X 1 e 2 x y 1 2 1 xx 2 x 2 2 1 xx y

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Unformatted text preview: Y ⎡ ( y − μY ) 2 ⎤ −1⎢ ⎥ 2 2 Y ⎣ ⎦ σ ⎤ ⎥ dy ⎥ ⎦ and each of the last two integrals is recognized as the integral of a normal probability density function from −∞ to ∞. That is, each integral equals one. Since f XY( x, y ) = f ( x) f ( y ) then X and Y are independent. 5-52 Let f XY ( x, y ) = − 1 2πσ x σ y 1 − ρ 2 ⎡⎛ X − μ X ⎢⎜ ⎢⎜ σ X ⎣⎝ 2 ⎞ 2ρ( X −μ X )(Y −μ X ) ⎛ Y −μY ⎟− +⎜ ⎟ ⎜σ σ X σY ⎠ ⎝Y ⎞ ⎟ ⎟ ⎠ 2⎤ ⎥ ⎥ ⎦ 2(1− ρ 2 ) e Completing the square in the numerator of the exponent we get: ⎡⎛ X −μ X ⎢⎜ ⎢⎜ σ X ⎣⎝ 2 ⎞ 2 ρ ( X − μ X )(Y − μ X ) ⎛ Y − μ Y ⎟− +⎜ ⎟ ⎜σ σ XσY ⎠ ⎝ Y ⎞ ⎟ ⎟ ⎠ 2⎤ ⎡ ⎥ = ⎢ ⎛ Y − μY ⎜ ⎥ ⎢⎜ σ Y ⎦ ⎣⎝ ⎞ ⎛ X −μ X ⎟− ρ ⎜ ⎟ ⎜σ ⎠⎝ X ⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦ 2 2 ⎛ X −μ X + (1 − ρ ) ⎜ ⎜σ X ⎝ But, ⎛ Y −μ ⎜ Y ⎜σ Y ⎝ ⎞ ⎛ ⎟ − ρ ⎜ X −μ X ⎟ ⎜σ X ⎠ ⎝ ⎞ ⎟= 1 ⎟ σY ⎠ ⎡ ⎤ 1 σ ⎢ (Y − μ Y ) − ρ Y ( X − μ x ) ⎥ = σY ⎢ ⎥ σX ⎣ ⎦ Substituting into fXY(x,y), we get 5-35 ⎡ ⎤ σ ⎢ (Y − ( μ Y + ρ Y ( X − μ x )) ⎥ ⎢ ⎥ σX ⎣ ⎦ ⎞ ⎟ ⎟ ⎠ 2 ∞ ∫∫ ∞ f XY ( x, y ) = −∞ −∞ =∫ 1 e 2πσ x ∞ −∞ 1 e 2πσ xσ y 1 − ρ 2 1 ⎛ x−μx ⎞ −⎜ ⎟ 2⎝ σ x ⎠ 2 2 ⎡1⎡ ⎛ x−μx ⎞ ⎤ ⎤ σ ⎢ ⎢ y − ( μY + ρ Y ( x − μ x )) ⎥ + (1− ρ 2 )⎜ ⎟⎥ 2 σX ⎢σ Y ⎣ ⎦ ⎝ σx ⎠ ⎥ ⎦ −⎣ 2(1− ρ 2 ) 2 dx × ∫ 1 ∞ −∞ 2πσ y 1 − ρ 2 e 2 ⎡⎛ σ ⎞⎤ ⎢ ⎜ y − ( μ y + ρ y ( x − μ x )) ⎟ ⎥ σx ⎢ ⎠⎥ −⎢⎝ ⎥ 2 2σ x (1− ρ 2 ) ⎢ ⎥ ⎢ ⎥ ⎢ ⎥...
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This document was uploaded on 02/09/2014.

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