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Considering the following MA(3) process: y!u! 0.7u!"# - 0.2u!"$+ 0.4u!"% Where u! is a white noise process with variance equal to 1. What is the value y& = cov(y!, y!"&) for k = 0? Provide the correct answer along with the working steps and underlying assumptions used to calculate the value of y& = cov(y!, y!"&) for k = 0. i. y = 1.69 ii. Y' = 0.63 iii. y' = 1

Considering the following MA(3) process: y!u! 0.7u!"# - 0.2u!"$+ 0.4u!"% Where u! is a white noise process with variance equal to 1. What is the value y& = cov(y!, y!"&) for k = 0? Provide the correct answer along with the working steps and underlying assumptions used to calculate the value of y& = cov(y!, y!"&) for k = 0. i. y = 1.69 ii. Y' = 0.63 iii. y' = 1

MATLAB: An Introduction with Applications
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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Considering the following MA(3) process: y! = u! 0.7u!"# 0.2u!"$+0.4u!"% Where u! is a white noise process with variance equal to 1. What is the value y& = cov(y!, y!"&) for k = 0? Provide the correct answer along with the working steps and underlying assumptions used to calculate the value of y& = cov(y!, y!"&) for k = 0. i. y = 1.69 ii. Y' = 0.63 iii. y = 1 iv. Y' = 0
Transcribed Image Text:Considering the following MA(3) process: y! = u! 0.7u!"# 0.2u!"$+0.4u!"% Where u! is a white noise process with variance equal to 1. What is the value y& = cov(y!, y!"&) for k = 0? Provide the correct answer along with the working steps and underlying assumptions used to calculate the value of y& = cov(y!, y!"&) for k = 0. i. y = 1.69 ii. Y' = 0.63 iii. y = 1 iv. Y' = 0
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Why do we get cov(y0, y0) instead of cov(yt, yt) ?

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