(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)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
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 above
and 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
-4pip2
Pi + P2
Pi – P2
None of the above
N/A
(Select One)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6d1bdda-b7a7-4c4f-bf66-52ea450cc4fa%2F117ce55f-47db-4530-9274-92403e122068%2Fkzv9mx_processed.png&w=3840&q=75)
Transcribed Image Text:(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 above
and 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
-4pip2
Pi + P2
Pi – P2
None of the above
N/A
(Select One)
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