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Let X₁, X₂, , Xₙ be a sequence of random variables (r. v. 's) and put Sₘ = ᵐ = ₁ X_, 1 m n. It is well-known that equation* (1) E|Sₙ|ʳ n^r - 1 ⁿ = ₁ E|X_|ʳ r > 1, equation* E|Sₙ|ʳ ⁿ = ₁ E|X_|nu|ʳ, r 1. However, if the r. v. 's satisfy the relations equation*2E (X₌ + ₁ Sₘ) = 0 a. s. 1 m n - 1, equation* it is possible to improve the first inequality considerably. The case r > 2 with independent r. v. 's will be treated elsewhere by one of the authors, von Bahr. If r = 2, we have, under (2), equation*3ES²ₙ = ⁿ = ₁ EX²_. equation* In the case 1 r 2, we will show that under (2) equation* (4) E|Sₙ|ʳ C (r, n) ⁿ = ₁e|X_|ʳ, equation* where C (r, n) is a bounded function of r and n. In Theorem 2 we show that (4) is true with C (r, n) = 2. If the distribution of each X₌ + ₁ conditioned by Sₘ is symmetric about zero, one can put C (r, n) = 1 (Theorem 1). Further, if the r. v. 's satisfy the following conditions equation* (5) E (Xᵢ R₌₈) = 0a. s. 1 i m + 1 n, equation* where R₌₈ = ^m + 1 = ₁, ₈ X_ it is possible to put C (r, n) = 2 - n^-1. The conditions (2) and (5) are fulfilled if the r. v. 's are independent and have zero means. In this case, however, it is possible to make C (r, n) dependent on r, so that C (r, n) 1 as r 2. It is possible to show by an example, that (4) is not generally true with C (r, n) = 1 even in this case. If 1 r < s 2 and E|X_|ˢ <, 1 n, it is generally better not to use (4) directly, but to use it together with E|Sₙ|ʳ (E|Sₙ|ˢ) ^r/s, so that E|Sₙ|ʳ (C (s, n) ⁿ = ₁ E|X_|ˢ) ^r/s. The case r < 1 is by (1) trivial.
Bahr et al. (1965) studied this question.