--- title: "Specification of the Truncated Cauchy calibration density in MCMCTree" author: "Gustavo A. Ballen and Sandra Reinales" date: "`r Sys.Date()`" output: rmarkdown::html_vignette vignette: > %\VignetteIndexEntry{Specification of the Truncated Cauchy calibration density in MCMCTree} %\VignetteEngine{knitr::rmarkdown} %\VignetteEncoding{UTF-8} --- It is possible to fix `p` to a given value describing the location of the distribution, that is, how close to the minimum age is the mode in the truncated Cauchy, and then estimate `c` so that the area under the curve attains a certain maximum age, allowing for some probability to be allocated to the right of it. The function `c_truncauchy` can be used provided that we have both age minimum and maximum, and define some probability `pr` to be allocated to the right of the maximum age. In this example, our minimum age is e.g. $1$ in units of $100 Ma$, and our maximum age is $4.93$ also in units of $100 Ma$. The minimum is a soft bound allowing $0.025$ of the density to be allocated to the left of it, whereas $0.975$ is the percentile at which we observe the maximum age. The quantity `p=0.001` has been chosen so that the mode is closer to the minimum age. ```{r} # load the package library(tbea) # estimate the c parameter for the L distribution cparam <- c_truncauchy(tl=0.4094, tr=0.4160, p=0.001, pr=0.975, al=0.025) cparam ``` The function gives us an estimated of approximately $2.0$ for `c`. Thus we can specify this distribution in `MCMCTree` as `L(0.4094, 0.1, 0.0009289821, 0.025)`. We can use the packages `mcmc3r` ^[dos Reis, M. et al. (2018). Using phylogenomic data to explore the effects of relaxed clocks and calibration strategies on divergence time estimation: Primates as a test case. Systematic Biology, 67(4):594-615.] to plot the `L` density. The following code should do the trick. ```{r, eval=FALSE, fig.align="center", out.width="55%"} # load the package library(mcmc3r) # using the function dL to plot the L density curve(dL(x, tL=0.4094, p=0.001, c=cparam), from=0.4094, to=0.4160) ```