ARMA forecasting - forecast error variance

In summary, the person is struggling to theoretically quantify the benefit of using an ARMA model for generating forecasts compared to using the mean level of the process. They believe that the ratio of variance of forecast error using ARMA to variance of forecast error using the mean will give them what they need. However, they have misunderstood the Box and Jenkins equation for the variance of ARMA forecast errors for various lead times and are seeking clarification. They have also overlooked a key component of the theory in their dissertation analysis and would like a better understanding of it.
  • #1
renucrew
6
0
Hi I've got an ARMA model, and I am 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 can't 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
 
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  • #2
renucrew said:
the mean square error is var((0-Xt+1)^2) ≈ 2.

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" ?

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

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

Is that supposed to be Xt+1 or Xt+1 ? If it is the "mean square", why use the notation "var" ?

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!
 
  • #4
renucrew said:
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.

[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.
 
  • #5


Dear researcher,

Your understanding of the B&J equation for the variance of ARMA forecast errors is correct. However, it is important to note that the equation assumes a stationary process, which may not always be the case in real-world scenarios. Additionally, the ratio of forecast error variance between an ARMA model and the mean level of the process may not always be a reliable measure of the benefit of using an ARMA model for forecasting.

In order to better quantify the benefit of using an ARMA model, it may be helpful to consider other metrics such as mean absolute error or root mean square error. These metrics take into account the magnitude of the forecast errors, rather than just their variance, and can provide a more comprehensive evaluation of the model's performance.

Furthermore, it is important to carefully select the appropriate ARMA model for your data and to validate its assumptions before using it for forecasting. This can help to ensure that the model is accurately capturing the correlation and dynamics of the process.

I hope this information helps. If you need further assistance, please do not hesitate to reach out. Good luck with your research.

Best,
 

1. What is ARMA forecasting?

ARMA (AutoRegressive Moving Average) forecasting is a statistical analysis method used to make predictions about future values of a time series data. It combines two models - the autoregressive (AR) model, which takes into account the relationship between current and past values, and the moving average (MA) model, which considers the relationship between current and past error terms.

2. How is ARMA forecasting different from other forecasting methods?

ARMA forecasting is different from other forecasting methods in that it takes into account both the trend and seasonality of a time series data, while also considering the impact of past values and error terms. This makes it a more comprehensive and accurate forecasting method compared to others like simple moving average or exponential smoothing.

3. What is the forecast error variance in ARMA forecasting?

The forecast error variance in ARMA forecasting is a measure of the accuracy of the predicted values. It is the average squared difference between the predicted values and the actual values, and is used to evaluate the performance of the ARMA model. A lower forecast error variance indicates a better fit of the model to the data.

4. How is the forecast error variance calculated in ARMA forecasting?

The forecast error variance is calculated by taking the mean squared error (MSE) of the predicted values. The MSE is the average of the squared differences between the predicted values and the actual values. It is a commonly used metric to evaluate the performance of time series forecasting models.

5. Can the forecast error variance in ARMA forecasting be reduced?

Yes, the forecast error variance in ARMA forecasting can be reduced by selecting the appropriate ARMA model that best fits the data, and by optimizing the model parameters. It is also important to regularly review and update the model as new data becomes available, to ensure that it continues to provide accurate predictions.

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