# GAM with unevenly distributed data

I am currently working with GAM from the MGCV package to develop a model for my system and I'm in the unfortunate situation where my observations are unevenly distributed in the covariate space. The situation is probably best illustrated with an example:

I create a GAM with a call that looks something like this:

myGam <- gam(y ~ te(x1, x2, bs=c("bs", "bs"), k=10), data=myData)


Plotting the surface gives me the following plot:

Where the surface is the fitted model, the green dots are the observations and the blue dots are the knots.
Zooming into the range where my observations are gives the following plot: which shows that I have a nice match where the function is supported by data. But in areas where there is no data to support the function, the model is a bit wild.

Now my question: Does anybody have an idea how to tame my function so that it has less steep changes when there is no data to support it?

I played a bit with manually setting the smoothing parameter. When I use the following code, I get a model that does more or less what I would expect:

myGam <- gam(y ~ te(x1, x2, bs=c("bs", "bs"), k=10, sp=c(10,10)), data=myData)


Which looks like this when plotted:

Unfortunately, it is quite time consuming and, to some extent, arbitrary if I just fiddle with the smoothing parameter until I have a model that I like.

You don't want to fix the smoothness parameters at some values. What you want to do is to try to control the behaviour of the spline at locations beyond the data.

Wood (2016) sets out one way to achieve this. The general idea is to place most of the knots for the spline within the limits of the data, but to place outer and inner boundary knots to bracket the region where you want the penalty to operate over but where you have no data.

Simon shows how to do this in the 1-d case, where he places knots specifically

b <- gam(y ~ s(x, bs = "bs", m = c(3,1), k = 20),
knots = list(x = c(-.5,0,1,1.5)),
method = "ML")


which has a second order penalty (implied by the c(3,1) passed to m), outer knots at -0.5 and 1.5, inner knots and 0 and 1, and then remaining knots (implied) spread evenly over the range 0–1.

The full example in ?smooth.construct.bs.smooth.spec produces the following plot (code from the example shown below)

• The red line is the result of a thin plate spline with penalty on the first derivative. The CI is constant because the slope of the line must be linear otherwise the penalty explodes to infinity. The constant CI is not a good description of our uncertainty as we extrapolate however; we would expect uncertainty to increase as we move further from the support of the data.

• The blue line shows what happens in the usual case where knots are restricted to the range of the data and we use a b spline with second derivative penalty. The CI explodes because there is no penalty outside the range of the data.

• The black line shows the effect of pushing the penalty out to -0.5 and 1.5. Because the penalty acts beyond the range of the data the CI is constrained. Notice however that in the example Simon has predicted a little bit beyond the limits of the region where the penalty applies; -0.7--1.7. This accounts for the change in the rate of increase in the spread of the CI. It explodes quickly again once the penalty is turned off beyond the specified interval -0.5--1.5.

So, in your example you want to follow the same idea, to set inner knots covering the range of the data in the two coordinates, and then add outer boundary knots to cover the interval you want to extrapolate over.

### Code for the example. All from ?smooth.construct.bs.smooth.spec

f3 <- function(x) 0.2 * x^11 * (10 * (1 - x))^6 + 10 * (10 * x)^3 * (1 - x)^10 ## test function
n <- 100
x <- runif(n)
y <- f3(x) + rnorm(n)*2

## first a model with first order penalty over whole real line (red)
b0 <- gam(y~s(x,m=1,k=20),method="ML")

## now a model with first order penalty evaluated over (-.5,1.5) (black)
b <- gam(y~s(x,bs="bs",m=c(3,1),k=20),knots=list(x=c(-.5,0,1,1.5)),
method="ML")

## and the equivalent with same penalty over data range only (blue)
b1 <- gam(y~s(x,bs="bs",m=c(3,1),k=20),method="ML")

## predict at new data extrapolating
pd <- data.frame(x=seq(-.7,1.7,length=200))

## fitted values for model b - the black line in the plot
fv <- predict(b,pd,se=TRUE)
## create a confidence band for the original fit, upper first, then lower
ul <- fv$fit + fv$se.fit*2
ll <- fv$fit - fv$se.fit*2

plot(x,y,xlim=c(-.7,1.7),ylim=range(c(y,ll,ul)),
main= "1st order penalties: red tps; blue bs over (0,1); black bs over (-.5,1.5)")
lines(pd$x,fv$fit)
lines(pd$x,ul,lty=2) lines(pd$x,ll,lty=2)

## fitted values and CI for the b0 model, red line
fv <- predict(b0,pd,se=TRUE)
ul <- fv$fit + fv$se.fit*2
ll <- fv$fit - fv$se.fit*2

## penalty defined on whole real line gives constant width intervals away
## from data, as slope there must be zero, to avoid infinite penalty:
lines(pd$x,fv$fit,col=2)
lines(pd$x,ul,lty=2,col=2) lines(pd$x,ll,lty=2,col=2)

## fitted values for the b1 model, where penalty is only over the
## range of the data, blue line
fv <- predict(b1,pd,se=TRUE)
ul <- fv$fit + fv$se.fit*2
ll <- fv$fit - fv$se.fit*2

## penalty defined only over the data interval (0,1) gives wild and wide
## extrapolation since penalty has been turned off' outside data range:
lines(pd$x,fv$fit,col=4)
lines(pd$x,ul,lty=2,col=4) lines(pd$x,ll,lty=2,col=4)
`

Wood, S. N. 2016. P-splines with derivative based penalties and tensor product smoothing of unevenly distributed data. Stat. Comput. 1–5. doi:10.1007/s11222-016-9666-x

• did you cite the right paper? as far as I can tell there's no reference to this idea there. or am I missing something? Nov 26 '19 at 6:05
• @sirallen Not sure what you mean? The entire paper is about the basis I illustrate and, as far as I understood it, the OP's Q. The green dots in the OP's question are the equivalent of the sample points in Simon's paper that fall into a restricted part of the space possibly spanned by the two coordinates. The paper shows a 2d example of this, but a brief re-read suggests that the paper is describing what I cover in the answer. Nov 26 '19 at 17:03