Time Series: stationary AR(1) -> MA(infinity)

[kingwinner]

Theorem:
A stationary AR(1) model can be expressed in terms of MA(infinity).

Proof:

Now I don't understand how they get from the second last line to the last line. Where did the term Y_t-m go?

I understand you can keep doing the substitution iteratively, but you always have to end up with a "Y" term no matter how many times you do it, but on the last line of the proof there is no "Y" term, so it magically disappared? I'm really confused.

Hopefully someone can explain this (in simple terms if possible). Thank you!

 

[pmsrw3]

They're taking the limit as [itex]\phi[/itex] goes to infinity. What happens to [itex]\phi^m[/itex] in that limit?

 

[kingwinner]

I think it's the limit as m->∞.

As m->∞,
φ^m ->0
BUT you also have to look at what happens to the other term, right? What if Y_t-m->∞? Then we have indeterminate form 0*∞, and we just can't conclude that the product->0.
(Also, another complication is that Y_t-m is a random variable, so I'm not sure if all of these make sense or not.)

 

[pmsrw3]

I think it's the limit as m->∞.

You're right -- sorry.

As m->∞,
φ^m ->0
BUT you also have to look at what happens to the other term, right? What if Y_t-m->∞? Then we have indeterminate form 0*∞, and we just can't conclude that the product->0.

(Also, another complication is that Y_t-m is a random variable, so I'm not sure if all of these make sense or not.)

I'll take those two complications in reverse order. First, the equation will in general be true "almost surely", meaning with probability 1. It may be possible for [itex]\phi^m y_{t-m}[/itex] to approach a constant or diverge, but this happens with probability 0. Second, you have to know SOMETHING about the y's. If this is the same book that your random walk problem came from, maybe they left out essential information. But the key fact is there in the word "stationary". It takes a little work to prove it, but If y is a decently-behaved RV (i.e., has a CDF) and is stationary and [itex]|\phi|<1[/itex], then [itex]\lim_{m\to\infty}\phi^m y_{t-m}=0[/itex] almost surely.

 

[blue_raver22]

Dear reader,

Thank you for your question. I understand your confusion and will try to explain the proof in simple terms.

First, it is important to understand that the proof is using a mathematical technique called "mathematical induction". This technique is commonly used to prove statements that involve a sequence of numbers or terms.

In this case, the proof is showing that the stationary AR(1) model can be expressed in terms of MA(infinity). This means that we can write the AR(1) model using an infinite number of MA terms.

The proof starts with the AR(1) model, which is written as Yt = a + φYt-1 + et. This is the second last line of the proof. The term Yt-m is not shown in this line, but it is included in the et term. Et represents the error term, which includes all the past Y terms up to Yt-m.

The next step of the proof is to substitute the AR(1) model into itself, using the Yt-1 term. This is the second line of the proof. This is where the Yt-m term appears. It is included in the et-1 term.

The proof continues to substitute the AR(1) model into itself, using the Yt-2, Yt-3, and so on, until we reach the last line of the proof. At this point, the Yt-m term has been substituted an infinite number of times, and it has disappeared completely. This is because the AR(1) model is stationary, which means that the values of Yt do not depend on the specific time period t. Therefore, the Yt-m term can be replaced by a constant value, which is represented by the term a in the last line of the proof.

In summary, the Yt-m term is not shown in the last line of the proof because it has been substituted an infinite number of times and has been replaced by a constant value. I hope this explanation helps to clarify the proof for you. If you have any further questions, please do not hesitate to ask. Thank you.