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L03_randomvectorsans

# L03_randomvectorsans - Answers to Exersices in Random...

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Answers to Exersices in: Random Vectors Stat 640 Answer 1 We get: E ( Y 1 ) = 0, E ( Y 2 1 ) = 1, so var( Y 1 ) = 1. Also, E ( Y 2 ) = 2, E ( Y 2 2 ) = 4 . 75, so var( Y 2 ) = 0 . 75. Now, cov( Y 1 , Y 2 ) = - . 25, so cov( Y ) = parenleftBigg 1 - . 25 - . 25 . 75 parenrightBigg Answer 2 The covariance of X and Y is zero. However, the random variables are not independent because clearly the value of Y completely depends on that of X . Answer 3 If c = (1 / 3 , 1 / 3 , 1 / 3) , then the variance of ¯ Y is c Σ c = 5 / 9. Answer 4 f ( x 1 , x 2 ) = 1 2 π ( σ 2 1 σ 2 2 - σ 2 12 ) 1 / 2 exp braceleftBigg - ( x 1 - μ 1 ) 2 σ 2 2 + ( x 2 - μ 2 ) 2 σ 2 1 - 2 σ 12 ( x 1 - μ 1 )( x 2 - μ 2 ) σ 2 1 σ 2 2 - σ 2 12 bracerightBigg Answer 5 The Cholesky decomposition is easy to obtain for 2 × 2 matrices. We simply write parenleftBigg 3 1 1 2 parenrightBigg = parenleftBigg a 0 b c parenrightBiggparenleftBigg a b 0 c parenrightBigg and get a = 3, b = 1 / 3, and c = radicalBig 5 / 3. This gives B = 3 0 1 / 3 radicalBig 5 / 3 , and X = BZ , where Z N (0 , I 2 ) will have the desired distribution. The Splus commands are: > z1_rnorm(1000) > z2_rnorm(1000) > x1_z1*sqrt(3) > x2_z1/sqrt(3)+z2*sqrt(5/3)

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An example of the result is shown below, where for these deviates, the sample variances are: var(x1)=2.98, var(x2)=2.03, and cov(x1,x2)=1.03.
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L03_randomvectorsans - Answers to Exersices in Random...

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