LEC_EAS305_F10_0927

# LEC_EAS305_F10_0927 - Functions of a Random Variable...

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Functions of a Random Variable Bivariate Random Variables Conditional Distributions Fall 2010 EAS 305 Lecture Notes Prof. Jun Zhuang University at Buﬀalo, State University of New York September 27, 29, . .. 2010 Prof. Jun Zhuang Fall 2010 EAS 305 Lecture Notes Page 1 of 52

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Functions of a Random Variable Bivariate Random Variables Conditional Distributions Agenda for Today 1 Functions of a Random Variable Discrete Case Continuous Case Inverse Transform Theorem 2 Bivariate Random Variables Discrete Case Continuous Case Bivariate CDF’s Marginal Distributions 3 Conditional Distributions Typical Problem Conditional Expectation Prof. Jun Zhuang Fall 2010 EAS 305 Lecture Notes Page 2 of 52
Functions of a Random Variable Bivariate Random Variables Conditional Distributions Discrete Case Continuous Case Inverse Transform Theorem Functions of a Random Variable Problem Statement Problem: You have a RV X and you know its PMF/PDF f ( x ). Deﬁne Y h ( X ) (some fn of X ). Find g ( y ), the PMF/PDF of Y . Prof. Jun Zhuang Fall 2010 EAS 305 Lecture Notes Page 3 of 52

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Functions of a Random Variable Bivariate Random Variables Conditional Distributions Discrete Case Continuous Case Inverse Transform Theorem Discrete Case Discrete Case: X discrete implies Y discrete implies g ( y ) = Pr( Y = y ) = Pr( h ( X ) = y ) = Pr( { x | h ( x ) = y } ) = X x | h ( x )= y f ( x ) Prof. Jun Zhuang Fall 2010 EAS 305 Lecture Notes Page 4 of 52
Functions of a Random Variable Bivariate Random Variables Conditional Distributions Discrete Case Continuous Case Inverse Transform Theorem Example Example: X is the # of H ’s in 2 coin tosses. Want PMF for Y = h ( X ) = X 2 - X . x 0 1 2 Pr( X = x ) 1 4 1 2 1 4 y = x 2 - x 0 0 2 g (0) = Pr( Y = 0) = Pr( X = 0 or 1) = 3 / 4 and g (2) = Pr( Y = 2) = Pr( X = 2) = 1 / 4. g ( y ) = ± 3 / 4 if y = 0 1 / 4 if y = 2 Prof. Jun Zhuang Fall 2010 EAS 305 Lecture Notes Page 5 of 52

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Functions of a Random Variable Bivariate Random Variables Conditional Distributions Discrete Case Continuous Case Inverse Transform Theorem More Example. .. Example: X is discrete with f ( x ) = 1 / 8 if x = - 1 3 / 8 if x = 0 1 / 3 if x = 1 1 / 6 if x = 2 0 otherwise Let Y = X 2 ( Y can only equal 0,1,4). g ( y ) = Pr( Y = 0) = f (0) = 3 / 8 Pr( Y = 1) = f ( - 1) + f (1) = 11 / 24 Pr( Y = 4) = f (2) = 1 / 6 0 , otherwise Prof. Jun Zhuang Fall 2010 EAS 305 Lecture Notes Page 6 of 52
Bivariate Random Variables Conditional Distributions Discrete Case Continuous Case Inverse Transform Theorem Continuous Case Continuous Case: X continuous implies Y can be continuous or discrete. Example: Y = X 2 (clearly continuous) Example: Y = ± 0 if X < 0 1 if X 0 is not continuous. Prof. Jun Zhuang

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## This note was uploaded on 10/31/2010 for the course EE 423 taught by Professor Mitin during the Spring '10 term at SUNY Buffalo.

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LEC_EAS305_F10_0927 - Functions of a Random Variable...

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