Some characterizations of Aresine distribution M

Some characterizations of Aresine distribution M.Ahsanullah Department of Management Sciences Rider University, New Jersey, USA Abstract Some distributional properties of the symmetric arcsine distribution in (-1.1) is presented. Based on the distributional properties several characterizations of the arcsine distribution are given. 1.Introduction A continuous random variable is said to be arcsine distribution (asd(-1,1)) if its probability density function (pdf) fX (x) is of the following form fC (x) = p11 x2 ; 1 < x < 1 (1.1) The corresponding distribution function (d.f.) F(x) is (1.2). FX (x) = 21 ( + 2 arcsin x) If the random variable X has asd (-1,1), then Y = X2 has asd (0,1) with the pdf fY (y) as fY (y) = p 1 ; 0 < y < 1, (1.3) y(1 y)

Arnold and Groeneveld( 1980) showed that If X is a symmetric random variable, then X 2 and 1+X 2 are identically distributed if and only if X has the pdf as given in (1.1).Arcsine distribution arises in random walk. Feller (1962) showed that the limiting distribution of the proportion of spent on the positive side of the x-axis by a symmetric random walk has the pdf as given in (1.3). In this paper some distributional properties of asd(-1.1) and characterizations based on the distributional properties will be given. 2. Main results The percentile points p= F(Xp );p= 0.01,...,0.09 of asd(-1,1) are as follows: p Xp 0.1 -0.95106 0.2 -0.80902 0.3 -0.58779 0.4 -0.30902 0.5 0.0 0.6 0.30902 0.7 0.58779 0.8 80902 0.9 0.95106 The mean and variance of asd (-1.1) are respectively 0 and 1/2.

1

p

The hazard rate h(x) is given by

1 x2

1

arcsin x :

2

The moment generating function MX (t) of asd(-1,1) is given by R1 M(t) = 1 etx p11 x2 dx P1 t2n R 1 2n px dx = n=0 (2n)! 1 1 x2 P1 t2n = n=0 4n (n!)2 The odd moments are zeros. The even moments 2n E(X 2n ):for passive integers are given by (2n)! 2n = 44 (n!)2 : The inverse function F 1 (x) = sin( (x = 0:5));We have 1 (1 c) F 1 (1 2c) 2 :If X1 :::; Xn are n independent limc!0 FF 1 (1 2c) F 1 (1 4c) = 2 observations from asd(-1,1) and Xn;n = max(X1 :::; Xn );then it follows from Theorem 2.1.5 of Ahsanullah and Nevzorov(2001) that Xn:n when standardize will converse to Type III (max) extreme value 1=2 distribution with F (x) = e ( x) ; x 0: It can be shown that 1 (c) F 1 (2c) 2 limc!0 FF 1 (2c) :Thus if X1;n = min(X1 :::; Xn );then it follows F 1 (4c) = 2 from Theorem 2.1.6 of Ahsanullah and Nevzorov (2001) that X1:n when standardized will converse to Type III (min) extreme value 1=2 distribution with d.f. F(x) = 1-e x ; x 0 Theorem 2.1. Suppose that the random variable X is absolutely continuous with d.f. F(x) and pdf f(x). with F(-1)=0, F(x) >0 all for all 0-1. E(XjX< x) = Suppose that E(XjX< x) = E(XjX Thus

x) =

Rx

1

p

u

1

u2

p

du

2 +arcsin x

1 x2 1 2 +arcsin x

=

p 1 x2 +arcsin x :we R x2 uf (u)dx 1 F (x)

have

Rx

p

2

1 x uf (u)dx = F (x) +arcsin (2.2) x 2 Di¤erentiating both sides of the above equation with respect to x„we obtain p 1 x2 x p xf (x) = f (x) +arcsin x + F (x) ( ( 1 x2 )( +arcsin x) ) 1

2

2

1 +F (x) ( 1 + 1 arcsin ) u)2 2 On simpli…cation, we have

2

f (x) F (x)

x p p ( 1 x2 )(x( 2 +arcsin x)+ 1 x2 ) 1 + ( +arcsin x)(x( +arcsin x)+p1 x2 ) 2 2 = ( +arcsin1 x)p1 x2 2

=

On integrating , we obtain F (x) = c( 2 + arcsin x), where c is a constant. Using the boundary conditions, F(-1) =0 and F(1)=1, we have F (x) = 12 + 1 arcsin x: Let X1 ; X2 ; :::; Xn are n independent copies of the random variable X having absolutely continuous distribution function F(x) and pdf f(x). Suppose that X1;n X2;n ; ; ; : Xn;n are the corresponding order statistics. It is known ( see Ahsanullah et al chapter 1and Arnold et al chapter 1) that the Xj;n jXk:n = x for 1 k < j n; is distributed as the j-k th order statistics from n-k independent observations from the random variable Y having the pdf 1 f F(x)(x) ; 0 f (x) < 1:further Xi:n jXk;n ; = x1 i < k n is distributed as ith order statistics from k observations from the random variable W having the pdf fW (wjx) where fW (wjx) = Ff (w) (x); w < x: Remark 2.1. If Sk 1 =(X1;n + X2;n + ::: + Xk 1 )::1 i < k n:If X has the pdf , fC (x) = p11 x2 ; 1 < x < 1;then E( Skk

1

1

p

2

1 x )= ;x > 2 +arcsin x = 0 for x 1:

charaterizes asd(-1,1). Proof. If X has the pdf fC (x) = E( Skk

Rx

wf (w)dw F (x)

1

p1 ; 1 x2 p

1-1.we have 1 x2 wf (w)dw == F (x) +arcsin x 1 2 and the result follows by Theorem 2.1. The following theorem given a characterization of asd (-1,1) based on Tk;n = n 1 k (Xk+1;n + Xk+2;n + ::: + Xn:n ):1 k < n: Theorem 2.2. Suppose the random variable X is absolutely continues with d.f. F(x) and pdf f(x). with F(-1)=0, F(x) >0 all for all x,-1U(n+1),Xj > XU (n 1) ; n > 1g and U(1)=1.We will denote the nth upper record X(n) = XU (n) : Theorem 2.3. Suppose the random variable X is absolutely continues with d.f. F(x) and pdf f(x). with F(-1)=0, F(x) >0 all for all -1