Questions: Central Limit Theorem (Rigorous via Characteristic Functions)

Question 1

The Cauchy distribution has a well-defined median but no finite variance. What does the CLT predict for the behavior of the standardized sum of n i.i.d. Cauchy variables as n → ∞?

• A. By the CLT, the standardized sum converges to a standard normal distribution for large enough n
• B. The standardized sum does not converge to a normal; it converges to another Cauchy distribution, because the finite variance condition of the CLT is violated
• C. The CLT applies as long as the distribution has a finite mean, regardless of variance
• D. The sum converges to a normal only if n exceeds some threshold that depends on the Cauchy scale parameter

Answer: (B)

The finite variance condition is not a technicality — it is load-bearing. The characteristic function proof requires the Taylor expansion φ_X(t/√n) = 1 − t²/(2n) + o(1/n), which depends on the second moment E[X²] being finite. For the Cauchy distribution, E[X²] is infinite; the second-order term in the Taylor expansion does not exist; and the argument that (1 − t²/2n)ⁿ → e^{−t²/2} fails entirely. Sums of Cauchy variables rescaled by n converge to another Cauchy, not to a normal. Options A and C state the misconception directly.

Question 2

A statistics instructor says: 'The CLT means that if we repeatedly measure the same person's height, those individual measurements will be approximately normally distributed for large samples.' A student objects. Who is correct?

• A. The instructor is correct — the CLT applies to any set of repeated measurements
• B. The student is correct — the CLT is a statement about the distribution of sample means or standardized sums across many samples, not about the distribution of individual observations, which remain distributed according to their own distribution
• C. Both are correct, since individual measurements have decreasing variance and thus approach normality
• D. The instructor is correct if and only if the measurement errors happen to be normally distributed

Answer: (B)

The CLT says: the standardized sum (Sₙ − nμ)/(σ√n) converges in distribution to N(0,1). This is a statement about the distribution of the SUM across many independent replications — the sampling distribution. Individual observations Xᵢ remain drawn from whatever distribution X follows, whether normal or not. If heights follow a skewed distribution, repeated measurements of one person still follow that skewed distribution. The CLT's normality emerges only at the level of the aggregate.

Question 3

If {Xₙ} are i.i.d. with finite variance, the Central Limit Theorem guarantees that for large n, each individual Xᵢ is approximately normally distributed.

• T. True
• F. False

Answer: False

The CLT is a statement about the distribution of the standardized SUM (or equivalently the sample mean), not about individual observations. Each Xᵢ retains the distribution it was drawn from — exponential, uniform, Poisson, or anything else — regardless of n. What converges to normality is the distribution of (Sₙ − nμ)/(σ√n) as n → ∞. This distinction is crucial: 'the sampling distribution of the mean is approximately normal' is very different from 'individual observations are approximately normal.'

Question 4

In the characteristic function proof of the CLT, the finite variance assumption is essential because the proof requires the second-order term in the Taylor expansion of φ_X(t/√n), which only exists when E[X²] is finite.

• T. True
• F. False

Answer: True

The proof expands φ_X(s) ≈ 1 − s²/2 + o(s²) around s = 0, using the fact that φ_X''(0) = −E[X²] = −σ² (after centering). This gives φ_X(t/√n) = 1 − t²/(2n) + o(1/n), which when raised to the n-th power yields e^{−t²/2}. If E[X²] is infinite, the second derivative of φ_X at 0 does not exist, the Taylor expansion has no t² term, and the entire argument collapses. Heavy-tailed distributions without finite variance produce sums that converge to stable distributions, not the normal.

Question 5

Explain the distinction between 'convergence in distribution' and 'almost sure convergence,' and describe what the CLT's mode of convergence means for statistical inference.

Model answer: Almost sure convergence means the actual values of the random variables converge to the limit with probability 1 — the sequence of numbers Xₙ(ω) → L(ω) for almost every sample path ω. Convergence in distribution is weaker: only the CDFs converge, F_n(x) → F(x) at all continuity points of F. The individual random variables need not converge at all. The CLT uses convergence in distribution: the distribution of (Sₙ − nμ)/(σ√n) approaches the standard normal CDF, but individual observations remain distributed according to whatever their original distribution is. For inference, this means: for large n, you can use normal critical values (z-tables) to construct confidence intervals or perform hypothesis tests about sample means, because the sampling distribution of the mean is approximately normal. The approximation improves with n but is never exact unless the original distribution is normal.

Understanding the mode of convergence prevents two common errors: (1) thinking the CLT implies individual observations become normal (they don't), and (2) thinking the CLT gives exact rather than asymptotic results. The correct use is always: 'For large n, the sampling distribution of the mean is approximately N(μ, σ²/n),' and inference using this approximation is valid asymptotically.