chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions107.a. Clearly f(x; λ 1 , λ 2 , p) ≥ 0 for all x, and∫ ∞ −∞∞−λx= pλe + ( − p)∫0f x;λ , , p)dx( 1 λ2∞− 2− 1[ 12 ] = 1 + ( − )1 λ xλ xλ e dx p λ e dx 1 p= p + (1 – p) = 1∞−λ2x1 ∫0 ∫ λ2edx0∫−λ1x−λ2xb. For x > 0, F(x; λ 1 , λ 2 , p) = f ( y;λ , λ , p)dy = p(1− e ) + (1 − p)(1− e ).∫ ∞0−λx−λ2xc. E(X) = x ⋅ [ pλ1 e ) + (1 − p)λ2e)]0x1 dx∞−λ1x∞−λ2x= p∫xλ01edx + ( 1 − p)∫ xλ02e12pdx = +λ1( 1−p)λ2d. E(X 2 ) =2pλ21( − p)2 1+ , so Var(X) =λ222pλ21( − p)2 1+λ22⎡ p− ⎢⎣λ1+( 1−p) ⎤2λ2⎥⎦λe. For an exponential r.v., CV = = 1. For X hyperexponential,⎡⎤⎢ 2p2( 1−p)⎥⎢+2 2 ⎥⎢λ1λ2−1⎥CV = ⎢2⎡ ( 1 )⎥⎢p − p ⎤⎢ + ⎥ ⎥⎢⎣ ⎣ λ⎥1 λ2⎦ ⎦= [2r – 1] 1/2 where r =1211λ22⎡2( pλ+ (1 − p)λ1)= 2⎢⎣( pλ+ (1 − p)λ )222( pλ+ (1 − p)λ )( pλ+ (1 − p)λ ) 21 provided λ1 ≠ λ2, so that CV > 1.221112⎤−1⎥⎦12. But straightforward algebra shows that r >f.µ = n ,λσ2n= , so σ2λn1= and CV = < 1 if n > 1.λn169