Linear Statistics with Random Coefficients and Characterization of Hyperbolic Secant Distribution
aa r X i v : . [ m a t h . G M ] M a y Linear Statistics with Random Coefficientsand Characterization of Hyperbolic SecantDistribution
Lev B. Klebanov a Abstract
There is given a characterization of hyperbolic secant distributionby the independence of linear forms with random coefficients. Alsowe provide a characterization by the identically distribution property.
Key words: hyperbolic secant distribution; characterization of dis-tributions; linear forms; random coefficients.
Many different characterizations of the normal distribution are known (see,for example [1]). Some of them looks very natural and suitable for appli-cations, especially for motivation of a distributions family choice. Let usmention four essential of them (from my point of view).1N. Normal distribution is only limit law for the sums of n independentidentically distributed (i.i.d.) random variables with finite variance.This is a consequence of well-known Central Limit Theorem.2N. Normal distribution is only one for which X is identically distributedwith ( X + X ) / √
2, where X , X are i.i.d. random variables. Thisstatement is known as D. Polya theorem. Of course, there exist manygeneralizations of this theorem. a Department of Probability and Mathematical Statistics, Charles University, Prague,Czech Republic. e-mail: lev.klebanov@mff.cuni.cz X + X and X − X are independent, where X , X are i.i.d. random variables.The statement is known as S.N. Bernstein theorem. Its generalizationis well-known Skitovich-Darmois theorem.4N. Normal distribution is only one for which sample variance has a con-stant regression on sample mean.One interesting (but not characteristic) property is that the density ofnormal distribution with zero mean has the same multiplicative type as itscharacteristic function up to normalizing constant.Here I like to mention previous and new result connected to some similarproperties of the hyperbolic secant distribution. These facts may be modifiedfor characterizations of generalized hyperbolic secant distribution, however,I will not concentrate myself on any generalizations of such kind. They willbe a subject of a separate publication. Let us mention some known results on characterization of the hyperbolicsecant distribution. For simplicity, I consider the symmetric case only.Firstly, for ”the standard” case , the density of hyperbolic secant distri-bution has the following form p ( x ) = 1 π · x ) . (2.1)Corresponding characteristic function is f ( t ) = 1cosh( πt/ . (2.2)Now we see, that the density of hyperbolic secant distribution has the samemultiplicative type as its characteristic function up to normalizing constant.This is similar to corresponding property of normal distribution.Let us mention other properties. In general case the density contains a scale parameter. We use term hyperbolic secantdistribution for this general case. X , . . . , X n , . . . be a sequence of i.i.d. random variableswith finite variance and zero first moment. Suppose that ν n , n ∈ N is positive integer–valued random variate with probability generatingfunction P n ( z ) = 1 T n (1 /z ) . (2.3)Then the sums (1 / √ n ) P ν n j =1 X j converge in distribution to the hyper-bolic secant distribution as n → ∞ . This fact was obtained in [2].4HS. There is a quadratic form of two (or more) i.i.d. random variables suchthat its regression on the sample mean is constant. This fact was firstlyobtained in [3] (see also [4] for more general facts).Now we see that the properties 1HS and 4HS are very similar to that ofnormal law 1N and 4N. Below there are given analogs of 2N and 3N. Let us start with equally distribution property.
Theorem 3.1.
Let X , X , X be i.i.d. random variables having a symmetricdistribution. Suppose that ε is Bernoulli random variable taking the values and with equal probabilities. Suppose also that ε is independent with X , X , X . The relation X d = X + X εX (3.1) holds if and only if X has hyperbolic secant distribution.Proof. The relation (3.1) may be written in term of characteristic function f ( t ) of random variable X as f ( t ) = f ( t/ f ( t ) + 12 . (3.2)3t is easy to see that if f ( t ) satisfies (3.2) then f ( at ) is also a solution of (3.2)for any positive a . Substituting (2.2) into (3.2) shows that the characteristicfunction of hyperbolic secant distribution is a solution of equation (3.2).Now we have to show that there are no other solutions of this equation.For that aim we use the method of intensively monotone operators (see [5]).First of all, let us mention that if symmetric characteristic function f is asolution of (3.2) then it has no zeros on the real line. Really, if f ( t o ) = 0then (3.2) shows f ( t o /
2) = 0 and, therefore f ( t o / k ) = 0 for k = 1 , , . . . . inview of continuity of f we must have f (0) = 0 in contrary with f (0) = 1.Define now the following operator( A f )( t ) = f ( t/ f ( t ) + 12 . Let E be a set of non-negative functions, which are continuous in [0 , T ] ( T > A is intensivelymonotone from E to the space of continuous functions on [0 , T ]. We useterminology from [5].Consider the family { ϕ ( at ) , a > } , where ϕ ( t ) = 1 / cosh( πt/ E -positive.The statement of Theorem 3.1 follows now from [5] Theorem 1.1.1.Let us note that the property given by Theorem 3.1 is similar to 2N, thatis to D. Polya theorem.Let us now provide a characterization of hyperbolic secant distributionby the property of independence of linear statistics with random coefficients. Theorem 3.2.
Let X , X , X be i.i.d. random variables having a symmetricdistribution. Suppose that ε is Bernoulli random variable taking the values and with equal probabilities. Suppose also that ε is independent with X , X , X . Consider two liner forms L = ( X + X ) / εX ; L = ( X − X ) / − ε ) X . (3.3) The forms (3.3) are stochastic independent if and only if X has hyperbolicsecant distribution.Proof. Calculate mutual characteristic function of the forms (3.3). We haveIE exp { isL + itL } = IE exp { i (cid:16) s + t X + s − t X + (cid:0) εs + (1 − ε ) t (cid:1) X (cid:17) } =4 f (( s + t ) / f (( s − t ) / f ( s ) + f ( t )2 , (3.4)where f is characteristic function of X . The forms L and L are indepen-dent if and only ifIE exp { isL + itL } = IE exp { isL } IE exp { itL } , which is equivalent to f (( s + t ) / f (( s − t ) / f ( s ) + f ( t )2 = h ( s ) h ( t ) . (3.5)Because X has a symmetric distribution we have h ( s ) = h ( s ) = f ( s/ f ( s ) + 12 . (3.6)This expression can be obtained by direct calculations or by substituting t = 0 (or s = 0) into (3.5). However, substituting t = s into (3.5) we find f ( s ) = (cid:16) f ( s/ f ( s ) + 12 (cid:17) . (3.7)Similarly to the proof of Theorem 3.1 we see that f has no zeros on real line.Therefore, equation (3.7) is equivalent to (2.2). Theorem 3.1 shows now thatthe condition of independence of the forms (2.3) implies hyperbolic secantdistribution of X .Direct calculations lead to the fact that (3.4) holds for characteristicfunction of hyperbolic secant distribution.It is clear that Theorem 3.2 is similar to Theorem by S.N. Bernstein,mentioned above as 3N. ACKNOWLEDGEMENT
The work was partially supported by Grant GA ˇCR 19-04412S.
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