Statistics - Estimate an MA(2) parameters process using GMM

[theend3r]

I'm a̶ ̶b̶i̶t̶ completely lost here and would welcome any help. I have data in this dataset which can be appoximately described by a simple MA(2) equation (according to the ACF of the data) with zero mean:

x(t) = ε(t) + θ1 ε(t-1) + θ2 ε(t-2).

What I need to do is to find the moments that would help me estimate the θ1 and θ2 using generalized method of moments.

Full instructions would be invaluable but any help is appreciated.

Edited November 28, 2014 by theend3r

[MBobrik]

moments won't help you, because they don't deal with time delayed functions. I would simply go for least squares method.

E = sum (1/2 *(e(t) + f1 *e(t-1) + f2*e(t-2) - x(t) )^2) = min

dE/df1 = sum ( (e(t) + f1 *e(t-1) + f2*e(t-2) - x(t) ) *e(t-1) ) = 0

dE/df2 = sum ( (e(t) + f1 *e(t-1) + f2*e(t-2) - x(t) ) *e(t-2) ) = 0

sum( x(t)*e(t-1) ) - sum( (e(t)*e(t-1) ) = f1 * sum( e(t-1) * e(t-1) ) + f2 * sum( e(t-1) * e(t-2) )

sum( x(t)*e(t-2) ) - sum( (e(t)*e(t-2) ) = f1 * sum( e(t-1) * e(t-2) ) + f2 * sum( e(t-2) * e(t-2) )

assuming

sum(e(t) ^ 2) = sum( e(t-1) * e(t-1) ) = sum( e(t-2) * e(t-2) )

sum( (e(t)*e(t-1) ) = sum( (e(t-1)*e(t-2) )

thus

sum( x(t)*e(t-1) ) - sum( (e(t)*e(t-1) ) = f1 * sum( e(t)^2 ) + f2 * sum( e(t) * e(t-1) )

sum( x(t)*e(t-2) ) - sum( (e(t)*e(t-2) ) = f1 * sum( e(t) * e(t-1) ) + f2 * sum( e(t)^2 )

substitute

A = sum( x(t)*e(t-1) ) - sum( (e(t)*e(t-1) )

B = sum( x(t)*e(t-2) ) - sum( (e(t)*e(t-2) )

C = sum( e(t)^2 )

D = sum( (e(t)*e(t-1) )

we get

A = C*f1+D*f2

B = D*f1+C*f2

and thus

(A*C-B*D)/(C^2-D^2) = f1

(B*C-A*D)/(C^2-D^2) = f2

[theend3r]

It definitely is possible. I found a paper touching the subject: http://annales.ensae.fr/anciens/n5556/vol5556-13.pdf, page 321, although I'm not entirely sure how to apply it to my case.

Thanks for your input.

Edited November 28, 2014 by theend3r

[MBobrik]

This is just juggling with labels. If you look at my derivation you will notice terms like sum( (e(t)*e(t-1) ) or sum( x(t)*e(t-1) ) which are just mixed moments of normal and time delayed variables. So at the end of the day the f1 and f2 estimators are expressed in terms of moments after all. I just forgot to call them that.

[theend3r]

It seemed that way to me but I wasn't entirely sure. I'll have a look at it, thanks a lot.

KSP forums really are awesome

[MBobrik]

Technically, I used another equivalent derivation of the same result. Moment method proper would take the regular and mixed moments and ask ( in the form of equations ) for what values of f1 and f2 we will get the same moments, thus arriving at the same formula by different means.