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Topic: Using an AR(1) model for time series forecasting with missing observations
Replies: 25   Last Post: Mar 9, 2007 9:24 AM

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 David Jones Posts: 637 Registered: 12/7/04
Re: Using an AR(1) model for time series forecasting with missing observations
Posted: Feb 23, 2007 11:16 AM

iwan2no@hotmail.com wrote:
>
> Usually, for an AR(1) process, you would estimate the value of phi_1
> by using:
>
> phi_1 = rho_1 (that's after solving the Yule-Walker equations)
>
> In this case however, I don't have phi_1 and only have phi_m. So as
> you mentioned, I could use:
>
> rho_m = (phi_1)^m, which gives me: phi_1 = (rho_m)^(1/m) in order to
> estimate phi_1.
>
> It seems to me that using the above equation, phi_1 = (rho_m)^(1/m),
> phi_1 can be computed using ANY lag m. However, this isn't usually
> done. Instead, we stick to the Yule-Walker equations where only

rho_1
> is requird.
>
> 1) My question is, what's the difference between estimating phi_1
> using rho_1 and estimating phi_1 using rho_m? I would think that

rho_1
> is preferable since the smaller the lag (i.e. m) the better your
> estimate will be? Would that be the correct explanation?
>

There are two separate issues here:

(i) for the case where all observations are avilable for estimating
both rho_1 and rho_m, the usual theory does suggest using rho_1 as the
estimate of phi_1. This seems to be related to the result that the
maximum-likelihood estimate is very close to the result of regressing
X(t+1) on X(t), not X(t+2) on X(t).

(ii) for your case where you are contemplating not making step-1
observations, you need to work out whether you want to compare rho_1
estimated from N observations at step-1 with rho_m estimated either
from
(a) N/m observations at step-m (so the time period is the same) or
(b) N observations at step-1 (so the time period is Nm).
For case (a) you get a dramatic increaese in variance because the
sample size is smaller. Case (b) is more difficult and would need
looking at with more care.

>
> 2) Also, if rho_m is negative, that means that the sign of the
> autocorrelations shows an alternating pattern beginning with a
> negative value. We can also say for certain that -1 < phi_1 < 0. (Am

I
> right to say this?)
>
> But if this is the case, then if rho_m is negative, shouldn't I be
> able to compute phi_1 using: phi_1 = -(|rho_m|)^(1/m). Why is this
> wrong?

It may well be OK if you are happy that negative phi_1 is possible.
But it only makes sense of applied when m is odd. The theory says that
negative rho_m can't occur when is even.
>
>> negative phi_1. However this is unlikely to be a case you are
>> interested in. If you do get a negative estimate of rho_m , you

might
>> want to attribute this to sampling error. Either reduce m, or try
to
>> get a reasonable guess for phi_1 by replacing the estimate for
rho_m
>> by a small positive number.
>
> 3. Why do you say that it's unlikely that I'll be interested in a
> nagative phi_1?

A negative phi_1 means that succesive observations would tend to
alternate in sign which is a particular type of behaviour that you
would tend to notice. My impression was that you were talking about a
moderatly slow drift up and down. You are the best judge of whether
you want to allow a model which corresponds to the alternating
behaviour.

> And how do you justify attributing a negative phi_1 to
> a sampling error?

This should be rho_1. A negative estimate of rho_1 or rho_m is
certainly possible even of the true values are positive. But if you
know the true values must be positive you would be wise not to accept
the negative estimate.

> Are you trying to imply that a negative phi_1 is not
> possible? Is this because you feel that for a gradually changing
> attribute, for small values of m, you're most likely to get a

positive
> autocorrelation (i.e. rho) and so phi in turn will be positive?

For the AR(1), the theoretical autocorelation is either always
positive or alternates as the lag inceases by each 1-step change in
lag. If there is good evidence for other behaviours in your sample
correlations this would suggest that AR(1) is not a good model.

>
> 4. What's your justification for "try to get a reasonable guess for
> phi_1 by replacing the estimate for rho_m by a small positive

number"?

None really. But if you are trying to develop an automatic system you
will need to think what to do for exceptional cases. It is also
somewhat informative. For example if rho_m is 0.01 for m=10, this
corresponds to phi_1=0.63, which is quite large. You need a very large
sample to estimate a corration with an accuracy of plus or minus 0.01
(of the order of 10,000 pairs going into the correlation).

David Jones

Date Subject Author
2/15/07 Matlab Newbie
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