CERTAIN BOUNDS CONCERNING CHARACTERISTIC FUNCTIONS1

The largest value of the constant c for which holds over the class of random variables X with non-zero mean and finite second moment, is c=π. Let the random variable (r.v.) X with distribution function F(·) have non-zero mean and finite second moment. In studying a certain random walk problem (Daley, 1976) we sought a bound on the characteristic function of the form for some positive constant c. Of course the inequality is non-trivial only provided that . This note establishes that the best possible constant c =π. The wider relevance of the result is we believe that it underlines the use of trigonometric inequalities in bounding the (modulus of a) c.f. (see e.g. the truncation inequalities in §12.4 of Loève (1963)). In the present case the bound thus obtained is the best possible bound, and is better than the bound (2) |1-?(θ)| ≥ |θEX|-θ²EX²\2 obtained by applying the triangular inequality to the relation which follows from a two-fold integration by parts in the defining equation (*). The treatment of the counter-example furnished below may also be of interest. To prove (1) with c=π, recall that sin u > u(1-u/π) (all real u), so Since |E sinθX|-|E sin(-θX)|, the modulus sign required in (1) can be inserted into (4). Observe that since sin u > u for u < 0, it is possible to strengthen (4) to (denoting max(0,x) by x₊) To show that c=π is the best possible constant in (1), assume without loss of generality that EX > 0, and take θ > 0. Then (1) is equivalent to (6) c < θEX²/{EX-|1-?(θ)|/θ} for all θ > 0 and all r.v.s. X with EX > 0 and EX². Consider the r.v. where 0 < x < 1 and 0 < γ < ∞. Then EX=1, EX²=1+γx², From (4) it follows that |1-?(θ)| > 0 for 0 < |θ| <π|EX|/EX² but in fact this positivity holds for 0 < |θ| < 2π|EX|/EX² because by trigonometry and the Cauchy-Schwartz inequality, |1-?(θ)| > |Re(1-?(θ))| = |E(1-cosθX)| = 2|E sin²θX/2| (10) >2(E sinθX/2)² (11) >(|θEX|-θ²EX²/2π)²/2 > 0, the inequality at (11) holding provided that |θEX|-θ²EX²/2π > 0, i.e., that 0 < |θ| < 2π|EX|/EX². The random variable X at (7) with x= 1 shows that the range of positivity of |1-?(θ)| cannot in general be extended. If X is a non-negative r.v. with finite positive mean, then the identity shows that (1-?(θ))/iθEX is the c.f. of a non-negative random variable, and hence (13) |1-?(θ)| < |θEX| (all θ). This argument fans if pr{X < 0}pr{X> 0} > 0, but as a sharper alternative to (14) |1-?(θ)| < |θE|X||, we note (cf. (2) and (3)) first that (15) |1-?(θ)| < |θEX| +θ²EX²/2. For a bound that is more precise for |θ| close to 0, |1-?(θ)|²= (Re(1-?(θ)))²+ (Im?(θ))² <(θ²EX²/2)²+(|θEX| +θ²EX²_-/π)², so (16) |1-?(θ)| <(|θEX| +θ²EX²_-/π) + |θ|³(EX²)²/8|EX|.