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01-05-2015 07:16 AM

Hi,

I have an estimated ARIMA model, and based on that I have the forecast for the next x periods.

I need the prediction and **prediction (forecast) intervals** for the following:

*0.5* forecast_lead_2+forecast_lead_7*

In the below example the prediction (forecast) is:

*0.5* 118440.4517 + 119798.8033*

How can I calculate intervals?

**proc** **arima** data=sashelp.citimon;

i var=CCIUAC(**1**);

e p=**6** q=**1**;

f lead=**12** out=forecast;

**run**;

**quit**;

In a general case I need prediction for the linear combination of forecasts with **prediction intervals**.

Thx

Accepted Solutions

Solution

01-06-2015
09:17 AM

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Posted in reply to gergely_batho

01-06-2015 09:17 AM

I tried to use the** Inpulse Respoonse Function** of the model to get the mentioned forecast intervals:

Each random shock is independent, so I can add the variance (sigma**2) *weighted* by the IRF. Then take the square root.

This is also how the simple forecast intervals (Standard Errors) are calculated.

Now I only need to be careful with the linear combination. For example the effect of the random shock **at lead(2)** - remember, we are calculating std error for lead(7) – is: **0.5*irf(0) + irf(5) **

**Anybody can see a theoretical error in this approach?**

PROC VARMAX can be used, to get the IRF.

Here's a code:

proc varmax data=sashelp.citimon;

model CCIUAC / dif(CCIUAC(1)) q=1 p=4 /*6*/ method=ml print=(impulse=(simple accum )) lagmax=6 ;

output out=vforecast lead=12;

run;

data _null_;

sigma=sqrt(586153.48852); /* number copied here from the output of VARMAX*/

array irf[0:7] (1, 0.33369, 0.29371, 0.46347, 0.23512, 0.21355, 0.22610, 0.14592);/*Impulse Response - numbers copied here from the output of VARMAX*/

array airf[0:7];/*Accumulated Impulse Response*/

airf[0]=1; do i=1 to 7; airf*=airf[i-1]+irf ;end;/*calculating airf from irf*/*

/*Calculating StdErr for forecast lead 2*/

do i=0 to 1;

std_2+airf***2;*

end;

std_2=sqrt(std_2)*sigma;

/*Calculating StdErr for forecast lead 7*/

do i=0 to 6;

std_7+airf***2;*

end;

std_7=sqrt(std_7)*sigma;

/*Calculating StdErr for the linear combination*/

/*It is similar to a convolution operator, but you sum the squares*/

do i=0 to 6;

if 0<=i<=4 then do;/**/

** lincomb+airf **2;**

end;

else do;/*5 and 6*/

** lincomb+(airf +0.5*airf[i-5])**2;**

end;

end;

**lincomb=sqrt(lincomb)*sigma;**

putlog sigma= std_2= std_7=** lincomb=**;

run;

Perhaps SAS/IML can do this with less coding.

Thx

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Posted in reply to gergely_batho

01-05-2015 08:08 AM

Post it at SAS Forecasting and Econometrics

Udo@sas is good at it .

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Posted in reply to Ksharp

01-05-2015 10:32 AM

Hello -

This paper might be of interest: http://support.sas.com/resources/papers/proceedings12/341-2012.pdf

Thanks,

Udo

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Posted in reply to udo_sas

01-05-2015 04:35 PM

Thank you Udo,

The mentioned paper is about combining forecasts of different models. But I have only one model, and I want to forecast (calculate) the linear combination of forecasts of two different periods.

On the other hand the suggested paper contains a description about calculating forecast confidence intervals using the **sample cross correlation**. Maybe that idea could be useful in my situation.

I still think it should be possible (somehow) to derive the cross correlation between *forecast_lead_2* and *forecast_lead_7* simply by using the model parameter estimates.

Or maybe this (linear combination) forecast interval could be directly calculated using the forecast intervals of the original forecasts*.*

Thx

Solution

01-06-2015
09:17 AM

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Posted in reply to gergely_batho

01-06-2015 09:17 AM

I tried to use the** Inpulse Respoonse Function** of the model to get the mentioned forecast intervals:

Each random shock is independent, so I can add the variance (sigma**2) *weighted* by the IRF. Then take the square root.

This is also how the simple forecast intervals (Standard Errors) are calculated.

Now I only need to be careful with the linear combination. For example the effect of the random shock **at lead(2)** - remember, we are calculating std error for lead(7) – is: **0.5*irf(0) + irf(5) **

**Anybody can see a theoretical error in this approach?**

PROC VARMAX can be used, to get the IRF.

Here's a code:

proc varmax data=sashelp.citimon;

model CCIUAC / dif(CCIUAC(1)) q=1 p=4 /*6*/ method=ml print=(impulse=(simple accum )) lagmax=6 ;

output out=vforecast lead=12;

run;

data _null_;

sigma=sqrt(586153.48852); /* number copied here from the output of VARMAX*/

array irf[0:7] (1, 0.33369, 0.29371, 0.46347, 0.23512, 0.21355, 0.22610, 0.14592);/*Impulse Response - numbers copied here from the output of VARMAX*/

array airf[0:7];/*Accumulated Impulse Response*/

airf[0]=1; do i=1 to 7; airf*=airf[i-1]+irf ;end;/*calculating airf from irf*/*

/*Calculating StdErr for forecast lead 2*/

do i=0 to 1;

std_2+airf***2;*

end;

std_2=sqrt(std_2)*sigma;

/*Calculating StdErr for forecast lead 7*/

do i=0 to 6;

std_7+airf***2;*

end;

std_7=sqrt(std_7)*sigma;

/*Calculating StdErr for the linear combination*/

/*It is similar to a convolution operator, but you sum the squares*/

do i=0 to 6;

if 0<=i<=4 then do;/**/

** lincomb+airf **2;**

end;

else do;/*5 and 6*/

** lincomb+(airf +0.5*airf[i-5])**2;**

end;

end;

**lincomb=sqrt(lincomb)*sigma;**

putlog sigma= std_2= std_7=** lincomb=**;

run;

Perhaps SAS/IML can do this with less coding.

Thx