You don't give an example so I'm not sure what you mean by an "interval" (how big?, etc), but you run the risk of having a biased estimate of the derivative of the smooth if your interval is long relative to the wiggliness of the estimated smooth.
It is quite easy to compute a finite difference-based estimate of the derivative of an estimated smooth using existing tools in mgcv. I'll summarise the approach below but this is cribbed from some blog posts I wrote that show how to do this and consider simultaneous intervals, not just the across-the-function intervals I'll show below.
I'll also use some functions I wrote to do this analysis that are in my gratia package.
library("mgcv")
library("gratia")
set.seed(2)
dat <- gamSim(1, n = 400, dist = "normal", scale = 2)
mod <- gam(y ~ s(x0) + s(x1) + s(x2) + s(x3), data = dat, method = "REML")
The basic idea is to use the type = "lpmatrix" option of the predict() method for gam objects. This returns the linear predictor matrix for a set of locations at which we evaluate the spline, which, when multiplied by the coefficients of the model yields predicted values for the smooth. We compute two of these matrices; one for a fine grid of points over the range of the covariate for the smooth, and the second at the same locations but shifted a tiny amount. We difference these two matrices to get a linear predictor matrix for the derivative at the grid of points. The variance-covariance matrix of the model coefficients is then used to created a confidence interval. As the model is additive we can ignore the other covariates and smooths and do this for each smooth in turn.
All of this is done by the fderiv() function in gratia:
## first derivatives of all smooths...
fd <- fderiv(mod)
## ...and a selected smooth
fd2 <- fderiv(mod, term = "x1")
Now fd and fd2 contain the finite difference-estimates of the derivative of all or one smooth from the fitted model. The blog posts linked below go into a lot more detail of what is going on under the hood.
We can generate a confidence interval for the derivative using the confint method
ci <- confint(fd, type = "confidence")
head(ci)
> head(ci)
term lower est upper
1 x0 -0.8496032 4.112256 9.074116
2 x0 -0.8489453 4.112287 9.073519
3 x0 -0.8448850 4.112468 9.069821
4 x0 -0.8329612 4.112988 9.058936
5 x0 -0.8108548 4.113933 9.038721
6 x0 -0.7769721 4.115360 9.007693
This has info for all the smooths, and by default is a 95% interval. First, add on a column of values where the derivatives were evaluated:
ci <- cbind(ci, x = as.vector(fd[['eval']]))
Then we can plot:
library("ggplot2")
ggplot(ci, aes(x = x, y = est, group = term)) +
geom_ribbon(aes(ymin = lower, ymax = upper), alpha = 0.3) +
geom_line() +
facet_wrap( ~ term)
Giving:
Blog posts that contain more detail:
The model you fitted with {mgcv} is
$$
y_i = \alpha + f(x_i) + \varepsilon_i
$$
not
$$
y_i = \alpha + \beta \cdot f(x_i) + \varepsilon_i
$$
So that makes me a little unsure as to what $\beta$ you want. In the {mgcv} formulation, the smooth function $f$ is parameterised as a penalised spline and represented as a sum of one or more ($K$) basis functions, each of which has a coefficient $\beta_k$
$$
f(x_i) = \sum_{k=1}^{K} \beta_k b_k(x_i)
$$
Hence, if you want those $\beta_k$, use the coef() method
library(mgcv)
set.seed(2) ## simulate some data...
dat <- gamSim(1, n = 400, dist = "normal", scale = 2)
m <- gam(y ~ s(x0) + s(x1) + s(x2) + s(x3), data = dat)
coef(m)
The last line yields:
> coef(m)
(Intercept) s(x0).1 s(x0).2 s(x0).3 s(x0).4
7.833279e+00 2.173964e-01 3.805544e-01 -1.356960e-01 -3.768858e-01
s(x0).5 s(x0).6 s(x0).7 s(x0).8 s(x0).9
-4.175098e-02 3.566379e-01 -1.460338e-02 1.662533e+00 -1.107835e-02
s(x1).1 s(x1).2 s(x1).3 s(x1).4 s(x1).5
-5.169824e-02 -3.356803e-01 -7.802932e-02 2.992721e-01 -7.940761e-02
s(x1).6 s(x1).7 s(x1).8 s(x1).9 s(x2).1
-2.829178e-01 -1.044189e-01 -1.378039e+00 1.520851e+00 -6.550020e+00
s(x2).2 s(x2).3 s(x2).4 s(x2).5 s(x2).6
1.005427e+01 2.136084e+00 1.614333e+00 -2.352140e+00 1.495629e+00
s(x2).7 s(x2).8 s(x2).9 s(x3).1 s(x3).2
-1.550494e+00 9.859330e+00 5.834315e+00 1.083155e-10 5.828135e-11
s(x3).3 s(x3).4 s(x3).5 s(x3).6 s(x3).7
-5.924120e-11 9.849940e-11 4.198290e-11 1.014248e-10 3.309607e-11
s(x3).8 s(x3).9
4.894399e-10 -2.082265e-01
Best Answer
This kind of smooth-linear interaction is known as a varying coefficient model. To fit these kinds of effects in {mgcv} you use the
byargument and pass it a continuous variable. SO I would expect something like this to do what you want:where:
s()term accounts for the spatial variation inY, andti()term accounts for the spatial variation in the effect ofYearonY(the spatially-varying linear trend)