Consider two random variables X, Y (not necessarily independent), and two real numbers a and b. Prove that Var(aX+bY) = a² Var(X)+b² Var(Y)+2ab Cov(X, Y). (Hint: Use the properties of Cov() reviewed in class.) - Prove that E[{X – E(X)}*] = E(X²) – {E(X)}², i.e., the two expressions (b) for Var(X) are indeed equivalent. (c) Prove that E[{X– E(X | Y)}² | Y] = E(X² | Y) – {E(X | Y)}², i.e., the %3D two expressions for Var(X | Y) are indeed equivalent.
Consider two random variables X, Y (not necessarily independent), and two real numbers a and b. Prove that Var(aX+bY) = a² Var(X)+b² Var(Y)+2ab Cov(X, Y). (Hint: Use the properties of Cov() reviewed in class.) - Prove that E[{X – E(X)}*] = E(X²) – {E(X)}², i.e., the two expressions (b) for Var(X) are indeed equivalent. (c) Prove that E[{X– E(X | Y)}² | Y] = E(X² | Y) – {E(X | Y)}², i.e., the %3D two expressions for Var(X | Y) are indeed equivalent.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Question 1
Consider two random variables X, Y (not necessarily independent), and two
(a)
real numbers a and b. Prove that Var(aX+bY) = a² Var(X)+b² Var(Y)+2ab Cov(X, Y).
(Hint: Use the properties of Cov() reviewed in class.)
Prove that E[{X - E(X)}²] = E(X²) – {E(X)}², i.e., the two expressions
(b)
for Var(X) are indeed equivalent.
Prove that E[{X – E(X | Y)}² | Y] = E(X² | Y) – {E(X | Y)}², i.e., the
two expressions for Var(X | Y) are indeed equivalent.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0d90012-896b-4f52-8f0b-e38034636c35%2F2acda21b-0a3b-429b-bc99-4a7aee403781%2F64ui4ab_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 1
Consider two random variables X, Y (not necessarily independent), and two
(a)
real numbers a and b. Prove that Var(aX+bY) = a² Var(X)+b² Var(Y)+2ab Cov(X, Y).
(Hint: Use the properties of Cov() reviewed in class.)
Prove that E[{X - E(X)}²] = E(X²) – {E(X)}², i.e., the two expressions
(b)
for Var(X) are indeed equivalent.
Prove that E[{X – E(X | Y)}² | Y] = E(X² | Y) – {E(X | Y)}², i.e., the
two expressions for Var(X | Y) are indeed equivalent.
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