12-Joint_Cov_Corr

# 12-Joint_Cov_Corr - IE 111 Fall Semester 2009 Note Set#12...

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IE 111 Fall Semester 2009 Note Set #12 Joint Distributions, Covariance and Correlation We begin this section of notes with another example: Example Consider the following continuous joint density function: f XY (x,y) = k(6 - x - y) for 0<x<2 and 2<y<4 a) Find k. We seek k such that the volume under the surface defined by f XY (x,y) over the region 0<x<2 and 2<y<4 is equal to one. k 2 4 ( ) 6 0 2 - - x y dxdy = k [ / 6 2 2 4 x xx xy - - ] 0 2 dy = k ( ) 12 2 2 2 4 - - y dy = k[10y - y 2 ] 2 4 = k[(40-16) - (20-4)] = 8k Thus k = 1/8 b) Find P(x<1 and y<3) = k 2 3 ( ) 6 0 1 - - x y dxdy = k [ / 6 2 2 3 x xx xy - - ] 0 1 dy = k ( . ) 6 05 2 2 3 - - y dy =k[5.5y - 0.5y 2 ] 2 3 = (1/8)[(33/9 - 9/2) - (11 - 2)] = 3/8 c) Find P(X+Y < 4) First we must determine the region and integration limits: 2 2 4 x y

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k 2 4 ( ) 6 0 4 - - - x y dxdy y = k [ / 6 2 2 4 x xx xy - - ] 0 4 - y dy = k 2 4 [6(4-y) - (0.5)(4-y) 2 - (4-y)y ]dy = k 2 4 [16 - 6y + (0.5)y 2 ]dy = k[ 16y - 3y 2 + (1/6)y 3 ] 2 4 = (1/8)[(64 - 48 +64/4) - (32 - 12 + 8/6)] = 2/3 d) Find the marginal distributions of X and Y f X (x) = k ( ) 6 2 4 - - x y dy = k[ 6y - xy - (0.5)y 2 ] 2 4 = k[(24-4x-8) - (12-2x-2)] = (3-x)/4 for 0<x<2 f Y (y) = k ( ) 6 0 2 - - x y dx = k[6x - 0.5x 2 - xy ] 0 2 = (1/8)(12-2-2y) = (5-y)/4 for 2<y<4 e) Find P(X<1.5) We can find this from the marginal density of X = [( ) / ] . 3 4 0 1 5 - x dx = 0.25[3x - x 2 /2 ] . 0 1 5 = 0.84375 f) Find the conditional distribution f X|y (X|Y=3) Conditional is the joint over the marginal. Also note that Y=3. f X|y (x|y) = (1/8)(6-x-y)/[(5-y)/4] = (6-x-y)/(10-2y) f X|y (x|Y=3) = [(1/8)(6-x-3)]/[(5-3)/4] = (3-x)/4 for 0<x<2 g) Find E(X|Y=3) 6 / 5 12 8 8 12 12 8 3 4 3 2 0 3 2 2 0 = - = - = - = = x x dx x x x In this case the domain of x is the same as in the joint density. In some cases, this will not be true. Suppose the region in the (x,y) plane over which the density is defined is: x y 2 2
Then the conditional density of X given Y=3 will have a domain of 0<x<1 even though the domain of x in the joint density is 0<x<2. Conditional Expectation We have learned already the definition of expected value: E(X) = Rx xf X (x)dx where Rx tells us to integrate over the domain of x. Suppose we want to find the expected value of X given that Y takes on some value? It is easy; we just use the conditional distribution of X given Y=y. E(X|Y) = Rx y | xf X|y (x|y)dx where Rx|y tells us to integrate over the domain of x in the conditional distribution. Covariance and Correlation Given two random variables X and Y, and a Joint distribution, we know how to find E(X), E(Y), V(X), and V(Y). We simply calculate them as before from the marginal distributions of X and Y. You might think that knowing the means and variances of the two random variables does not capture an important feature, namely the degree to which the two random variables are related. i.e. are they independent, or how dependent are

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## This note was uploaded on 02/21/2010 for the course IE 111 taught by Professor Storer during the Spring '07 term at Lehigh University .

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12-Joint_Cov_Corr - IE 111 Fall Semester 2009 Note Set#12...

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