3. Recall that (Corollary 3.8 in the textbook) if two random variables X and Y are inde- pendent, then they must be uncorrelated i.e. Cov(X,Y)= 0. However, the converse is not true in general and this problem provides an example. Let X be a random variable with continuous uniform distribution on the interval [-1,1], i.e. its probability density function is given by f(x) = [1/2, if x € [-1,1], otherwise. a) Show that Cov(X, X²) = 0. b) Prove mathematically (not just argue by intuition) that X and X² are not inde- pendent. One way to do this is by showing that they do not satisfy the property: P(X € A, X² € B) = P(X E A) P(X² = B) for all A, BCR. You may also use other equivalent definitions of independence.
3. Recall that (Corollary 3.8 in the textbook) if two random variables X and Y are inde- pendent, then they must be uncorrelated i.e. Cov(X,Y)= 0. However, the converse is not true in general and this problem provides an example. Let X be a random variable with continuous uniform distribution on the interval [-1,1], i.e. its probability density function is given by f(x) = [1/2, if x € [-1,1], otherwise. a) Show that Cov(X, X²) = 0. b) Prove mathematically (not just argue by intuition) that X and X² are not inde- pendent. One way to do this is by showing that they do not satisfy the property: P(X € A, X² € B) = P(X E A) P(X² = B) for all A, BCR. You may also use other equivalent definitions of independence.
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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![3. Recall that (Corollary 3.8 in the textbook) if two random variables X and Y are inde-
pendent, then they must be uncorrelated i.e. Cov(X,Y)= 0. However, the converse is
not true in general and this problem provides an example.
Let X be a random variable with continuous uniform distribution on the interval [-1,1],
i.e. its probability density function is given by
f(x) =
(1/2, if xe [-1, 1],
0,
otherwise.
a) Show that Cov(X, X²) = 0.
b) Prove mathematically (not just argue by intuition) that X and X² are not inde-
pendent. One way to do this is by showing that they do not satisfy the property:
P(X E A, X² € B) = P(X € A) · P(X² € B)
.
for all A, BCR. You may also use other equivalent definitions of independence.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1652026a-6ba6-42d7-9e35-ad3e7516e1ec%2F0081e515-0f99-4257-9a03-18bb70da1d81%2Fll5pyh_processed.png&w=3840&q=75)
Transcribed Image Text:3. Recall that (Corollary 3.8 in the textbook) if two random variables X and Y are inde-
pendent, then they must be uncorrelated i.e. Cov(X,Y)= 0. However, the converse is
not true in general and this problem provides an example.
Let X be a random variable with continuous uniform distribution on the interval [-1,1],
i.e. its probability density function is given by
f(x) =
(1/2, if xe [-1, 1],
0,
otherwise.
a) Show that Cov(X, X²) = 0.
b) Prove mathematically (not just argue by intuition) that X and X² are not inde-
pendent. One way to do this is by showing that they do not satisfy the property:
P(X E A, X² € B) = P(X € A) · P(X² € B)
.
for all A, BCR. You may also use other equivalent definitions of independence.
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