solution to quiz1 - Bora Kim(2924798272(Quiz 1 Econ 318...

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Bora Kim (2924798272) ( Quiz 1 Econ 318 ) Problem 1 Consider the following bivariate probability distribution for X and Y: X=1 X=3 Y=0 0.20 0.15 Y=2 0.15 0.20 Y=4 0.25 0.05 (a) E ( X ) = 1 × (0 . 2 + 0 . 15 + 0 . 25) + 3 × (0 . 15 + 0 . 2 + 0 . 05) (b) E ( Y ) = 0 × (0 . 2 + 0 . 15) + 2 × (0 . 15 + 0 . 2) + 4 × (0 . 25 + 0 . 05) (c) V ( X ) = E ( X 2 ) - E ( X ) 2 where E ( X 2 ) = 1 2 (0 . 2 + 0 . 15 + 0 . 25) + 3 2 (0 . 15 + 0 . 2 + 0 . 05) (d) V ( Y ) = E ( Y 2 ) - E ( Y ) 2 where E ( Y 2 ) = 0 2 × (0 . 2 + 0 . 15) + 2 2 × (0 . 15 + 0 . 2) + 4 2 × (0 . 25 + 0 . 05) (e) E ( Y | X = 3) = 0 × 0 . 15 0 . 4 + 2 × 0 . 2 0 . 4 + 4 × 0 . 05 0 . 4 . Note that Pr( Y = 0 | X = 3) + Pr( Y = 2 | X = 3) + Pr( Y = 4 | X = 3) = 1. If it does not sum up to 1, you are calculating wrong. (f) V ( Y | X = 3) = E ( Y 2 | X = 3) - E ( Y | X = 3) 2 where E ( Y 2 | X = 3) = 0 2 × 0 . 15 0 . 4 + 2 2 × 0 . 2 0 . 4 + 4 2 × 0 . 05 0 . 4 . Again (0.15+0.2+0.05)/0.4=1. (g) Cov ( X, Y ) = E ( XY ) - E ( X ) E ( Y ) where E ( XY ) = 0 × 1 × 0 . 2 + 0 × 3 × 0 . 15 + 2 × 1 × 0 . 15 + 2 × 3 × 0 . 2 + 4 × 1 × 0 . 25 + 4 × 3 × 0 . 05. Again 0.2+0.15+0.15+0.2+0.25+0.05=1! (g) V ( W ) = V (2 X + 3 Y ) = 4 V ( X ) + 9 V ( Y ) + 12 Cov ( X, Y ). Note: unless X and Y are independent, we always have the covariance term: V ( X + Y ) = V ( X )+ V ( Y )+2 · Cov ( X, Y ). Think of the formula for ( a + b ) 2 . Remark For this type of question, please explicitly write down the formula you’re using. Also calculate them and get the final answer. Problem 2 X N (1 , 4), Y N (6 , 16) and X and Y are independent. (a) W = 2 X + Y follows N (2 × 1 + 6 , 4 × 4 + 16) = N (8 , 32). Pr( W > 0) = Pr( W - 8 32 > 0 - 8 32 ) = Pr( Z > - 8 32 ) Use the normal table to find the value. Page 1 of 4
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Bora Kim (2924798272) ( Quiz 1 Econ 318 ) (b) If V = ( X - 1) 2 4 + ( Y - 6) 2 16 , V is the sum of two independent squared normal random variables. Therefore it will follow χ 2 (2). Use the χ 2 table to find Pr( V < 5 . 99).
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