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ARMA forecasting - forecast error variance

  1. Apr 23, 2012 #1
    Hi ive got an ARMA model, and im struggling to theoretically quantify the benefit of using it to generate forecasts for various lead times, compared to using the mean level of the process. I think the ratio of variance of forecast error using ARMA to variance of forecast error using the mean will give me what i need. I really need help though!

    If the process is standard normal then the best estimate of the process without an ARMA model is simply 0, and the variance of the mean square error is var((0-Xt+1)^2) ≈ 2.

    The equation given in box and jenkins for the variance of ARMA forecast errors for various lead times, l:

    var(e(l)) = (1 + (ψ1)^2 + (ψ2)^2 + ... +ψ(l-1)^2) (σa)^2

    is always 1 for lead time 1, irrespective of the parameters or level of correlation of the model so according to my misunderstanding of the B&J all standard normal ARMA processes would result in a 50% decrease in forecast error variance for lead time 1. This cant be correct because surely the variance of the forecast errors would depend on the amount of correlation of the process.

    Any tips would be really useful!!!

    many thanks
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 24, 2012 #2

    Stephen Tashi

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    Is that supposed to be [itex] X_{t+1} [/itex] or [itex] X_t + 1 [/itex] ? If it is the "mean square", why use the notation "var" ?

    You shoud explain the quantities that appear in this equation if you expect anyone to interpret it.
  4. Apr 25, 2012 #3
    Hi sorry for the lack of clarity!

    Oops that is just meant to be Xt, and the expression just says that the variance of the E(Xt)=0 (that is using the mean as the best estimate) square forecast errors is around 2 if the process is standard normal.

    The var(e(l)) is correct, the expression gives the variance of the forecast error for lead time, l, as a function of the psi weights of the general linear filter form of the ARMA model.

    Box and Jenkins use this expression for the variance of forecast errors to derive the confidence intervals of forecasts for a given lead time, since a reduction in the variance of forecast errors represents an increase in forecast precision. But from my understanding of this expression the variance of the lead 1 forecast error is 1 irrespective of the psi weights of the model so i my understanding must be wrong since confidence interval of a lead 1 forecast using an ARMA model must be dependent on the strength of the lag 1 autocorrelation.

    Ive kind of glissed over this bit of theory in my dissertation analysis by simply saying that the accuracy of the forecast is dependent on the psi weights and lead time, and by evaluating the forecast accuracy of my model empirically but it would be good to understand the theory properly!
  5. Apr 25, 2012 #4

    Stephen Tashi

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    [itex] E(X_t) [/itex] is the expected value of a random variable. It would be some constant. Thus it would have zero variance.

    Are you trying to make a statement about [itex] E( (X_t - 0)^2 ) [/itex] ? If [itex] X_t [/itex] is a normal random variable with mean 0 and variance 1 then [itex] E( (X_t - 0)^2) [/itex] is the variance of [itex] X_t [/itex], which is 1.
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