Autokovarianz eines ARMA (2,1) -Prozesses - Ableitung eines analytischen Modells für

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Ich muss analytische Ausdrücke für die Autokovarianzfunktion γ(k) eines ARMA (2,1) -Prozesses ableiten , die bezeichnet werden durch:

yt=ϕ1yt1+ϕ2yt2+θ1ϵt1+ϵt

Also, ich weiß das:

γ(k)=E[yt,ytk]

damit ich schreiben kann:

γ(k)=ϕ1E[yt1ytk]+ϕ2E[yt2ytk]+θ1E[ϵt1ytk]+E[ϵtytk]

Um dann die analytische Version der Autokovarianzfunktion abzuleiten, muss ich Werte von k - 0, 1, 2 ... einsetzen, bis ich eine Rekursion erhalte, die für alle k größer als eine ganze Zahl gültig ist .

Daher setze ich k=0 und arbeite dies durch, um zu erhalten:

γ(0)=E[yt,yt]=ϕ1E[yt1yt]+ϕ2E[yt2yt]+θ1E[ϵt1yt]+E[ϵtyt]

Jetzt kann ich die ersten beiden Begriffe vereinfachen und dann wie bisher durch ersetzen yt:

γ(0)=ϕ1γ(1)+ϕ2γ(2)+θ1E[ϵt1(ϕ1yt1+ϕ2yt2+θ1ϵt1+ϵt)]+E[ϵt(ϕ1yt1+ϕ2yt2+θ1ϵt1+ϵt)]

dann multipliziere ich die acht Terme, die sind:

+θ1ϕ1E[ϵt1yt1]+θ1ϕ2E[ϵt1yt2]+θ12E[(ϵt1)2]=θ12σϵ2+θ1E[ϵt1ϵt]=θ1E[ϵt1]E[ϵt]=0+ϕ1E[ϵtyt1]+ϕ2E[ϵtyt2]+θ1E[ϵtϵt1]=θ1E[ϵt]E[ϵt1]=0+E[(ϵt)2]=σϵ2

Daher muss ich die vier verbleibenden Bedingungen noch klären. Ich möchte für die Zeilen 1, 2, 5 und 6 dieselbe Logik verwenden wie für die Zeilen 4 und 7 - zum Beispiel für Zeile 1:

θ1ϕ1E[ϵt1yt1]=θ1ϕ1E[ϵt1]E[yt1]=0 because E[ϵt1]=0.

Similarly for lines 2, 5 and 6. But I have a model solution that suggests the expression for γ(0) simplifies to:

γ(0)=ϕ1γ(1)+ϕ2γ(2)+θ1(ϕ1+θ1)σϵ2+σϵ2

This suggests my simplification as described above would miss the term with the coefficient ϕ1 - which under my logic should be 0. Is my logic at fault, or is the model solution I found incorrect?

The worked solution also suggest that "analogously" γ(1) can be found as:

γ(1)=ϕ1γ(0)+ϕ2γ(1)+θ1σϵ2

and for k>1:

γ(k)=ϕ1γ(k1)+ϕ2(k2)

I hope the question is clear. Any assistance will be much appreciated. Thank you in advance.

This is a question related to my research, and is not in preparation for any exam or coursework.

hydrologist
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Antworten:

8

If the ARMA process is causal there is a general formula that provides the autocovariance coefficients.

ARMA(p,q)

yt=i=1pϕiyt1+j=1qθjϵtj+ϵt,
where ϵt is a white noise with mean zero and variance σϵ2. By the causality property, the process can be written as
yt=j=0ψjϵtj,
where ψj denotes the ψ-weights.

The general homogeneous equation for the autocovariance coefficients of a causal ARMA(p,q) process is

γ(k)ϕ1γ(k1)ϕpγ(kp)=0,kmax(p,q+1),
with initial conditions
γ(k)j=1pϕjγ(kj)=σϵ2j=kqθjψjk,0k<max(p,q+1).
QuantIbex
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2

Your calculation mistake in your original question lies in

θ1ϕ1E[ϵt1yt1]=θ1ϕ1E[ϵt1]E[yt1]=0(mistaken)

You cannot separate the expectation E[ϵt1yt1] - ϵt1 and yt1 are not independent.

Alecos Papadopoulos
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As you can see from my update (below) I realised this soon after completing the post - but many thanks for your help!
hydrologist
1

OK. So the process of writing the post actually pointed me to the solution.

Consider the Expectation terms 1, 2, 5 and 6 from above that I thought should be 0.

Immediately for terms 5 - E[ϵtyt1] - and 6 - E[ϵtyt2]: these terms are definitely zero, because yt1 and yt2 are independent of ϵt and E[ϵt]=0.

However, terms 1 and 2 look as though the Expectation is of two correlated variables. So, consider the expressions for yt1 and yt2 thus:

yt1=ϕ1yt2+ϕ2yt3+θ1ϵt2+ϵt1yt2=ϕ1yt3+ϕ2yt4+θ1ϵt3+ϵt2

And recall term 1 - ϕ1θ1E[ϵt1yt1]. If we multiply both sides of the expression for yt1 by ϵt1 and then take Expectations, it is clear that all terms on the right hand side except the last become zero (because the values of yt2, yt3, and ϵt2 are independent of ϵt1 and E[ϵt1]=0) to give:

E[ϵt1yt1]=E[(ϵt1)2]=σϵ2

So term 1 becomes +ϕ1θ1σϵ2. For term 2, it should be clear that, by the same logic, all terms are zero.

Hence the original model answer was correct.

However, if anyone can suggest an alternative way to obtain a general (even if messy) solution, I would be very pleased to hear it!

hydrologist
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