CODEWITHSUNDEEP
rforecasting
enk
enk

Reputation: 41

forecast differenced series in R

I'm new on R, I've always use e-views and now I have to do a regression

I have Xt stationary and Yt not stationary so I need to difference

yt=Yt-Y(t-1)

then the regression is

yt = a + bXt

how can I do the forecast on R and obtain the "real" value and not the differences?

in e-views is enouth to write d(Yt) but in R is impossible

Upvotes: 1

Views: 3455

Answers (3)

Stéphane
Stéphane

Reputation: 207

If you forecast the first difference h many days ahead, you have an estimate of how higher or lower the series will be in h many days. You just to add the last observed value to this difference to get the level:

dy_{t}:= y_{t} - y_{t-1}

You forecast dy_{T+h}, you know the last observation, y_{T}, so y_{T+h} = dy_{T+h} + y_{T}. Just augment all forecasts of the future value by the last observation and you get a level forecast.

If you used the first difference of logarithms, dlny_{t} := lny_{t} - lny_{t-1}, you have a first order Taylor approximation to a growth rate, so you'd forecasting growth rates. In that case, level forecasts would be y_{T+h} = (1+dlny_{T+h}) y_{T}.

You don't need any complicated function to do those transformations. In fact, because y_{T} is a scalar, if you have a vector for dy_{T+h} or dlny_{T+h} for all h, R will allow you to write the expressions I wrote above using scalars and vectors and R will manage dimension problems correctly on its own.

Upvotes: 0

Rob Hyndman
Rob Hyndman

Reputation: 31820

First, if you think X affects y, then you should difference both variables. Differencing only one of them leads to a model where X affects the change in y rather than y itself.

You could do this with the arima() function (and differencing both variables):

fit <- arima(y, xreg=X, order=c(0,1,0))

Then predictions on the undifferenced scale are obtained using

fcast <- predict(fit, n.ahead=10, newxreg=futureX)

where futureX contains the next 10 values of X.

If you really wanted to model the affect of X on d(y), then create a new variable

sumX <- cumsum(X)

and use that instead of X in the fit (and similarly modify futureX).

Upvotes: 4

mrip
mrip

Reputation: 15163

I think what you are looking for is cumsum, which is the inverse operation of diff (almost). You can recover a vector from its differences like this:

> z<-sample(20)
> dz<-diff(z)
> z0<-cumsum(c(z[1],dz))
> all(z==z0)
[1] TRUE

In your case, it would look something like this:

dY<-diff(Y)
dYhat<-lm(dY ~ X[-1])$fitted
Yhat<-cumsum(c(Y[1],dYhat))

Upvotes: 5

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