Almost Sure Central Limit Theory for Self-Normalized Products of Sums of Partial Sums

Theorem 1.1. Let {X, X_n} _n∈ℕ be a sequence of i.i.d. positive random variables in the domain of attraction of the normal law with 𝔼X = μ > 0. Suppose

For 0 ≤ α < 1/2, set

Then for any  x ∈ ℝ, where F is the distribution function of the random variable $$.

Corollary 1.2. Theorem 1.1 remains valid if we replace the weight sequence {d_k} _k∈ℕ by $$ such that $$, $$.

Remark 1.3. If 𝔼X² = σ² < ∞, then X is in the domain of attraction of the normal law and $$, thus (1.6) holds. Therefore, the class of random variables in Theorems 1.1 is of very broad range.

Remark 1.4. Whether Theorem 1.1 holds for 1/2 ≤ α < 1 remains open.

Lemma 2.1. Let X be a random variable with 𝔼X = μ, and denote l(x) = 𝔼(X − μ) ²I{|X − μ | ≤ x}. The following statements are equivalent:

• (i)

  l(x) is a slowly varying function at ∞;

• (ii)

  X is in the domain of attraction of the normal law;

• (iii)

  x²ℙ(|X − μ| > x) = o(l(x));

• (iv)

  x𝔼(|X − μ | I(|X − μ | > x)) = o(l(x));

• (v)

  𝔼(|X − μ|^αI(|X − μ | ≤ x)) = o(x^α−2l(x)) for α > 2.

Lemma 2.2. Let {ξ, ξ_n} _n∈ℕ be a sequence of uniformly bounded random variables. If there exist constants c > 0 and δ > 0 such that

then where d_k and D_n are defined by (1.7).

Proof. We can easily apply the similar arguments of (2.1) in Wu [16] to get Lemma 2.2, and we omit the details here.

Lemma 2.3. (i) c_n,i = 2(b_n,i − d_n,i);

• (ii)

  $$;

• (iii)

  $$;

• (iv)

  $$.

Lemma 2.4. Suppose that the assumptions of Theorem 1.1 hold. Then

where d_k and D_n are defined by (1.7) and f is a nonnegative, bounded Lipschitz function.

Proof. By 𝔼(X − μ) = 0, Lemma 2.1(iv), we have

Thus, by (1.5) and Lemma 2.3 (iv), By (1.5) and Lemma 2.3 (i), Thus by combining Lemma 2.3 (iv), (1.6), and (2.8), Lindeberg condition hold.

Hence, it follows that

This implies that for any g(x), which is a nonnegative, bounded Lipschitz function, Hence, we obtain from the Toeplitz lemma.

On the other hand, note that (2.4) is equivalent to

from Theorem  7.1 of Billingsley [17] and Section  2 of Peligrad and Shao [18]. Hence, to prove (2.4), it suffices to prove for any g(x) which is a nonnegative, bounded Lipschitz function.

Let

Clearly, there is a constant c > 0 such that

For any 1 ≤ k < l, note that

For any 1 ≤ k < j, note that $$ and $$ are independent, g(x) is a nonnegative, bounded Lipschitz function, $$, $$, and ln x ≤ 4x^1/4, x ≥ 1. By the definition of η_j, we get

By Lemma 2.2, (2.15) holds.

Now we prove (2.5). Let

It is known that I(A ∪ B) − I(B) ≤ I(A) for any sets A and B; then for 1 ≤ k < j, by Lemma 2.1 (iii) and (1.5), we get Hence for 1 ≤ k < j, By Lemma 2.2, (2.5) holds.

Finally, we prove (2.6). Let

For 1 ≤ k < j, By Lemma 2.2, (2.6) holds. This completes the proof of Lemma 2.4.

Proof of Theorem 1.1. Let Z_j = T_j/(j(j + 1)μ/2); then (1.8) is equivalent to

where Φ(x) is the distribution function of the standard normal random variable 𝒩.

Let q ∈ (4/3,2), then 𝔼 | X|^q < ∞. Using Marcinkiewicz-Zygmund strong large number law, we have

Thus, Hence by |ln (1 + x) − x | = O(x²) for |x | < 1/2, for any 0 < ε < 1, from 3/2 − 2/q > 0, l(x) is a slowly varying function at ∞, and η_k ≤ k + 1.

Hence for almost every event ω and any δ > 0, there exists k₀ = k₀(ω, δ, x) such that for k > k₀,

Note that under condition |X_j − μ | ≤ η_k, 1 ≤ j ≤ k,

Thus, for any given 0 < ε < 1, δ > 0, combining (2.29), we have for k > k₀

Therefore, to prove (2.25), it suffices to prove for any 0 < ε < 1 and δ₁ > 0.

Firstly, we prove (2.32). Let 0 < β < 1/2 and h(·) be a real function, such that for any given x ∈ ℝ,

By 𝔼(X_i − μ) = 0, Lemma 2.1 (iv) and (1.5), we have

This, combining with (2.4), (2.36) and the arbitrariness of β in (2.36), (2.32) holds.

By (2.5), (2.21) and the Toeplitz lemma,

That is, (2.33) holds.

Now we prove (2.34). For any given ε > 0, let f be a nonnegative, bounded Lipschitz function such that

From $$, $$ is i.i.d., $$, Lemma 2.1 (v), and (1.5),

Therefore, combining (2.6) and the Toeplitz lemma, Hence, (2.34) holds. By similar methods used to prove (2.34), we can prove (2.35). This completes the proof of Theorem 1.1.