Joint Continuous Distributions
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1 Joint Continuous Distributions Statistics 11 Summer 6 Copright c 6 b Mark E. Irwin
2 Joint Continuous Distributions Not surprisingl we can look at the joint distribution of or more continuous RVs. For eample, we could look at the amount of time it takes to get to the Science Center from home each morning for the remaining das this week (X = Thursda travel time and Y = Frida s travel time). Probabilities are based on the joint PDF f X,Y (, ). being in the event A is given b The probabilit of P [(X, Y ) A] = A f X,Y (, )dd for an A R. Joint Continuous Distributions 1
3 Note that a joint densit must satisf Ω f(, ) f(, )dd = 1 where Ω is the sample space for the combination of RVs. f(,) Joint Continuous Distributions
4 For rectangular regions, the joint CDF is useful for calculating probabilities. F X,Y (, ) = P [X, Y ] = f X,Y (, )dd P [ 1 < X, 1 < Y ] = F (, ) F (, 1 ) F ( 1, ) + F ( 1, 1 ) Joint Continuous Distributions 3
5 As with the univariate case, the joint PDF is given b f X,Y (, ) = F X,Y (, ) wherever the partial derivative is defined. For small and, if f is continuous at (, ), P [ X +, Y + ] f X,Y (, ) so the probabilit of getting in a small region around (, ) is proportional to f X,Y (, ) so the densit is giving information about how likel and observation at the point is. Joint Continuous Distributions 4
6 The marginal distribution of each component is easil determined from the joint densit as f X () = f X,Y (, )d This is the analogue to the discrete case, where we are integrating out instead of adding over it. Joint Continuous Distributions 5
7 For eample, let f X,Y (, ) = 9 ; 1, z Then f X () = 1 9 d = 3 ; 1 Similarl f Y () = 3. Joint Continuous Distributions 6
8 Also F X,Y (, ) = 3 3 ; 1, 1 z F X,Y (, ) = 9 dd; 1, 1 Joint Continuous Distributions 7
9 Not all joint RVs are defined on nice rectangles. For eample z f X,Y (, ) = e (+) ;, is defined on an infinite triangle. You need to be careful in determining the marginal distributions. Joint Continuous Distributions 8
10 f X () = e e d = e ; and f Y () = e e d = e (1 e ); Joint Continuous Distributions 9
11 The concept of conditional distribution is a bit more comple in the continuous case. P [ X + Y + ] = P [ X +, Y + ] P [ Y + ] f X,Y (, ) f Y () { } fx,y (, ) = f Y () So conditional on Y [, + ], X has, approimatel, a densit given b the epression in { }. Note that this densit does not depend on as long as it is small. Joint Continuous Distributions 1
12 Thus we write the conditional PDF of X Y = as f X Y ( ) = f X,Y (, ) f Y () X Y = 1 Y X = f(,).1 f(,) Joint Continuous Distributions 11
13 For the eamples seen so far f X,Y (, ) = 9 f X Y ( ) = 9 3 = 3 ; 1 f Y X ( ) = 9 3 = 3 ; 1 Joint Continuous Distributions 1
14 f X,Y (, ) = e (+) f X Y ( ) = e (+) e (1 e ) = e 1 e ; This is an eample of a truncated distribution. X has an eponential distribution ecept that values larger than are removed. f Y X ( ) = e (+) e = e ; This is an eample an eample of a shifted eponential. Y is eponential on the interval [, ). Joint Continuous Distributions 13
15 Independent Continuous Random Variables A set of continuous RVs X 1, X,..., X n are independent if and onl if for all 1,,..., n. f( 1,,..., n ) = f X1 ( 1 )f X ( )... f Xn ( n ) Note that the tet defines independence in terms of CDFs instead of the densities. These are equivalent definitions since n 1... n F X1 ( 1 )F X ( )... F Xn ( n ) = f X1 ( 1 )f X ( )... f Xn ( n ) and Independent Continuous Random Variables 14
16 1... n f X1 (u 1 )f X (u )... f Xn (u n )du 1 du... du n n = i=1 [ i f Xi (u i )du i ] = F X1 ( 1 )F X ( )... F Xn ( n ) Both of these are equivalent to requiring P [X A, Y B] = P [X A]P [Y B] for all sets A and B (with the obvious etension to 3 or more RVs) The proof of this is similar to that for the discrete case. Replace the sums b integrals. Independent Continuous Random Variables 15
17 This factorization of densities (or CDFs) gives an eas wa to check whether RVs are independent. If the joint densit can be written as (using RVs for eample where X and Y) f X,Y (, ) = g()h()for all X, Y with g() and h(), X and Y are independent. Note that in this factorization g and h don t have to be densities (the will be proportional to the marginal densities). For eample, with f(, ) = 9, this can be decomposed with g() = 9 and h() =. However an eample discussed in section 3.3, f(, ) = + 4, X and Y are not independent since there is no decomposition of the valid form. Independent Continuous Random Variables 16
18 The condition for all X, Y is important. The eample where the sample space was defined on the triangle f X,Y (, ) = e (+) ;, appears that it can be factored in the desired form (e (+) = e e ). However it doesn t account for the region with probabilit. This result isn t usuall used to show dependence. To do this, usuall ou will show that the joint densit is not the product of the marginals or in terms of conditional distributions. There is the similar result with the CDF. F (, ) = G()H() with G and H both nondecreasing, non-negative functions. Independent Continuous Random Variables 17
19 Two continuous RVs are independent iff or f Y X ( ) = f Y () for all f X Y ( ) = f X () for all Actuall if one holds, the other has to as well. Technical point: Actuall needs to be for all but a countable number of points. Independent Continuous Random Variables 18
20 Dependent Continuous Random Variables As with the discrete case, joint distributions can be built up with the use of conditional distributions. The joint densit of two RVs can be written as f X,Y (, ) = f X ()f Y X ( ) = f Y ()f X Y ( ) There is the obvious etension to three variables of f X,Y,Z (,, z) = f X ()f Y X ( )f Z X,Y (z, ) Of course there are versions with all 6 possible orderings of X, Y, and Z. Dependent Continuous Random Variables 19
21 Eample: A model for SAT like scores Let X be the results of test 1 (e.g. math) and Y be the results of test (e.g. English). A possible model for this this is X N(µ X, σ X) Y X = N(µ Y + ρ σ Y σ X ( µ X ), (1 ρ )σ Y ) where 1 ρ 1. If ρ > this model suggests that if X is bigger than its mean, Y tends to be bigger than its mean. For those of ou that have seen linear regression E[Y X = ] = µ Y + ρ σ Y σ X ( µ X ) is the population regression line and ρ is the population correlation. Dependent Continuous Random Variables
22 The joint densit of X and Y is f X,Y (, ) = = ( 1 ep 1 σ X π ( µ X ) ) σ X 1 σ Y π(1 ρ ) ep 1 ( µ Y ρ σ Y σ X ( µ X ) σ Y (1 ρ ) ( [ 1 πσ X σ ep 1 ( µx ) Y 1 ρ (1 ρ ) + ( µ Y ) σ Y σ X ρ( µ X)( µ Y ) σ X σ Y ) ]) This is known as the bivariate normal densit. Dependent Continuous Random Variables 1
23 ρ = 3 ρ = f(,) ρ =.5 3 ρ = f(,) Dependent Continuous Random Variables
24 ρ =.5 3 ρ = f(,) ρ =.75 3 ρ = f(,) Dependent Continuous Random Variables 3
25 Useful properties of the bivariate normal Marginals are univariate normal X N(µ X, σ X) and Y N(µ Y, σ Y ) Conditional distributions are univariate normal Y X = N(µ Y + ρ σ Y σ X ( µ X ), (1 ρ )σ Y ) X Y = N(µ X + ρ σ X σ Y ( µ Y ), (1 ρ )σ X) Sums of normals are normal. If Z = X + Y, then Z N(µ X + µ Y, σ X + σ Y + ρσ X σ Y ) Dependent Continuous Random Variables 4
26 Note that for all the eamples presented µ X = µ Y = and σ X = σ Y = 1, so the pairs of marginal distributions is the same in all 4 cases. However the joint distributions, and thus the conditional distributions, are all different. To prove that a sum of normals is normal, ou can use the following lemma. Lemma. If X and Y are two continuous RVs and Z = X + Y, then the PDF of Z is f Z (z) = = X Y f X,Y (, z )d f X,Y (z, )d Eample: Let X 1, X,... X n be independent Ep(λ) RVs. Then S n = X 1 + X X n Gamma(n, λ). Dependent Continuous Random Variables 5
27 For S = X 1 + X f S (s) = = s f X1 ()f X (s )d ( λe λ ) ( λe λ(s )) d s = λ e λs d = λ se λs Then for S 3 = S + X 3 f S3 (s) = f S ()f X3 (s )d = s = λ 3 e λs d = λ3 s e λs s ( λ e λ) ( λe λ(s )) d which is a Gamma(3, λ) densit. The general case follows b induction. Dependent Continuous Random Variables 6
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