Question

Let $\epsilon_t \sim i.i.d.N(-2, 9)$. In addition, $y_t$ follows the process in (1): $y_t = 1 + 0.75y_{t-1} + 0.25y_{t-2} + \epsilon_t$ 1. What does it mean for a process to be stationary? Answer: 2. Is $y_t$ in eqn (1) stationary? Explain. Answer: (1) 

          Let $\epsilon_t \sim i.i.d.N(-2, 9)$. In addition, $y_t$ follows the process in (1):
$y_t = 1 + 0.75y_{t-1} + 0.25y_{t-2} + \epsilon_t$
1. What does it mean for a process to be stationary?
Answer:
2. Is $y_t$ in eqn (1) stationary? Explain.
Answer:
(1)
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Let ∼ i.i.d.N(-2, 9). In addition, yt follows the process in (1): yt = 1 + 0.75yt-1 + 0.25yt-2 + 1. What does it mean for a process to be stationary? Answer: 2. Is yt in eqn (1) stationary? Explain. Answer: (1)

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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Let $t$ be i.i.d. $N(-2,9)$. In addition, $y_t$ follows the process in (1): $y_t = 1 + 0.75y_{t-1} + 0.25y_{t-2} + t$ (1)
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Transcript

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00:01 In this question we have been given an expression and we need to simplify the expression and also we are required to verify the answer.
00:08 So the expression is y t plus 1 minus 2y t and that is equal to 2 raise to the power t and then here we can say that this expression can be written as y of t minus 2y of 1 and this is equals to 2 t minus 1 and hence from it will be like 2 y t and this is minus 2 raised to the power 2 y of t minus 1 and that is equal to 2 multiplied with t and then here we can say it is going to be over here like 2 raised to the power 2 y of t minus 1 and that is minus 2 raised to the power 3 and then here we can say it will be y t minus 2 t minus 2 and that is equal to here 2 t and then we can say that if we proceed further what will happen is we can say it will go on as such and then it will be 2 y2 and then it is minus here it will be 2 t and then it will be y1 and that is equals to 2 raise to the power t so this equation has been written in continuation to what we were earlier doing right so if we add all these equations what will it be it will be basically y, t plus 1 minus 2t, and then it will be y1, and that is equal to t multiplied with 2t...
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