clr {FDboost}

Clr and inverse clr transformation

Description

clr computes the clr or inverse clr transformation of a vector f with respect to integration weights w, corresponding to a Bayes Hilbert space B^2(\mu) = B^2(\mathcal{T}, \mathcal{A}, \mu).

Usage

clr(f, w = 1, inverse = FALSE)

Arguments

f

a vector containing the function values (evaluated on a grid) of the function f to transform. If inverse = TRUE, f must be a density, i.e., all entries must be positive and usually f integrates to one. If inverse = FALSE, f should integrate to zero, see Details.

w

a vector of length one or of the same length as f containing positive integration weights. If w has length one, this weight is used for all function values. The integral of f is approximated via \int_{\mathcal{T}} f \, \mathrm{d}\mu \approx \sum_{j=1}^m w_j f_j, where m equals the length of f.

inverse

if TRUE, the inverse clr transformation is computed.

Details

The clr transformation maps a density f from B^2(\mu) to L^2_0(\mu) := \{ f \in L^2(\mu) ~|~ \int_{\mathcal{T}} f \, \mathrm{d}\mu = 0\} via

\mathrm{clr}(f) := \log f - \frac{1}{\mu (\mathcal{T})} \int_{\mathcal{T}} \log f \, \mathrm{d}\mu.

The inverse clr transformation maps a function f from L^2_0(\mu) to B^2(\mu) via

\mathrm{clr}^{-1}(f) := \frac{\exp f}{\int_{\mathcal{T}} \exp f \, \mathrm{d}\mu}.

Note that in contrast to Maier et al. (2021), this definition of the inverse clr transformation includes normalization, yielding the respective probability density function (representative of the equivalence class of proportional functions in B^2(\mu)).

The (inverse) clr transformation depends not only on f, but also on the underlying measure space \left( \mathcal{T}, \mathcal{A}, \mu\right), which determines the integral. In clr this is specified via the integration weights w. E.g., for a discrete set \mathcal{T} with \mathcal{A} = \mathcal{P}(\mathcal{T}) the power set of \mathcal{T} and \mu = \sum_{t \in T} \delta_t the sum of dirac measures at t \in \mathcal{T}, the default w = 1 is the correct choice. In this case, integrals are indeed computed exactly, not only approximately. For an interval \mathcal{T} = [a, b] with \mathcal{A} = \mathcal{B} the Borel \sigma-algebra restricted to \mathcal{T} and \mu = \lambda the Lebesgue measure, the choice of w depends on the grid on which the function was evaluated: w_j must correspond to the length of the subinterval of [a, b], which f_j represents. E.g., for a grid with equidistant distance d, where the boundary grid values are a + \frac{d}{2} and b - \frac{d}{2} (i.e., the grid points are centers of intervals of size d), equal weights d should be chosen for w.

The clr transformation is crucial for density-on-scalar regression since estimating the clr transformed model in L^2_0(\mu) is equivalent to estimating the original model in B^2(\mu) (as the clr transformation is an isometric isomorphism), see also the vignette "FDboost_density-on-scalar_births" and Maier et al. (2021).

Value

A vector of the same length as f containing the (inverse) clr transformation of f.

Author(s)

Eva-Maria Maier

References

Maier, E.-M., Stoecker, A., Fitzenberger, B., Greven, S. (2021): Additive Density-on-Scalar Regression in Bayes Hilbert Spaces with an Application to Gender Economics. arXiv preprint arXiv:2110.11771.

Examples

### Continuous case (T = [0, 1] with Lebesgue measure):
# evaluate density of a Beta distribution on an equidistant grid
g <- seq(from = 0.005, to = 0.995, by = 0.01)
f <- dbeta(g, 2, 5)
# compute clr transformation with distance of two grid points as integration weight
f_clr <- clr(f, w = 0.01)
# visualize result
plot(g, f_clr , type = "l")
abline(h = 0, col = "grey")
# compute inverse clr transformation (w as above)
f_clr_inv <- clr(f_clr, w = 0.01, inverse = TRUE)
# visualize result
plot(g, f, type = "l")
lines(g, f_clr_inv, lty = 2, col = "red")

### Discrete case (T = {1, ..., 12} with sum of dirac measures at t in T):
data("birthDistribution", package = "FDboost")
# fit density-on-scalar model with effects for sex and year
model <- FDboost(birth_densities_clr ~ 1 + bolsc(sex, df = 1) + 
                   bbsc(year, df = 1, differences = 1),
                 # use bbsc() in timeformula to ensure integrate-to-zero constraint
                 timeformula = ~bbsc(month, df = 4, 
                                     # December is followed by January of subsequent year
                                     cyclic = TRUE, 
                                     # knots = {1, ..., 12} with additional boundary knot
                                     # 0 (coinciding with 12) due to cyclic = TRUE
                                     knots = 1:11, boundary.knots = c(0, 12), 
                                     # degree = 1 with these knots yields identity matrix 
                                     # as design matrix
                                     degree = 1),
                 data = birthDistribution, offset = 0, 
                 control = boost_control(mstop = 1000))
# Extract predictions (clr-transformed!) and transform them to Bayes Hilbert space
predictions_clr <- predict(model)
predictions <- t(apply(predictions_clr, 1, clr, inverse = TRUE))