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By Alexander D. Poularikas

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2 = E{X 2} − E 2 {X} Example P{X = k ) = e − λ λk , k = 0, 1, L = Poisson distribution. k! ∞ ∑ ke E{X ) = −λ k =0 ∞ ∞ λk λk λk = e−λ ∑ k = e−λ ∑ k . k! k! k! k =0 k =1 but d λ d e = dλ dλ ∞ ∑k k =1 λk −1 1 ∞ λk = ∑k = eλ k! λ k =1 k! or ∞ λ = e−λ ∑ k k =1 λk k! and hence, E{X} = λ . 13 Generalized Moments a µ k′ = E{( X − a) k }, © 1999 by CRC Press LLC Mk′ = E{ X − a } k a Example 1 E{X 2 n} = 1 2a ∫ a −a x 2 n dx = a2n a2 , σ 2 = E{x 2} = 2n + 1 3 for X uniformly distributed in (-a,a). Example 2 E{X n} = a b +1 Γ (b + 1) ∫ ∞ 0 x n x b e − ax dx = a b +1Γ (b + n + 1) a b +n+1Γ (b + 1) for a gamma density f ( x ) = {a b +1 / Γ (b + 1)]x b e − ax u( x ), u( x ) = unit step function.

With E{A} = E{B} = 0 and E{A 2} = E{B 2} = σ 2 and a = constant. Hence E{X (t )} = E{A}cos at + E{B}sin at = 0, R(t1 , t2 ) = E{( A cos at1 + B sin at1 )( A cos at2 + B sin at2 )} = E{A 2 cos at1 cos at2 + E{B 2}sin at1 sin at2 = σ 2 cos ω (t2 − t1 ) implies that X(t) has mean value zero and variance σ 2 which and,hence f ( x, t ) = (1 / σ 2π )exp( − x / 2σ 2 ). 1) 2 f ( x1 , x 2 ; t1 , t2 ) = [1 /(2πσ 2 1 − cos 2 aτ )]exp[−( x12 − 2 x1 x 2 cos aτ + x 22 ) /(2σ 2 (1 − cos 2 aτ ))] where τ = t1 − t2 .

V. are indedz −∞ f xy ( x, y) = f x ( x ) f y ( y) and hence fz ( z ) = ∫ ∫ ∞ −∞ f x ( z − y) f y ( y)dy = ∫ ∞ −∞ f x ( x ) f y ( z − x )dx = f x ( z ) ∗ f y ( z ) = convolution of densities. Example 2 Z = X 2 + Y 2 , if z > 0 so then x2 + y2 ≤ z = circle with radius z , Fz ( z ) = ∫∫ f ( x, y)dxdy, if z < 0, x 2 + y2 ≤ z z Fz ( z ) = 0. f xy ( x, y) = (1 / 2πσ 2 )exp[−( x 2 + y 2 ) / 2σ 2 t h e n z > 0 and fz ( z ) = © 1999 by CRC Press LLC 1 − z / 2σ 2 e ,z ≥ 0 2σ 2 Fz ( z ) = 2 2 2 1 2πre − r / 2σ dr = 1 − e − z / 2σ , 2πσ 2 0 ∫ Example 3 z f xy ( x, y) = (1 / 2πσ 2 )exp[−( x 2 + y 2 ) / 2σ 2 ], Z = + X 2 + Y 2 , Fz ( z ) = 2 2 2 2 1 2πre − r / 2σ dr = 1 − e − z / 2σ , 2πσ 2 0 ∫ z > 0, fz ( z ) = ( z / σ 2 )exp( − z 2 / 2σ 2 0, z > 0 ≡ Rayleigh distributed, E{Z} = σ π / 2 , E{Z 2} = 2σ 2 ,σ z2 = (2 − (π / 2))σ 2 Example 4 If f xy ( x, y) = f xy ( − x, − y) then Fz ( z ) = 2 ∞ ∫∫ yz −∞ 0 ∫ ∞ f xy ( x, y)dxdy, fz ( z ) = 2 yf xy ( zy, y)dy.

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Handbook of Formulas and Tables for Signal -8579-cilt-2 by Alexander D. Poularikas


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