(9) When X, Y have a bivariate normal density with respective means ux, HY , respective variables o, of and correlation p, we have E(X|Y = y) = µx + oy po X (y - HY), Var(X|Y = y) = o3 (1 – p'). Verify the theorem of total expectation (TTE). The following verifications are proposed. (a) Since Var(X|Y = y) = o}(1 – p²) does not depend on y, therefore, the TTE holds. (b) Since E(XY (c) Since E(X|Y) = 4x + X (Y – µy), we see that E(E(XY))= 4x = E(X). Therefore, the TTE holds. y) depends only on y, the TTE holds. oy (d) TTE does not hold. (e) None of the above The correct verification is (a) (b) (c) (d) (e) N/A (Select One)

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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(9) When X, Y have a bivariate normal density with respective means ux: HY, respective variables o, o and correlation p, we have
po x
(y – HY),
oy
Var(X|Y = y) = ož(1 – p*).
E(X|Y = y) = µx +
Verify the theorem of total expectation (TTE).
The following verifications are proposed.
(a) Since Var(X|Y = y) = o} (1 p) does not depend on y, therefore, the TTE holds.
(b) Since E(XY = y) depends only on y, the TTE holds.
(c) Since E(X|Y) = µx + X (Y
Hy), we see that E(E(X|Y)) = ux = E(X). Therefore, the TTE holds.
(d) TTE does not hold.
(e) None of the above
The correct verification is
(a)
(b)
(c)
(d)
(e)
N/A
(Select One)
Transcribed Image Text:(9) When X, Y have a bivariate normal density with respective means ux: HY, respective variables o, o and correlation p, we have po x (y – HY), oy Var(X|Y = y) = ož(1 – p*). E(X|Y = y) = µx + Verify the theorem of total expectation (TTE). The following verifications are proposed. (a) Since Var(X|Y = y) = o} (1 p) does not depend on y, therefore, the TTE holds. (b) Since E(XY = y) depends only on y, the TTE holds. (c) Since E(X|Y) = µx + X (Y Hy), we see that E(E(X|Y)) = ux = E(X). Therefore, the TTE holds. (d) TTE does not hold. (e) None of the above The correct verification is (a) (b) (c) (d) (e) N/A (Select One)
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