Simple Question Seasonal Autoregressive Model

[Time Serious]

Hi, I hope I'm posting this in an appropriate place. It's not exactly linear regression.

If you have a seasonal moving average model like \(\displaystyle ARIMA(0,0,0)(0,0,1)_{12}\), and you want to find the expectation of \(\displaystyle y_t\), then you just use the fact that expectation is a linear operator and that the expected value of any error term is zero. However if you have a seasonal auto-regressive model like \(\displaystyle ARIMA(0,0,0)(2,0,0)_{12}\) for example, it will look like \(\displaystyle y_t=cy_{t-24} + \varepsilon_t\). So trying to compute \(\displaystyle E(y_t),\) you get \(\displaystyle E(y_t)=cE(y_{t-12})+\displaystyle E(\varepsilon_t)=cE(y_{t-12})\) but where do you go from there?

 

[vinux]

There is a small correction ARIMA(0,0,0)(2,0,0) is

\( y_t = c_0 + c_1 y_{t-12} + c_2 y_{t-24} +\epsilon_t\)
The constant part is zero in pure SARIMA model.
Hint:
The derivation of expectation is similar to the derivation of expectation of AR(2).
Use repeated substitution or stationary property

 

[Time Serious]

OK, I have two attempts.

[TEX]\displaystyle E(y_t)=c_1E(y_{t-12})\ plus\ c_2E(y_{t-24})=c_1^2(y_{t-24})\ plus\ c_2^2(y_{t-36})=c_1^3(y_{t-36})\ plus\ c_2^3(y_{t-48})=\ldots[/TEX]
Unfortunately I don't know when to stop. Can you tell me where I'd find a derivation for AR(2)?

However, perhaps this will work.
[TEX]\displaystyle E(y_t)=c_1E(y_{t-12})\ plus\ c_2E(y_{t-24})=c_1E(y_t)\ plus\ c_2E(y_t)[/TEX] so that
[TEX]\displaystyle E(y_t)(1-c_1-c_2)=0[/TEX] so that [TEX]\displaystyle E(y_t)=0[/TEX] unless [TEX]\displaystyle 1=c_1 \ plus\ c_2[/TEX]? Is this right?

(I use the word "plus" because the + does not appear in TeX for some reason.)

 

[vinux]

Use \( insted of [tex] i can't really read your post.\)

 

[Time Serious]

\( works much better. So here is the post in readable form:

OK, I have two attempts.

\( E(y_t)=c_1E(y_{t-12})+ c_2E(y_{t-24})=\) \(c_1^2E(y_{t-24})+ c_2^2E(y_{t-36})=c_1^3E(y_{t-36})+c_2^3E(y_{t-48})=\ldots\)
Unfortunately I don't know when to stop. Can you tell me where I'd find a derivation for AR(2)?

However, perhaps my second attempt will work:
\(\displaystyle E(y_t)=c_1E(y_{t-12})+ c_2E(y_{t-24})=c_1E(y_t)+ c_2E(y_t)\) so that
\(\displaystyle E(y_t)(1-c_1-c_2)=0\) so that \(\displaystyle E(y_t)=0\) unless \(\displaystyle 1=c_1 + c_2\)? Is this right?\)

 

[vinux]

You are right. If c0 =0, the expectation of yt is zero.

\(
Unfortunately I don't know when to stop. Can you tell me where I'd find a derivation for AR(2)?
\)

\(


Substitution you need to int terms of white noise.\)

 

[Time Serious]

Substitution you need to int terms of white noise.

Thanks for your help. I'm afraid I don't know what you mean by the quoted text. I'm not sure if "int" should be "in" but even then I don't know what to do.

 

[vinux]

White noise is [tex]\epsilon_t [/tex] in your model. The assumption on [tex]\epsilon_t [/tex] are, there are uncorrelated with mean 0 and constant variance.

You have considered this in your derivation( E[et] =0). You need to use the property [tex]0<c_1,c_2 <1 [/tex] and [tex] c_1^n, c_2^n \rightarrow 0 [/tex] as n large.

 

[Time Serious]

Right, how about this
\(E(y_t)=c_1^n(E(y_t))+c_2^n(E(y_t))\ \forall n\)
so that
\(E(y_t)=\lim_{n\to\infty}E(y_t)=E(y_t)\lim_{n\to\infty}(c_1^n+c_2^n)=E(y_t).0=0\)?

Is there something else I should do?

 

[BGM]

Basically you want to invert an \( \text{AR}(2) \) representation into \( \text{MA}(\infty) \) representation. See

http://faculty.chicagobooth.edu/john.cochrane/research/papers/time_series_book.pdf

p. 14 - 16 for the example, using the lag/backshift operator, with partial fraction and the geometric series, you can derive the proof. Finally you should obtain the result on the top of p. 16

 

[Time Serious]

Basically you want to invert an \( \text{AR}(2) \) representation into \( \text{MA}(\infty) \) representation. See

http://faculty.chicagobooth.edu/john.cochrane/research/papers/time_series_book.pdf

p. 14 - 16 for the example, using the lag/backshift operator, with partial fraction and the geometric series, you can derive the proof. Finally you should obtain the result on the top of p. 16

Thanks, I'll check that out.