bayou (version 2.1.1)

dsb: Probability density functions for bayou

Description

This function provides a means to specify the prior for the location of shifts across the phylogeny. Certain combinations are not allowed. For example, a maximum shift number of Inf on one branch cannot be combined with a maximum shift number of 1 on another. Thus, bmax must be either a vector of 0's and Inf's or a vector of 0's and 1's. Also, if bmax == 1, then all probabilities must be equal, as bayou cannot sample unequal probabilities without replacement.

Usage

dsb(sb, ntips = ntips, bmax = 1, prob = 1, log = TRUE)
rsb(k, ntips = ntips, bmax = 1, prob = 1, log = TRUE)

Arguments

sb

A vector giving the branch numbers (for a post-ordered tree)

ntips

The number of tips in the phylogeny

bmax

A single integer or a vector of integers equal to the number of branches in the phylogeny indicating the maximum number of shifts allowable in the phylogeny. Can take values 0, 1 and Inf.

prob

A single value or a vector of values equal to the number of branches in the phylogeny indicating the probability that a randomly selected shift will lie on this branch. Can take any positive value, values need not sum to 1 (they will be scaled to sum to 1)

log

A logical indicating whether the log probability should be returned. Default is 'TRUE'

k

The number of shifts to randomly draw from the distribution

Value

The log density of the particular number and arrangement of shifts.

Details

dsb calculates the probability of a particular arrangement of shifts for a given set of assumptions.

Examples

# NOT RUN {
n=10
tree <- sim.bdtree(n=n)
tree <- reorder(tree, "postorder")
nbranch <- 2*n-2
sb <- c(1,2, 2, 3)

# Allow any number of shifts on each branch, with probability 
# proportional to branch length
dsb(sb, ntips=n, bmax=Inf, prob=tree$edge.length)

# Disallow shifts on the first branch, returns -Inf because sb[1] = 1
dsb(sb, ntips=n, bmax=c(0, rep(1, nbranch-1)), prob=tree$edge.length)

# Set maximum number of shifts to 1, returns -Inf because two shifts 
# are on branch 2
dsb(sb, ntips=n, bmax=1, prob=1)

# Generate a random set of k branches
rsb(5, ntips=n, bmax=Inf, prob=tree$edge.length)
# }