore a.s. \(\frac{S_n}{a_n} = T_n - \sum_{k=0}^{n-1} T_k w_{k,n} \to T - T = 0.\)
So \(\frac{\sum_{k=1}^{n} (Y_k - E(Y_k))}{a_n} \to 0 \quad\text{a.s.}\)
Finally, apply part (a)(ii): since we have shown exactly the condition required there, it follows that \(\frac{\sum_{k=1}^{n} X_k - E(Y_k)}{a_n} \xrightarrow[n\to\infty]{\text{a.s.}} 0.\)
Conclusion
The properly normalized sum of the original $X_k$ behaves like its truncated expectation: \(\frac{\sum_{k=1}^{n} X_k - E(Y_k)}{a_n}\to 0\quad\text{a.s.}\)
Key Takeaways
- Convergence of the series $\sum (Y_k - E(Y_k))/a_k$ plus monotonicity of $a_k$ yields $S_n/a_n\to0$ via summation by parts.
- Part (a)(ii) then bridges from truncated variables back to the original $X_k$.
- This is a “generalized law of large numbers” with non standard normalization $a_n$ chosen to control the tails.
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