Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e",
using the function "lag"? If so, how? If not, of what use is the
function "lag"? I get the same answer from y~x as y~lag(x), whether
using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider
the following:
> set.seed(1)
> x <- rep(c(rep(0, 4), 9), len=9)
> y <- (rep(c(rep(0, 5), 9), len=9)+rnorm(9)) # y[t] = x[t-1]+e
>
> lm(y~x)
(Intercept) x
1.2872 -0.1064
> lm(y~lag(x))
(Intercept) lag(x)
1.2872 -0.1064
> arima(y, xreg=x)
intercept x
1.2872 -0.1064
s.e. 0.9009 0.3003
sigma^2 estimated as 6.492: log likelihood = -21.19, aic = 48.38
> arima(y, xreg=lag(x))
intercept lag(x)
1.2872 -0.1064
s.e. 0.9009 0.3003
> arima(y, xreg=c(NA, x[-9]))
intercept c(NA, x[-9])
0.4392 0.8600
s.e. 0.2372 0.0745
sigma^2 estimated as 0.3937: log likelihood = -7.62, aic = 21.25
> arima(ts(y), xreg=lag(ts(x)))
arima(x = ts(y), xreg = lag(ts(x)))
intercept lag(ts(x))
1.2872 -0.1064
s.e. 0.9009 0.3003
sigma^2 estimated as 6.492: log likelihood = -21.19, aic = 48.38
Thanks for your help.
Spencer Graves
Spencer, You may want to peruse the list archive for posts that match 'ts' and are written by Brian Ripley -- these issues have come up before. The ts class is designed for arima and friends (like Kalman filtering), and very useful in that context, but possibly not so much anywhere else. lag() only shifts the _reference dates_ attached to the object. So in a data.frame context (as for lm()) .... nothing happens. Personally, I use its as my main container for daily or weekly data. There is also zoo, which I have meant to examine more closely for a while now. You want want to use one of those for shifting, matching, intersecting, ... and then use one of the exporter functions (core() for its) to pass to lm(), say. Hope this helps, Dirk -- Better to have an approximate answer to the right question than a precise answer to the wrong question. -- John Tukey as quoted by John Chambers
Spencer Graves <spencer.graves <at> pdf.com> writes:
:
: Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e",
: using the function "lag"? If so, how? If not, of what use is the
: function "lag"? I get the same answer from y~x as y~lag(x), whether
: using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider
: the following:
:
: > set.seed(1)
: > x <- rep(c(rep(0, 4), 9), len=9)
: > y <- (rep(c(rep(0, 5), 9), len=9)+rnorm(9)) # y[t] = x[t-1]+e
: >
: > lm(y~x)
: (Intercept) x
: 1.2872 -0.1064
: > lm(y~lag(x))
: (Intercept) lag(x)
: 1.2872 -0.1064
: > arima(y, xreg=x)
: intercept x
: 1.2872 -0.1064
: s.e. 0.9009 0.3003
: sigma^2 estimated as 6.492: log likelihood = -21.19, aic = 48.38
: > arima(y, xreg=lag(x))
: intercept lag(x)
: 1.2872 -0.1064
: s.e. 0.9009 0.3003
: > arima(y, xreg=c(NA, x[-9]))
: intercept c(NA, x[-9])
: 0.4392 0.8600
: s.e. 0.2372 0.0745
: sigma^2 estimated as 0.3937: log likelihood = -7.62, aic = 21.25
: > arima(ts(y), xreg=lag(ts(x)))
: arima(x = ts(y), xreg = lag(ts(x)))
: intercept lag(ts(x))
: 1.2872 -0.1064
: s.e. 0.9009 0.3003
: sigma^2 estimated as 6.492: log likelihood = -21.19, aic = 48.38
:
Here is some sample code:
R> # following 3 lines are from your post
R> set.seed(1)
R> x <- rep(c(rep(0, 4), 9), len=9)
R> y <- (rep(c(rep(0, 5), 9), len=9)+rnorm(9)) # y[t] = x[t-1]+e
R>
R> # here are some examples using ts class - first one uses no lag
R> lm(y ~ x, cbind(y = ts(y), x = ts(x)))
Call:
lm(formula = y ~ x, data = cbind(y = ts(y), x = ts(x)))
Coefficients:
(Intercept) x
1.2872 -0.1064
R> # now lets redo it with a lag.
R> lm(y ~ lagx, cbind(y = ts(y), lagx = lag(ts(x), -1)) )
Call:
lm(formula = y ~ lagx, data = cbind(y = ts(y), lagx = lag(ts(x), -1)))
Coefficients:
(Intercept) lagx
0.4392 0.8600
R> # here is arima without a lag
R> b <- cbind(ts(y), ts(x))
R> arima(b[,1], order = c(1,1,1), xreg = b[,2])
Call:
arima(x = b[, 1], order = c(1, 1, 1), xreg = b[, 2])
Coefficients:
ar1 ma1 b[, 2]
0.3906 -1.0000 -0.3803
s.e. 0.4890 0.4119 0.3753
sigma^2 estimated as 7.565: log likelihood = -20.2, aic = 48.4
R> # and now we redo arima with a lag
R> bb <- cbind(ts(y), lag(ts(x),-1))
R> arima(bb[,1], order = c(1,1,1), xreg = bb[,2])
Call:
arima(x = bb[, 1], order = c(1, 1, 1), xreg = bb[, 2])
Coefficients:
ar1 ma1 bb[, 2]
-0.2991 -0.8252 0.8537
s.e. 0.4516 1.0009 0.0838
sigma^2 estimated as 0.444: log likelihood = -7.9, aic = 23.8
R> # you can alternately use the I notation with lm and ts objects
R> # if you load zoo first
R> library(zoo)
R> yt <- ts(y); xt <- ts(x)
R> lm(I(yt ~ xt))
Call:
lm(formula = I(yt ~ xt))
Coefficients:
(Intercept) xt
1.2872 -0.1064
R> lm(I(yt ~ lag(xt, -1)))
Call:
lm(formula = I(yt ~ lag(xt, -1)))
Coefficients:
(Intercept) lag(xt, -1)
0.4392 0.8600
I use the following two function for a lagged regression: lm.lag=function(y,lag=1) summary(lm(embed(y,lag+1)[,1]~embed(y,lag+1)[,2:(lag+1)])) lm.lag.x=function(y,x,lag=1) summary(lm(embed(y,lag+1)[,1]~embed(x,lag+1)[,2:(lag+1)])) for, respectively, y_t=a+b_1*y_t-1+...+b_lag*y_t-lag y_t=a+b_1*x_t-1+...+b_lag*x_t-lag I am not quite sure whether this an answer to your question, but here are two examples: set.seed(7) ar1=arima.sim(n=300,list(ar=0.8)) lm.lag(ar1) lm.lag.x(ar1,ar1) set.seed(8) ar3=arima.sim(n = 200, list(ar = c(0.4, -0.5, 0.7))) lm.lag(ar3,3) lm.lag.x(ar3,ar3,3) Best wishes, Rem Saturday, March 5, 2005, 6:14:15 PM, you wrote: SG> Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e", SG> using the function "lag"? If so, how? If not, of what use is the SG> function "lag"? I get the same answer from y~x as y~lag(x), whether SG> using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider SG> the following: >> set.seed(1) >> x <- rep(c(rep(0, 4), 9), len=9) >> y <- (rep(c(rep(0, 5), 9), len=9)+rnorm(9)) # y[t] = x[t-1]+e >> >> lm(y~x) SG> (Intercept) x SG> 1.2872 -0.1064 >> lm(y~lag(x)) SG> (Intercept) lag(x) SG> 1.2872 -0.1064 >> arima(y, xreg=x) SG> intercept x SG> 1.2872 -0.1064 SG> s.e. 0.9009 0.3003 SG> sigma^2 estimated as 6.492: log likelihood = -21.19, aic = 48.38 >> arima(y, xreg=lag(x)) SG> intercept lag(x) SG> 1.2872 -0.1064 SG> s.e. 0.9009 0.3003 >> arima(y, xreg=c(NA, x[-9])) SG> intercept c(NA, x[-9]) SG> 0.4392 0.8600 SG> s.e. 0.2372 0.0745 SG> sigma^2 estimated as 0.3937: log likelihood = -7.62, aic = 21.25 >> arima(ts(y), xreg=lag(ts(x))) SG> arima(x = ts(y), xreg = lag(ts(x))) SG> intercept lag(ts(x)) SG> 1.2872 -0.1064 SG> s.e. 0.9009 0.3003 SG> sigma^2 estimated as 6.492: log likelihood = -21.19, aic = 48.38 SG> Thanks for your help. SG> Spencer Graves SG> ______________________________________________ SG> R-help at stat.math.ethz.ch mailing list SG> https://stat.ethz.ch/mailman/listinfo/r-help SG> PLEASE do read the posting guide! SG> http://www.R-project.org/posting-guide.html
Remigijus Lapinskas <remigijus.lapinskas <at> maf.vu.lt> writes: : : I use the following two function for a lagged regression: : : lm.lag=function(y,lag=1) summary(lm(embed(y,lag+1)[,1]~embed(y,lag+1)[,2: (lag+1)])) : lm.lag.x=function(y,x,lag=1) summary(lm(embed(y,lag+1)[,1]~embed(x,lag+1)[,2: (lag+1)])) : : for, respectively, : : y_t=a+b_1*y_t-1+...+b_lag*y_t-lag : y_t=a+b_1*x_t-1+...+b_lag*x_t-lag One could combine these into one function like this: lm.lag <- function(y, x = y, lag = 1) summary( lm( embed(y, lag+1)[,1] ~ embed(x, lag+1)[,-1] ) )