smoothSurv (version 2.6)

derivative.expAD: Work Function for 'smoothSurvReg', currently nowhere used

Description

Function to compute derivatives of exp(a) w.r.t. exp(d) where d stands for a shorter vector of 'a' G-spline coefficients.

Usage

derivative.expAD(knots, sdspline, last.three, all = TRUE)

Value

A matrix with $\omega$ coefficients.

Arguments

knots

A vector of G-spline knots $\mu$.

sdspline

Standard deviation ${\sigma }_0$ of the basis G-spline.

last.three

Indeces of the three 'a' G-spline coefficients which are expressed as a function of the remaining (g-3) 'a' G-spline coefficients such that the three constraints are satisfied. This must be a vector of length 3 with three different numbers from 1:length(mu). Consequently, a[last.three[1]] <- 0.

all

If TRUE, matrix (g - 2) x g (there is one zero column) is returned. If FALSE, matrix (g - 2) x 2 is returned. The first row is always an intercept. See details.

Author

Arnošt Komárek

Details

To satisfy the three constraints $$\sum _{j=1}^g{c}_j=1,$$ $$\sum _{j=1}^g{c}_j{\mu }_j=0,$$ $$\sum _{j=1}^g{c}_j{\mu }_j^2=1-{\sigma }_0^2$$ imposed on the G-spline we use the following parametrization: $$c_j=\frac{\exp (a_j)}{\sum _{l=1}^g\exp (a_l)},j=1,\dots ,g.$$ The constraints can be solved such that a[last.three[1]] = 0 and a[last.three[2:3]] are expressed as a function of a[-last.three] in the following way: $$a_k=\log \{{\omega }_{0,k}+\sum _{j\ne last.three}{\omega }_{j,k}\exp (a_j)\},k=last.three[2],last.three[3],$$ where $\omega$ coefficients are a function of knots and G-spline standard deviation. If we denote $d$ the vector a[-last.three] this function computes derivatives of $\exp (a)$ w.r.t. $\exp (d)$ together with the intercept term used to compute $\exp (a)$ from $\exp (d)$. This is actually a matrix of $\omega$ coefficients. If we denote it as $\mathrm{\Omega }$ then if all == TRUE $$\exp (a)={\mathrm{\Omega }}_{1,\cdot }^T+{\mathrm{\Omega }}_{-1,\cdot }^T\exp (d)$$ and if all == FALSE $$\exp (a[last.three[2:3]])={\mathrm{\Omega }}_{1,\cdot }^T+{\mathrm{\Omega }}_{-1,\cdot }^T\exp (d).$$