ECE313.Lecture36

ECE313.Lecture36 - ECE 313 Probability with Engineering...

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Functions of Many Random Variables II Professor Dilip V. Sarwate Department of Electrical and Computer Engineering © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign. All Rights Reserved ECE 313 Probability with Engineering Applications
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 2 of 39 X and Y are jointly continuous, and Z is continuous? Find the pdf of Z = g( X , Y ) as follows: Figure out where joint pdf is nonzero Sketch the curve g(u, v) = α in u-v plane Find F Z ( α ) = P{ Z α } = P{g( X , Y ) ≤ α } = P{( X , Y ) region g(u, v) ≤ α in plane} Repeat for other values of α Differentiate F Z ( α ) = P{ Z α } with respect to α to find f Z ( α ) Function of jointly continuous RVs
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 3 of 39 Example C: Z = X – Y The joint pdf of X and Y is given by f X , Y (u,v) = exp(–u) for 0 ≤ v ≤ u < and f X , Y (u,v) = 0 otherwise Find the pdf of Z = X Y X is Γ (2,1) RV; Y is Γ (1,1) = exponential u v Line u=v u v 0 1
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 4 of 39 Example C: P{Z ≤ α } = P{X – Y ≤ α } P{ X Y α } = volume under pdf surface in green shaded region = integral of f X , Y (u,v) over green shaded region = 0 if α < 0 u v Line u=v α Line u–v= α
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 5 of 39 Example C: Finding P{X – Y ≤ α } For any fixed value of v, 0 ≤ v < , u varies from v to α +v P{ X Y α } = exp(–u) du dv v=0 u=v α +v = 1 – exp(– α ) u v Line u=v α Line u–v= α v
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 6 of 39 CDF of Z = X Y is 1 – exp(– α ), α > 0 f Z ( α ) = exp(– α ), α > 0 Z is an exponential RV with parameter 1 Z is a gamma random variable with parameters (1,1), that is, Z is a Γ (1,1) RV Remember that Y is a Γ (1,1) RV If Y and Z were independent Γ (1,1) RVs, then their sum Y + Z would be a Γ (2,1) RV But, Y + Z = X is a Γ (2,1) RV So, are Y and Z independent Γ (1,1) RVs? Example C: pdf of Z = X – Y
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved Slide 7 of 39 P{ Y β , Z α } = exp(–u) du dv v=0 β u=v α +v = [1 – exp(– β )]•[1 – exp(– α )] = P{ Y β }•P{ Z α } since Y , Z are Γ (1,1) RV Hence, Y and Z are independent Γ (1,1) RVs u v Line u=v α Line u–v= α v β
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ECE 313 - Lecture 36 © 2000 Dilip V. Sarwate, University of Illinois at Urbana-Champaign, All Rights Reserved
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ECE313.Lecture36 - ECE 313 Probability with Engineering...

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