11 & 12 (11) Show that if E(X), E(Y) and E(XY)are well defined then Cov(X, Y) = E{(X – E(X))(Y – E(Y))}.The following answers are proposed:(a) First expand (X – E(X))(Y – E(Y) and then take its expectation. Now linearity should give the result.(b) First note that E(XY) = E(X)E(Y)(c) The stated result is, in fact, false.(d) Since X, Y are independent, E { (X – E(X))(Y – E(Y))} = 0 = Cov(X, Y).(e) None of the aboveand then take the right side to the left side. This gives the result.(а)(b)(c)(d)(e)N/A(Select One)(12) If (X, Y) ~ Trinomial(4, p1, p2) then find Cov(X, Y).4pip2-4pip2Pi + P2Pi – P2None of the aboveN/A(Select One)

“Since you have asked multiple questions, we will solve the first question for you. If you want any…

(1) Show that if E(X), E(Y) and E(XY) are well defined then Cov(X, Y) = E{(X – E(X))(Y – E(Y))}. The following answers are proposed. ta) First expand (X – E(X))(Y – E(Y) and then take its expectation. Now linearity should give the result. (bi First note that E(XY) = E(X)E(Y) and then take the right side to the left side. This gives the result. (c) The stated result is, in fact, false. «dh Since X, Y are independent, E { (X – E(X))(Y – E(Y))} = 0 = Cov(X, Y). te) None of the above (а) (b) (c) (d) (e) N/A (Select One)

(1) Show that if E(X), E(Y) and E(XY) are well defined then Cov(X, Y) = E{(X – E(X))(Y – E(Y))}. The following answers are proposed. ta) First expand (X – E(X))(Y – E(Y) and then take its expectation. Now linearity should give the result. (bi First note that E(XY) = E(X)E(Y) and then take the right side to the left side. This gives the result. (c) The stated result is, in fact, false. «dh Since X, Y are independent, E { (X – E(X))(Y – E(Y))} = 0 = Cov(X, Y). te) None of the above (а) (b) (c) (d) (e) N/A (Select One)

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.6: Applications And The Perron-frobenius Theorem

Problem 70EQ

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