Question

为什么mgcv::gam.vcomp显示与mgcv::ti进行交互的两个方差组件？¨

我似乎无法在任何地方找到解释或行间解释。或许是因为交互中每个组成部分的差异吗？

require(mgcv)
test1 <- function(x,z,sx=0.3,sz=0.4) { 
  x <- x*20
  (pi**sx*sz)*(1.2*exp(-(x-0.2)^2/sx^2-(z-0.3)^2/sz^2)+
                 0.8*exp(-(x-0.7)^2/sx^2-(z-0.8)^2/sz^2))
}
n <- 500
old.par <- par(mfrow=c(2,2))
x <- runif(n)/20;z <- runif(n);
xs <- seq(0,1,length=30)/20;zs <- seq(0,1,length=30)
pr <- data.frame(x=rep(xs,30),z=rep(zs,rep(30,30)))
truth <- matrix(test1(pr$x,pr$z),30,30)
f <- test1(x,z)
y <- f + rnorm(n)*0.2
b3 <- gam(y~ ti(x) + ti(z) + ti(x,z))
b3s <- gam(y~ ti(x) + ti(z) + s(x,z)) # describing the itneraction with s().

我知道我们在这里混合苹果和橘子。

gam.vcomp(b3)
     ti(x)      ti(z)   ti(x,z)1   ti(x,z)2 
0.06609731 0.01476070 0.08834218 0.05700322 

gam.vcomp(b3s)
    ti(x)     ti(z)    s(x,z) 
0.1623056 2.4870344 7.7484987

Answer 1

You'll see the same behaviour with te(x, z)

> b <- gam(y ~ te(x,z))
> gam.vcomp(b)
  te(x,z)1   te(x,z)2 
0.08668107 0.04596708

and arises because tensor product smooths are defined by two, in this case, marginal bases, each of which have a smoothness parameter. Hence there are two variance components, one per smoothness parameter/marginal basis.

ti(x,z)1 is the variance component for the marginal basis of x,
ti(x,z)2 is the variance component for the marginal basis of z.

As these tensor product interactions smooth have had the main effects removed from them, physical interpretation is complicated, but in a practical sense, the values are the variance component interpretation of the smoothness parameters of the marginal bases.

The reason s(x, z) has just one variance component is that it is a 2-d thinplate spline basis. This basis is isotropic; there is the same smoothness in the to dimensions and hence a single smoothness parameter is required for the basis. Hence there is a single variance component.

R :: mgcv中张量相互作用的方差分量

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