Dr. Hackney STA Solutions pg 26

# Dr. Hackney STA Solutions pg 26 - Second Edition 2-11 EX 2...

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Second Edition 2-11 E X 2 = d 2 dt 2 M x ( t ) ± ± ± t =0 = λe t e λ ( e t - 1) λe t + λe t e λ ( e t - 1) ± ± ± t =0 = λ 2 + λ. Var X = E X 2 - (E X ) 2 = λ 2 + λ - λ 2 = λ. b. M x ( t ) = X x =0 e tx p (1 - p ) x = p X x =0 ((1 - p ) e t ) x = p 1 1 - (1 - p ) e t = p 1 - (1 - p ) e t , t < - log(1 - p ) . E X = d dt M x ( t ) ± ± ± ± t =0 = - p (1 - (1 - p ) e t ) 2 ² - (1 - p ) e t ³ ± ± ± ± ± t =0 = p (1 - p ) p 2 = 1 - p p . E X 2 = d 2 dt 2 M x ( t ) ± ± ± ± t =0 = ² 1 - (1 - p ) e t ³ 2 ² p (1 - p ) e t ³ + p (1 - p ) e t 2 ² 1 - (1 - p ) e t ³ (1 - p ) e t (1 - (1 - p ) e t ) 4 ± ± ± ± ± ± ± t =0 = p 3 (1 - p ) + 2 p 2 (1 - p ) 2 p 4 = p (1 - p ) + 2(1 - p ) 2 p 2 . Var X = p (1 - p ) + 2(1 - p ) 2 p 2 - (1 - p ) 2 p 2 = 1 - p p 2 . c. M x ( t ) = R -∞ e tx 1 2 πσ e - ( x - μ ) 2 / 2 σ 2 dx = 1 2 πσ R -∞ e - ( x 2 - 2 μx - 2 σ 2 tx + μ 2 ) / 2 σ 2 dx. Now com- plete the square in the numerator by writing x 2 - 2 μx - 2 σ 2 tx + μ 2 = x 2 - 2( μ + σ 2 t ) x ± ( μ + σ 2 t ) 2 + μ 2 = ( x - ( μ + σ 2 t )) 2 - ( μ + σ 2 t ) 2 + μ 2 = ( x - ( μ + σ 2 t )) 2 - [2 μσ 2 t + ( σ 2 t ) 2 ] . Then we have M x ( t ) = e [2 μσ 2 t +( σ 2 t ) 2 ] / 2 σ 2 1 2 πσ R -∞ e - 1 2 σ 2 ( x - ( μ + σ 2 t )) 2 dx = e μt + σ 2 t 2 2 . E X = d dt M x ( t ) ± ± t =0 = ( μ + σ 2 t ) e μt + σ 2 t 2 / 2 ± ± ± t =0 = μ. E X 2 = d 2 dt 2 M x ( t ) ± ± ± t =0 = ( μ + σ 2 t ) 2 e μt + σ 2 t 2 / 2 + σ 2 e μt + σ 2 t/ 2 ± ± ±
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