HW 4 Solutions

1. Given that X and Y are two discrete random variables, and the table below is the joint probability distribution for these variables: P (X, Y) Y P(X) 3 4 5 X 0 0.08 0.06 0.06 0.2 1 0.12 0.09 0.09 0.3 2 0.20 0.15 0.15 0.5 P(Y) 0.4 0.3 0.3 1 a) Using the data in the table, calculate the values for E(X), E(Y), Var(X), and Var(Y). (2 points) E ( X ) = ∑ X ∗ f ( X )= ( 0 ∗ 0.2 ) + ( 1 ∗ 0.3 ) + ( 2 ∗ 0.5 ) 0 + 0.3 + 1.0 = 1.30 E ( Y ) = ∑ Y ∗ f ( Y )= ( 3 ∗ 0.4 ) + ( 4 ∗ 0.3 ) + ( 5 ∗ 0.3 ) = 1.2 + 1.2 + 1.5 = 3.90 E ( X 2 ) = ∑ X 2 ∗ f ( X )= ( 0 2 ∗ 0.2 ) + ( 1 2 ∗ 0.3 ) + ( 2 2 ∗ 0.5 ) = 0 + 0.3 + 2.0 = 2.30 E ( Y 2 ) = ∑ Y 2 ∗ f ( Y )= ( 3 2 ∗ 0.4 ) + ( 4 2 ∗ 0.3 ) + ( 5 2 ∗ 0.3 ) = 15.90 Var ( X ) = E ( X 2 ) − ( E ( X ) ) 2 = 2.30 − 1.30 2 = 0.61 Var ( Y ) = E ( Y 2 ) − ( E ( Y ) ) 2 = 15.90 − 3.90 2 = 0.69 b) Using the data in the table, calculate the values of E(XY), COV(X,Y), and CORR(X,Y) . (2 points) E ( XY ) = ∑ x = 0 2 ∑ y = 3 5 ( X ∗ Y ) ∗ P ( X ,Y ) = ( 0 ∗ 3 ∗ 0.08 ) + ( 0 ∗ 4 ∗ 0.06 ) + ( 0 ∗ 5 ∗ 0.06 ) + ( 1 ∗ 3 ∗ 0.12 ) + ( 1 ∗ 4 ∗ 0.09 ) + ( 1 ∗ 5 ∗ 0.09 ) + ( 2 ∗ 3 COV ( X ,Y ) = E ( XY ) − E ( X ) ∗ E ( Y ) = 5.07 − 1.30 ∗ 3.90 = 0 CORR ( X ,Y ) = COV ( X ,Y ) σ X ∗ σ Y = COV ( X ,Y ) √ Var ( X )∗ √ Var ( Y ) = 0 √ 0.61 ∗ √ 0.69 = 0 c) Are X and Y independent of each other? Explain why using the result from previous parts. (1 point) Yes , X ∧ Y areindependent if P ( X ,Y ) = P ( X ) ∗ P ( Y ) for all X ∧ Y pairs. All of the X ,Y pairs followthis pattern,thus , yes,they areindependent .