# Revert AR process with constant

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

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

• you need to add self-study tag, and also see the guide on the tag – Aksakal Aug 28 at 16:16
• stats.stackexchange.com/tags/self-study/info – Aksakal Aug 28 at 16:18
• 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. – Michael Chernick Aug 28 at 16:28
• @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 – Lada Dudnikova Aug 28 at 16:34
• 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. – Michael Chernick Aug 28 at 17:16

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

• Edited question. I hope that works this way. – Lada Dudnikova Aug 28 at 16:35

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 $$\frac{2}{1-\rho}$$.