# chap5 - Joint probability distributions Chap 5 excluding...

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Joint probability distributions Chap 5, excluding 5-1.6 1

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Two or more r.v.’s Often interested in more than one r.v. Stock prices of eBay and Amazon GPA and salary • Joint probability distribution for multiple r.v.’s Relation between r.v.’s: correlation 2
Two discrete r.v.’s Example: Check airline schedules using a cell phone: 123 4 0.15 0.10 0.05 0.30 3 0.02 0.10 0.05 0.17 2 0.02 0.03 0.20 0.25 1 0.01 0.02 0.25 0.28 0.20 0.25 0.55 Y: # of times city name is stated X: # of bars of signal strength 3

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Joint probability mass function • Joint pmf of X,Y : f XY (1,2)=P(X=1,Y=2)=0.02, etc. Requirements: 1 ) , ( 0 1 ) , ( , = y x f y x f XY y x XY ) , ( ) , ( y Y x X P y x f XY = = = 4
Marginal probability distribution • Marginal pmf of X Marginal pmf of X,Y f X (1)= 0.20 , f X (2)= 0.25 , f X (3)= 0.55 f Y (1)= 0.28 , f Y (2)= 0.25 , f Y (3)= 0.17 , f Y (4)= 0.30 = = = y XY X y x f x X P x f ) , ( ) ( ) ( 5

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Mean and variance of X E(X)=1 x 0.20+2 x 0.25+3 x 0.55=2.35 , () (, ) XX X Y xx y x f f x y μ == 2 , 2 2 2 ) , ( ) ( ) ( X y x XY X X x y x f x x f x X Var = = 6
Conditional probability distribution The condition pmf of Y given X=x P(Y=4|X=1)=f XY (1,4)/f X (1)=0.15/0.2=0.75 ) ( / ) , ( ) ( / ) , ( ) | ( ) ( | x f y x f x X P x X y Y P x X y Y P y f X XY x Y = = = = = = = = 7

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Distribution of Y conditional on X=1: 1 4 0.15/ 0.2 =0.75 3 0.02/ 0.2 =0.10 2 0.02/ 0.2 =0.10 1 0.01/ 0.2 =0.05 Y: # of times city name is stated X: # of bars of signal strength 8
Conditional pmf Conditional mean and variance E(Y|X=1)=4 x 0.75+3 x 0.1+2 x 0.1+1 x 0.05=3.55 1 ) ( , 1 ) ( 0 | | = y x Y x Y y f y f = = = y x Y x Y y yf x X Y E ) ( ) | ( | | μ 2 | | 2 ) ( ) | ( x Y y x Y y f y x X Y Var = = 9

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Independence X and Y are independent if for any x and y f Y|x (y)=f Y (y) or equivalently f XY (x,y)= f X (x)f Y (y) since f Y|x (y)=f XY (x,y)/f X (x)= f Y (y), or equivalently, for any sets A, B , (, ) () ( ) PX AY B PX APY B ∈= 10
Are X (# of bars), Y (# of times) independent? to show independence, need to verify for all (x,y) to disprove independence, only need to find one pair (x,y) such that f XY (x,y) f X (x)f Y (y) f XY (1,4)=0.15, f X (1)=0.2, f Y (4)=0.3 Æ dependent 11

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More than 2 discrete r.v.’s The joint pmf of 3 discrete r.v.’s X,Y,Z • Marginal probability distributions ) , , ( ) , , ( z Z y Y x X P z y x f XYZ = = = = = = = z x XYZ Y z y x f y Y P y f , ) , , ( ) ( ) ( = = = = y XYZ XZ z y x f z Z x X P z x f ) , , ( ) , ( ) , ( 12
Knowing the marginal distribution of Y , compute its mean and variance as usual • Conditional probability distributions • Independence defined similarly | () ( , ) / Yx XY X fy f x yf x = ) , ( / ) , , ( ) , ( / ) , , ( ) , | ( ) ( | y x f z y x f y Y x X P z Z y Y x X P y Y x X z Z P z f XY XYZ xy Z = = = = = = = = = = = 13

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Two continuous r.v.’s • Joint probability density function f XY (x,y) Requirements: ∫∫ = A XY dxdy y x f A Y X P ) , ( ) ) , (( ∫∫ + + = 1 ) , ( , 0 ) , ( dxdy y x f y x f XY XY 14
Example: the joint pdf of X,Y is f XY (x,y)=2(x+y) for 0<x<1 and 0<y<x f=0 anywhere else Joint probability distribution 15

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What is P(X>1/2, Y<1/2) ∫∫ = + = < > 1 0 2 1 2 1 2 1 2 1 2 / 1 ) ( 2 ) , ( dydx y x Y X P 16
What is P(X>1/2) 8 / 7 ) ( 2 ) 2 / 1 ( 1 0 2 1 = + = > ∫∫ x dydx y x X P 17

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When necessary, need to decompose a region into small pieces: P(1/2
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## This note was uploaded on 03/16/2010 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.

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chap5 - Joint probability distributions Chap 5 excluding...

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