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2024.Q1 – Variance Asymptotics, Normalized Convergence, and Almost Sure Limits

2024 Probability Prelim Exam (PDF)

Problem Statement (verbatim)

Let ${X_n, n \ge 1}$ be a sequence of random variables such that
\(E(X_n)=0, \qquad E(X_n^2)=\frac{1}{n\ln(n+1)}, \qquad E(X_n X_{n+1})=\frac{2}{n^2},\) and
\(E(X_m X_n)=0 \quad \text{if } |m-n|\ge 2.\)

(a)

Let $\sigma_n^2 = \mathrm{Var}(S_n)$ where $S_n=\sum_{i=1}^n X_i$.
Prove that
\(\lim_{n\to\infty} \frac{\sigma_n^2}{\ln\ln n} = 1.\)

(b)

Prove that for every sequence ${a_n}$ of positive numbers with $a_n\to\infty$, \(\frac{S_n}{\sqrt{a_n \ln\ln n}} \longrightarrow 0 \quad \text{in probability and in } L^2.\)

(c)

For every $\varepsilon>0$ prove:

(i)
\(\sum_{n=1}^\infty E\!\left[\frac{S_n^2}{\,n\ln n (\ln\ln n)^{2+\varepsilon}}\right] < \infty, \quad\text{and hence } \sum_{n=1}^\infty \frac{S_n^2}{\,n\ln n (\ln\ln n)^{2+\varepsilon}} \text{ converges a.s.}\)

(ii)
\(\lim_{n\to\infty} \frac{|S_n|}{\sqrt{n\ln n (\ln\ln n)^{2+\varepsilon}}} = 0 \quad \text{a.s.}\)

Survival-Guide Writeup

Part (a)

Claim.

\(\lim_{n\to\infty} \frac{\sigma_n^2}{\ln\ln n} = 1.\)

Proof.

Start from the variance identity: $$ \sigma_n^2 = \mathrm{Var}(S_n) = \sum_{i=1}^n E(X_i^2)

  • 2\sum_{1\le i<j\le n} E(X_i X_j). $$

Given:

  • $E(X_i^2)=\frac{1}{i\ln(i+1)}$,
  • only adjacent pairs have nonzero covariance:
    $E(X_i X_{i+1})=\frac{2}{i^2}$.

Thus \(\sigma_n^2 =\sum_{i=1}^n \frac{1}{i\ln(i+1)} +\sum_{i=1}^{n-1} \frac{4}{i^2}.\)

The second sum converges: \(\sum_{i=1}^\infty \frac{4}{i^2}=O(1).\)

Hence the asymptotics are governed entirely by
\(\sum_{i=2}^n \frac{1}{i\ln(i+1)}.\)

Use the expansion $$ \frac{1}{i\ln(i+1)} = \frac{1}{i\ln i}

  • \frac{\ln(1+1/i)}{i\ln i \ln(i+1)} = \frac{1}{i\ln i} + O!\left(\frac1{i^2(\ln i)^2}\right). $$

Thus \(\sum_{i=2}^n \frac{1}{i\ln(i+1)} = \sum_{i=2}^n \frac{1}{i\ln i} + O(1).\)

By the integral test: \(\sum_{i=2}^n \frac{1}{i\ln i} = \ln\ln n + O(1).\)

Putting everything together: \(\sigma_n^2 = \ln\ln n + O(1).\)

Therefore \(\frac{\sigma_n^2}{\ln\ln n} \to 1.\)

Conclusion.

The dominating contribution to $\sigma_n^2$ is the harmonic–logarithmic term
$\sum 1/(i\ln i)$, and all other contributions are $O(1)$.
Thus $\sigma_n^2 \sim \ln\ln n$.

Key Takeaways

  • Use decomposition:
    $\mathrm{Var}(S_n)=\sum \mathrm{Var}(X_i)+2\sum\mathrm{Cov}(X_i,X_j)$.
  • Identify when covariance structure is “local” (only neighbors matter).
  • Integral test for $\sum 1/(i\ln i)$ leading to $\ln\ln n$.
  • Asymptotic manipulations of slowly varying functions.

Part (b)

Claim.

If $a_n\to\infty$, then \(\frac{S_n}{\sqrt{a_n\ln\ln n}} \to 0 \quad\text{in } L^2 \text{ and in probability}.\)

Proof.

Since $E(S_n)=0$, \(E\!\left[\left(\frac{S_n}{\sqrt{a_n\ln\ln n}}\right)^2\right] =\frac{\sigma_n^2}{a_n\ln\ln n}.\)

From part (a),
\(\sigma_n^2 \sim \ln\ln n.\)

Therefore \(\frac{\sigma_n^2}{a_n\ln\ln n} \sim \frac{1}{a_n}\to 0,\) because $a_n\to\infty$.

Thus \(\frac{S_n}{\sqrt{a_n\ln\ln n}} \to 0 \quad \text{in } L^2.\)

Since $L^2\to 0$ implies convergence in probability, the result follows.

Conclusion.

The scaling by $\sqrt{a_n\ln\ln n}$ dominates the growth of the variance of $S_n$, forcing the normalized sequence to converge to zero.

Key Takeaways

  • $L^2$-convergence can be checked by variance alone when means are zero.
  • If variance goes to zero, convergence in probability follows immediately.
  • The “natural size” of $S_n$ is $\sqrt{\ln\ln n}$, so any additional divergence in $a_n$ kills the ratio.

Part (c)(i)

Claim.

\(\sum_{n=1}^\infty E\!\left[\frac{S_n^2}{n\ln n (\ln\ln n)^{2+\varepsilon}}\right] <\infty,\) and therefore
\(\sum_{n=1}^\infty \frac{S_n^2}{n\ln n (\ln\ln n)^{2+\varepsilon}} \quad \text{converges a.s.}\)

Proof.

Compute the expectation term: \(E\!\left[\frac{S_n^2}{n\ln n(\ln\ln n)^{2+\varepsilon}}\right] = \frac{\sigma_n^2}{n\ln n (\ln\ln n)^{2+\varepsilon}}.\)

Using $\sigma_n^2 \sim \ln\ln n$: \(\frac{\sigma_n^2}{n\ln n (\ln\ln n)^{2+\varepsilon}} \sim \frac{1}{n\ln n (\ln\ln n)^{1+\varepsilon}}.\)

Test convergence via the integral test: \(\int^\infty \frac{dx}{x\ln x (\ln\ln x)^{1+\varepsilon}} < \infty \quad (\varepsilon>0).\)

Thus \(\sum_n E[Z_n] < \infty, \quad Z_n := \frac{S_n^2}{n\ln n(\ln\ln n)^{2+\varepsilon}} \ge 0.\)

By the Monotone Convergence Theorem, finite expectation of a nonnegative series implies the series converges a.s.

Conclusion.

The normalization kills the growth of $S_n^2$ strongly enough to give absolute almost sure convergence.

Key Takeaways

  • If $\sum E(Z_n)<\infty$ and $Z_n\ge0$, then $\sum Z_n$ converges a.s.
  • Integral test is a crucial tool for slowly varying functions.
  • Useful trick: replace $\sigma_n^2$ with its asymptotic equivalent.

Part (c)(ii)

Claim.

\(\lim_{n\to\infty} \frac{|S_n|}{\sqrt{n\ln n (\ln\ln n)^{2+\varepsilon}}}=0 \quad\text{a.s.}\)

Proof.

From part (c)(i), we know the series
\(\sum_{n=1}^\infty \frac{S_n^2}{n\ln n(\ln\ln n)^{2+\varepsilon}}\) converges a.s.

A necessary condition for convergence of the terms is: \(\frac{S_n^2}{n\ln n(\ln\ln n)^{2+\varepsilon}} \to 0 \quad\text{a.s.}\)

Taking square roots: \(\frac{|S_n|}{\sqrt{n\ln n(\ln\ln n)^{2+\varepsilon}}}\to 0 \quad\text{a.s.}\)

Conclusion.

Almost sure convergence of a series implies its terms must tend to zero. Applying this to $S_n^2$ with the given normalization yields the desired a.s. convergence.

Key Takeaways

  • “Series converges a.s.” $\implies$ “its terms $\to 0$ a.s.”
  • Classic method: reduce a strong a.s. statement to earlier summability results.
  • Normalization $\sqrt{n\ln n(\ln\ln n)^{2+\varepsilon}}$ is much larger than the natural scale $\sqrt{\ln\ln n}$.

END OF SURVIVAL GUIDE ENTRY

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