Dr. Hackney STA Solutions pg 59

Dr. Hackney STA Solutions pg 59 - Second Edition 4-15 4.45...

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Second Edition 4-15 4.45 a. We will compute the marginal of X . The calculation for Y is similar. Start with f XY ( x,y ) = 1 2 πσ X σ Y p 1 - ρ 2 × exp " - 1 2(1 - ρ 2 ) ± ² x - μ X σ X ³ 2 - 2 ρ ² x - μ X σ X ³² y - μ Y σ Y ³ + ² y - μ Y σ Y ³ 2 ´# and compute f X ( x ) = Z -∞ f XY ( x,y ) dy = Z -∞ 1 2 πσ X σ Y p 1 - ρ 2 e - 1 2(1 - ρ 2 ) ( ω 2 - 2 ρωz + z 2 ) σ Y dz, where we make the substitution z = y - μ Y σ Y , dy = σ Y dz , ω = x - μ X σ X . Now the part of the exponent involving ω 2 can be removed from the integral, and we complete the square in z to get f X ( x ) = e - ω 2 2(1 - ρ 2 ) 2 πσ X p 1 - ρ 2 Z -∞ e - 1 2(1 - ρ 2 ) [ ( z 2 - 2 ρωz + ρ 2 ω 2 ) - ρ 2 ω 2 ] dz = e - ω 2 / 2(1 - ρ 2 ) e ρ 2 ω 2 / 2(1 - ρ 2 ) 2 πσ X p 1 - ρ 2 Z -∞ e - 1 2(1 - ρ 2 ) ( z - ρω ) 2 dz. The integrand is the kernel of normal pdf with σ 2 = (1 - ρ 2 ), and μ = ρω , so it integrates to 2 π p 1 - ρ 2 . Also note that e - ω 2 / 2(1 - ρ 2 ) e ρ 2 ω 2 / 2(1 - ρ 2 ) = e - ω 2 / 2 . Thus, f X ( x ) = e - ω 2 / 2 2 πσ X p 1 - ρ 2 2 π p 1 - ρ 2 = 1 2 πσ X e - 1 2 ( x - μ X σ X ) 2 , the pdf of n( μ X 2 X ). b. f Y | X ( y | x ) = 1 2 πσ X σ Y 1 - ρ 2 e - 1 2(1 - ρ 2 ) h ( x - μ X σ X ) 2 - 2 ρ ( x - μ X σ X )( y - μ Y σ Y ) + ( y - μ Y σ Y ) 2 i 1 2 πσ
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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