David A. Stephens Department of Mathematics and Statistics McGill University. October 28, 2006

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1 556: MATHEMATICAL STATISTICS I COMPUTING THE HYPEBOLIC SECANT DISTIBUTION CHAACTEISTIC FUNCTION David A. Stephens Department of Mathematics and Statistics McGill University October 8, 6 Abstract We give two methods for computing the characteristic function of the hyperbolic secant (sech) distribution. The first utilizes Fourier series expansions, the second complex analysis. Introduction We aim to compute the characteristic function for pdf f X (x) = cosh(x) = e x e x = ( ) k exp{ (k ) x } x. () that corresponds to the hyperbolic secant distribution. Using series expansions. A series expansion for the characteristic function k= Note first that the expansion in equation () is generated as follows. Consider first x > ; we have e x e x = e x = e x ( ) k exp{ kx} = ( ) k exp{ (k )x} () e x k= But for x >, f X ( x) = f X (x), so equation () holds for x < also with x replaced by x. Thus f X (x) = ( ) k exp{ (k ) x } x. Note that for t <, so the mgf exists. Thus, C X (t) = = = 4 e itx k= ( ) k (k ) k= k= e tx e x dx < ex ( ) k exp{ (k ) x } dx = { t (k ) } k= ( ) k e itx exp{ (k ) x } dx k= ( ) k (k ) (k ) t (3) k= using the result from lectures that e itx e x dx = t. The lead factor of was inadvertently dropped in the previous calculations

2 . A Fourier series expansion for sech(x) A Fourier series expansion for function f(x) takes the form where for k a k = f(x) = a (a k cos(kx) b k sin(kx)) (4) f(x) cos(kx) dx b k = f(x) sin(kx) dx. Note that if f(x) is an even function, then f(x) sin(kx) is odd and f(x) cos(kx) is even, so a k = f(x) cos(kx) dx. and b k =. Conversely, if f(x) is odd, then f(x) sin(kx) is even and f(x) cos(kx) is odd, so and a k =. b k = f(x) sin(kx) dx. Consider first the expansion of f(x) = cos(θx), where θ is not integer valued. This is an even function, so we have b k = and a k = f(x) cos(kx) dx = cos(θx) cos(kx) dx = = [ ] sin((θ k)x) sin((θ k)x) (θ k) (θ k) cos((θ k)x) cos((θ k)x) dx = sin((θ k)) sin((θ k)) (θ k) (θ k) = (θ k) sin((θ k)) (θ k) sin((θ k)) (θ k)(θ k) = (θ k) sin(θ) cos(k) (θ k) cos(θ) sin(k) (θ k) sin(θ) cos( k) (θ k) cos(θ) sin( k) (θ k ) = (θ k) sin(θ)( ) k (θ k) sin(θ)( ) k (θ k ) = ( ) k θ sin(θ) (θ k ) Also, Hence, from equation (4) a = cos(θx) dx = sin(θ). θ cos(θx) = a a k cos(kx) = sin(θ) θ = sin(θ) θ = θ sin(θ) ( ) k θ sin(θ) (θ k ) cos(kx) θ sin(θ) ( ) k (θ k ) cos(kx) [ θ cos(x) (θ ) cos(x) ] (θ ) (5)

3 Now consider the expansion of f(x) = sin(θx), where θ is not integer valued. This is an odd function, so a k = and by similar calculation b k = = f(x) sin(kx) dx = = ( ) k k sin(θ) (θ k ) sin(θx) sin(kx) dx cos((θ k)x) cos((θ k)x) dx Hence, from equation (4) sin(θx) = b k sin(kx) = ( ) k k sin(θ) (θ k ) sin(kx) = sin(θ) ( ) k k sin(kx) (θ k ) = sin(θ) [ sin(x) (θ ) sin(x) (θ ) 3 sin(3x) ] (θ 3 ) Now, for any θ, so that But from equation (6), for any x and at x = /, Hence, from equation (7) and equivalently, at θ = x or at θ = ix and hence sin(θ/) sin(θ) sin(θ) = sin(θ/) cos(θ/) cos(θ/) = sin(θ/). (7) sin(θ) sin(θx) sin(θ) = ( ) k k sin(kx) (θ k ) = ( ) k k sin(k/) (θ = ( ) k (k ) k ) (θ (k ) ) cos(θ/) = 4 ( ) k (k ) (θ (k ) ) cos(x) = 4 ( ) k (k ) (4x (k ) ) cosh(x) = 4 ( ) k (k ) ( 4x (k ) ) = 4 ( ) k (k ) ((k ) 4x ) cosh(x) = 4 ( ) k (k ) ((k ) 4x ).3 The Characteristic Function of the sech(x) distribution Setting x = t/ in equation (8), we conclude that = 4 ( ) k (k ) cosh(t/) ((k ) t / ) k= = 4 ( ) k (k ) (k ) t and by comparison with equation (3). C X (t) = k= k= cosh(t/) = sech(t/) = (9) e t/ e t/ 3 (6) (8)

4 .4 A note on the Fourier series representation Technically, the Fourier series expansion in equation (4) is valid on x <, yet clearly the cf C X is defined for all t as e itx e x dx e x < t. The cf is not identically zero when t/ < ; in fact it does follow that the formula in equation (9) is valid for all t as the function sech is analytic at all values of t, and thus C X (t) = sech(t/) t by the technique of analytic continuation. 3 A Proof using Complex Analysis We now compute the result using complex analysis. This proof is adapted from the proof of Priestley, p4. Consider the integral of the complex-valued function f(z) = for a, and < a < ; f has simple poles at eaz cosh(z) = eaz e z e z z C z = (k )i k Z as { } exp (k )i exp { } ( ) (k )i = cos (k ) =. Technically, f is holomorphic inside and on the rectangular contour C defined (anti-clockwise) by the corners (, ), (, ), (, ), (, ), except at the pole z = i/. The residue at this covert pole is defined as where so that es(f(z), z ) = g(z ) h (z ) g(z) = e az h(z) = e z e z h (z) = e z e z es(f(z), z ) = e ai/ e i/ e = eai/ i/ i sin(/) = ieai/ Then by Cauchy s esidue Theorem, the integral around the contour is equal to the residue multiplied by i, that is f(z)dz = ies(f(z), z ) = e ai/ () For the line integral f(z)dz = C = C f( iy) dy f(x i) dx e a(iy) dy e (iy) e (iy) e a(iy) dy e (iy) e (iy) e a(xi) e (xi) e f( iy) dy f(x i) dx (xi) dx e a(xi) dx e (xi) e (xi) Taking the integrals in turn, and examining limiting behaviour as the, S e a(iy) dy e (iy) e (iy) e a(iy) e (iy) e (iy) dy e a e e dy Introduction to Complex Analysis, nd Edition, H. A. Priestley, 3, Oxford University Press. 4

5 as, as a <. Similarly e a(iy) e (iy) e (iy) dy as S, as a >. For the remaining integrals e a(iy) e (iy) e (iy) dy e as e e S dy e a(xi) dx = e (xi) eai e (xi) e ax e x dx = eai e x e a(xi) e ax dx = e (xi) e (xi) e x dx e x so neither of these integrands depend on or S. Thus lim f(z)dz = ( e ai e ax ),S C e x dx e x and hence, from equation (), we have ( e ai e ax ) e x dx = eai/ e x so that e ax e x dx e x e ax eai/ e x dx = e x ( e ai ) = e ai/ e = ai/ cos(a/) = sec(a/). Making the change of variable x x yields e ax e x dx = sec(a/) e x and setting t = a yields Thus and M X (t) = e tx e x dx = sec(t/). e x e tx e x dx = sec(t/) e x C X (t) = sec(it/) = sech(t/) = cosh(t/). 5