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Published online by Cambridge University Press: 01 December 2004
The central limit theorem for student's distribution—solution.
Consider the Lindeberg–Feller central limit theorem (CLT), which we state as follows. Let {xn} be a sequence of independent random variables with means {μn} and nonzero variances {σn2} (both existing), and c.d.f.s {Fn}. Define λn > 0 by

. Then, Lindeberg's condition

is equivalent to

where the latter limit is known as the uniform asymptotic negligibility (u.a.n.) condition. One can usually interpret λn2 as the variance of the numerator of zn. We shall see, however, that there are cases where asymptotic normality holds in spite of {xn} having infinite variances.
Let {xn} be a random sample from Student's t(ν). For ν < 2, no λn exists that can lead to

. This is because the tails of the density of t(ν) decay at a rate of u−ν−1 and the stable limit theorem tells us that a nonnormal stable law arises if the tails of the p.d.f. of xi decay at a rate of u−a where a < 3; e.g., see Loève (1977, §25) or Hoffmann-Jørgensen (1994, §5.25). For example, for ν = 1, the average of standard Cauchy variates is standard Cauchy too, so that there exists no λn achieving asymptotic normality of zn.
For ν > 2, both the mean and the variance exist, and the Lindeberg–Feller CLT applies, with λn2 = n var(xi). The interesting part is ν = 2, where we will show that asymptotic normality of zn holds, in spite of var(xi) being infinite, and we will derive the appropriate λn. We will require the additional assumption that λn2 → ∞ as n → ∞. In the standard CLT, this assumption was unnecessary, as it followed from λn2 = n var(xi). We will see subsequently that λn2 can be interpreted in terms of truncated variances for ν = 2.
To prove the asymptotic normality of zn, we need to show that the characteristic function φ(t) := E(exp(itxi)) satisfies

for some choice of λn, with λn → ∞ as n → ∞. Because the sequence {xn} is i.i.d., the uniform asymptotic negligibility condition

is satisfied for all ε > 0, thus implying

This allows us to take the leading term of the logarithmic expansion of the left-hand side of (1) as

where the linear term in t drops out because the sequence {xn} is centered around zero. Asymptotic standard-normality obtains if we can find the appropriate λn2 that makes the latter limit equal to 1 for all ε > 0. Notice that this limit is the complement of Lindeberg's condition, where

is replaced by n because {xn} is an i.i.d. sequence.
From Student's t(2) density,

tends to infinity as c → ∞. We need to solve

where we have dropped

that is dominated by sinh−1(c) → ∞. By using the logarithmic representation of the latter and simplifying,

is solved by

or any other function that is asymptotically equivalent to it (such as

). Therefore,
