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I have got this task at the Time Series course as a part of Statistical minor. I am math major, and have gone through basic Probability (read:measure theory) course.

Let us have $y_t = 0.4y_{t-1}+2+\epsilon_t$

This is AutoRegression process, and I am asked to find $E(y_t)$ and write the corresponding $MA(\infty)$ process (if it's possible).

If the process was $y_t = 0.4*y_{t-1}+\epsilon_t$ then the mean would be 0 (I don't know why) and the corresponding $MA$ would be $y_t = \epsilon_t+0.4 \epsilon_{t-1} + 0.16 \epsilon_{t-2} ....$.

Is the corresponding MA correctly defined in this case? What is the definition of equivalent AR and MA processes, by the way?

Here $y_t = 0.2y_{t-1}+3+e_t$ is written as $(1-0.2L)(y_t- 3.75) = e_t$

Why is it so? UPD ( because there is one more progression and $3 * {10 \over 8} = 3.75)$

After plugging the appropriate $y_{t-n}$ I get

$y_t = (1+0.4L+0.16L^2...)e_t + 2*(1/(1-0.4))= (1+0.4L+0.16L^2...)e_t + {10 \over 3}$

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    $\begingroup$ you need to add self-study tag, and also see the guide on the tag $\endgroup$ – Aksakal Aug 28 at 16:16
  • $\begingroup$ stats.stackexchange.com/tags/self-study/info $\endgroup$ – Aksakal Aug 28 at 16:18
  • $\begingroup$ The addition of +2 at each stage turns a stationary AR(1) process into a non-stationary one with the a growing mean (you are assuming e$_t$ has mean 0 and a constant variance). A stationary finite autoregressive process has an inifnite moving average representation. It is just a mathematical way to describe the exact same process. $\endgroup$ – Michael Chernick Aug 28 at 16:28
  • $\begingroup$ @MichaelChernick Why does the mean grow? If there wasn't any \varepsilon, then I would say that E(y) = 10/3, because 10/3 = 0.4 * 10/3 + 2 $\endgroup$ – Lada Dudnikova Aug 28 at 16:34
  • $\begingroup$ The noise term continues to have a 0 mean but there is a deterministic straight line component with a slope of 2 that will eventually cause the process to grow indefinitely. $\endgroup$ – Michael Chernick Aug 28 at 17:16
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Without giving you the solution, the hint is to keep plugging $y_t$ into the equation until you see what's going on:

$$y_t=2+\varepsilon_t+0.4\times(2+\varepsilon_{t-1}+0.4\times(2+\varepsilon_{t-2}+0.4\times \dots))$$

After re-arranging the terms, it should look almost like without constant, but with a change that should be easy to figure out how to express it in a short expression

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  • $\begingroup$ Edited question. I hope that works this way. $\endgroup$ – Lada Dudnikova Aug 28 at 16:35
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The method that Aksakal used is fine but this is where the use of geometric series makes things easier:

We have,

$ y_t = \rho y_{t-1} + 2 + \epsilon_{t} $

So, this can be re-written as $ y_{t}(1 - \rho L) = 2 + \epsilon_{t} $

Then, we can divide both sides by $(1- \rho L )$ but $\frac{1}{1-\rho L}$ can be written as an infinite series so we obtain

$ y_{t} = \sum_{i=0}^{\infty} \rho^{i}(2 + \epsilon_{t-i}) $

$ = \sum_{i=0}^{\infty} (2 \rho^{i} + \rho^{i} \epsilon_{t-i})$

Note that this is an MA($\infty$} process with a non-zero mean equal to $\frac{2}{1-\rho}$.

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