505 d p3x2y28 pz28 187705pz11396 0873 section

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ⎦ dydx dy The integrand in the second integral above is in the form of a normally distributed random variable. By definition of the integral over this function, the second integral is equal to 1: ∫ 1 ∞ 2πσ x −∞ =∫ e 1 ⎛ x−μx ⎞ −⎜ ⎟ 2⎝ σ x ⎠ dx × ∫ 1 ∞ −∞ 1 e 2πσ x ∞ −∞ 2 1 ⎛ x−μx ⎞ −⎜ ⎟ 2⎝ σ x ⎠ 2πσ y 1 − ρ 2 e 2 ⎡⎛ σ ⎞⎤ ⎢ ⎜ y − ( μ y + ρ y ( x − μ x )) ⎟ ⎥ σx ⎢ ⎠⎥ −⎢⎝ ⎥ 2 2σ x (1− ρ 2 ) ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ dy 2 dx ×1 The remaining integral is also the integral of a normally distributed random variable and therefore, it also integrates to 1, by definition. Therefore, ∞ ∞ ∫−∞ ∫−∞ f XY ( x, y ) =1 5-53 ⎡ 1 f X ( x) = ∫ ⎢ ⎢ 2πσ Xσ Y −∞ ⎣ ∞ 1− ρ 2 e −0.5 ( x − μ X ) = = 1 2π σe 1 1− ρ 2 σ 2 2 X X e 2π σ X 2 2 −0.5 ⎡ ( x − μ X ) − 2 ρ ( x − μ X )( y − μY ) + ( y − μY ) ⎤ ⎥ 2 2 2⎢ 1− ρ ⎣ X Y X Y ⎦ ∞ ∫ −∞ −0.5 ( x−μ X 2 X σ )2 σσ σ 2π σ 1 Y 1− ρ 2 e ∞ ∫ −∞ 2π σ 1 Y 1− ρ 2 e σ ⎤...
View Full Document

This document was uploaded on 02/09/2014.

Ask a homework question - tutors are online