log likelihood and derivatives for Weibull distribution
Value
data frame with fx
for the log pdf value of with dProb
that has the derivatives with respect to the prob
parameters at
the observation time-point
Details
In an rxode2()
model, you can use llikWeibull()
but you have to
use the x and rate arguments. You can also get the derivative of shape
or scale
with
llikWeibullDshape()
and llikWeibullDscale()
.
Examples
# \donttest{
llikWeibull(1, 1, 10)
#> fx dShape dScale
#> 1 -2.402585 -1.072327 -0.09
# rxode2 can use this too:
et <- et(seq(0.001, 1, length.out=10))
et$shape <- 1
et$scale <- 10
model <- function() {
model({
fx <- llikWeibull(time, shape, scale)
dShape<- llikWeibullDshape(time, shape, scale)
dScale <- llikWeibullDscale(time, shape, scale)
})
}
rxSolve(model, et)
#>
#>
#> ℹ parameter labels from comments are typically ignored in non-interactive mode
#> ℹ Need to run with the source intact to parse comments
#>
#>
#> using C compiler: ‘gcc (Ubuntu 11.4.0-1ubuntu1~22.04) 11.4.0’
#> ── Solved rxode2 object ──
#> ── Parameters (value$params): ──
#> # A tibble: 1 × 0
#> ── Initial Conditions (value$inits): ──
#> named numeric(0)
#> ── First part of data (object): ──
#> # A tibble: 10 × 6
#> time fx dShape dScale shape scale
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 0.001 -2.30 -8.21 -0.100 1 10
#> 2 0.112 -2.31 -3.44 -0.0989 1 10
#> 3 0.223 -2.32 -2.72 -0.0978 1 10
#> 4 0.334 -2.34 -2.29 -0.0967 1 10
#> 5 0.445 -2.35 -1.97 -0.0956 1 10
#> 6 0.556 -2.36 -1.73 -0.0944 1 10
#> # ℹ 4 more rows
# }