Chapter 4 Multiple Random Variables

Transcription

1 Review for the previous lecture Theorems and Examples: How to obtain the pmf (pdf) of U = g ( X Y 1 ) and V = g ( X Y) Chapter 4 Multiple Random Variables Chapter 43 Bivariate Transformations Continuous Case: Consider ( XY ) a continuous bivariate random vector with joint pdf f ( xy ) and define U = g ( x y 1 ) and V = g ( x y) Let A = {( x y ); f ( x y ) > 0} and B = {( uv ) : u= g1 ( xy ) and v= g( xy )} for some ( x y) A Assume that U and V are 1-1 transformation from A to B Define x = h1 ( u v) and y= h ( u v) be the inverse transformations The Jacobian J of the transformation is defined as the determinant of a matrix of partial derivatives: x x u v x y y x J = = y y u v u v u v Therefore the joint pdf of U and V is given by f ( uv ) = f ( h( uv ) h( uv )) J where J is the absolute value of the Jacobian UV XY 1 1

2 Example 433 (Distribution of the product of two independent beta variables): Let X ~ beta( α β ) and Y ~ beta(α + βγ ) then the joint pdf of ( XY ) is Γ ( α + β) Γ ( α + β + γ) f x y x x x x x y Γ( α) Γ( β) Γ ( α + β) Γ( γ) α 1 β 1 α+ β 1 γ 1 XY ( ) = (1 ) (1 ) 0< < 10< < 1 Consider U = XY and V = X then this an one to one transformation from {( x y) : 0 < x< 10 < y< 1} to {( uv ) :0 < u< v< 1} In addition X = V Y = U / V J = 1/ v therefore Γ( α + β) α 1 β 1 Γ ( α + β + γ) α+ β 1 γ 11 fuv ( u v) = fxy ( v u/ v) = v (1 v) ( u/ v) (1 u/ v) Γ( α) Γ( β) Γ ( α + β) Γ( γ) v and 1 f ( u) = f ( u v) dv U u U V Γ ( α + β + γ) u α + β v β γ (1 v) β ( v u) γ = dv Γ( α) Γ( β) Γ( γ) u = Γ ( α + β + γ) α 1 β+ γ 1 (1 ) 0 1 Γ( α) Γ ( β + γ) u u < u< Example 434 (Sum and difference of two independent normal random variables): Let X ~ n( α σ ) and Y ~ n( β σ ) and they are independent Consider U = X + Y and V = X Y then X = ( U + V) / Y = ( U V)/ and J = 1/ Therefore

3 1 fuv ( u v) = fxy (( u+ v)/( u v)/) 1 1 (( u+ v)/ α) + (( u v)/ β) = exp( ) πσ σ 1 1 u ( α + β) u+ v ( α β) v+ α + β = exp( ) πσ 4σ 1 ( u ( α + β)) 1 ( v ( α β)) = exp( ) exp( ) π σ 4σ π σ 4σ Theorem 435: Let X and Y be independent random variables Let g( x) be a function only of x and h( y) be a function only of y Then the random variables U = g( X) and V = h(y) are independent Proof: We only prove it for the continuous case For any u R and v R we define A = { x: g( x) u} and B = { y: h( y) v} then the joint cdf of ( UV ) is UV The joint pdf of ( UV ) is u F ( u v) = P( U u V v) = P( X A Y B ) = P( X A ) P( Y B ) v u v u ( uv ) = F ( uv ) = PX ( A) f UV UV u PY ( Bv) uv u v v Question: Why can t we use the transformation technique to prove this? 1 f ( uv ) = f ( g ( x) h 1 ( y)) J UV XY Notes: 3

4 1 If only one function is of interest say U = g ( x y) In this case choose a convenient 1 V = g ( X Y) so that ( UV ) is a 1-1 transformation from A onto B and obtain the joint pdf of U and V given by fuv ( uv ) Finally one can obtain the marginal of U by integrating over all values of V What is the transformation is not 1-1? We use a generalized version of Theorem 18 from univariate to many-to-one functions by finding partitions A0 A1 An of A where the transformation U = g ( x y) and 1 V = g ( X Y) is 1-1 from Ai ( i= 1 n) onto B and PA ( 0 ) = 0 Therefore the joint pdf of U and V is given by n f ( uv ) = f ( h( uv ) h ( uv )) J UV i= 1 XY 1i i i 3 Even the transformation is not 1-1 we can always use the method in the proof of Theorem of 435 to calculate the joint cdf and pdf of U and V Example 436: Show that the distribution of the ratio of two independent normal variables is a Cauchy random variable That is if X ~ n (01) Y ~ n (01) then U = X / Y ~ Cauchy(0) Solution 1 Let V = Y then A1 = {( x y) : y> 0} A = {( x y) : y< 0} A0 = {( x y) : y= 0} and 1 x + y fxy ( x y) = exp( ) Therefore π Solution For v ( u + 1) v fuv ( u v) = fxy ( uv v) v + fxy ( uv v) v = exp( )( < u< 0< v< ) π v ( u + 1) v 1 ( u + 1) v 1 fu ( u) = exp( ) dv= exp( ) 0 0 = < u< π π(1 + u ) π(1 + u ) u < 0 we have 4

5 F ( u) = P( U u) = P( X / Y u) U 0 x/ u 0 0 X y 0 x/ u X y = f ( x) f ( y) dydx+ f ( x) f ( y) dydx Take the derivative respect to u we have 0 1 x x 1 x 1 x x 1 x fu ( u) = exp( )( ) exp( ) dx+ exp( )( ) exp( ) dx 0 π u π u π u π u ux + x = xexp( ) dx 0 π u = u π u + 1 Similarly you can get f U ( u) for u 0 You also need to verify the conditions such that the integration and differentiation can be exchanged Chapter 44 Hierarchical Models and Mixture Distributions Examples: 1 Binomial-Poisson Hierarchy: X =number survived and Y =number of eggs laid Then we can use the models X Y ~ binomial( Y p ) and Y ~ Poisson( λ ) In this case X ~ Poisson( λ p) Poisson-Exponential Hierarchy: Let Y =number of eggs laid and Λ =variability across different mothers Y Λ ~Poisson( Λ ) and Λ ~exponential( β ) In this case Y ~ negative binomial( r = 1 p= 1/(1 + β )) 3 Binomial-Poisson-Exponential Hierarchy (Three-stage hierarchy): Let X =number survived Y =number of eggs laid and Λ = variability across different mothers X Y ~ binomial( Y p ) and Y Λ ~ Poisson( Λ ) and Λ ~exponential( β ) Equivalently we can look at this as a two-stage hierarchy where X Y ~binomial( Y p) and Y ~ negative binomial( r = 1 p= 1/(1 + β )) 5

6 4 Beta-Binomial Hierarchy: X P ~ binomial ( np ) and P ~ beta( α β ) Definition 444: A random variable X is said to have a mixture distribution if the distribution of X depends on a quantity that also has a distribution IMPORTANT RESULTS: These results are useful when one is interested only the expectation and variances of a random variable Theorem 443: If X and Y are any two random variables then E X = E( E( X Y) ) provided that the expectations exist Proof: We prove this theorem in the continuous situation Let f ( xy XY ) denote the joint pdf of X and Y Then we have EX = xfxy ( xydxd ) y= [ xf( x ydxf ) ] Y( ydy ) = EX ( y) fy( ydy ) = EEX ( ( Y)) Theorem 447: For any two random variables X and Y then VarX = E( Var( X Y )) + Var( E( X Y )) provided that the expectations exist Proof First we have 6

7 In addition and VarX = E( X EX) = E( X E( X Y) + E( X Y) EX) = E X E X Y + E E X Y EX + E X E X Y E X Y EX ( ( )) ( ( ) ) ([ ( )][ ( ) ]) E([ X E( X Y)][ E( X Y) EX]) = E( E{[( X E( X Y))( E( X Y) EX)] Y}) E{[ X E( X Y)][ E( X Y) EX] Y} = ( E( X Y) EX)( E{[ X E( X Y)] Y}) = ( E( X Y) EX)( E( X Y) E( X Y)) = 0 E X E X Y E E X E X Y Y ( ( )) = ( [ ( )] ) = EVar ( ( X Y)) Finally we have E EX Y EX VarEX Y ([ ( ) ] ) = ( ( ) VarX = E( Var ( X Y )) +Var( E ( X Y )) Illustrations: 1 Binomial-Poisson Hierarchy: models X Y ~ binomial(( Y p ) and Y ~ Poisson( λ ) (Recall that X ~ Poisson( λ p) ) Using Theorem 443 and 445 we get EX = E( E( X Y)) = E( py) = λ p and VarX = E( Var( X Y )) + Var( ( Y )) E( (1 p) ) + ( py) = (1 p) + p E X = Yp Var p λ λ Beta-Binomial Hierarchy: X P ~binomial ( np ) and P ~ beta( α β ) Find EX and VarX 7

8 3 Non-central chi-squared distribution with degrees of freedom k and noncentrality parameter λ p/+ k 1 x/ k λ x e λ e f( x λ p) = k= 0 p/+ k Note that X K ~ χ p + k and K ~Poisson( λ ) Therefore Γ ( p / + k) k! EX = E( E( X P)) = E( p+ K) = p+ λ 8