I will illustrate one such simulation for the simple tree depicted in figure 3.4b. We first set the ancestral character state to be $\bar{z}(0)$, which will then be the expected value for all the nodes and tips in the tree. This tree has three branches, so we draw three values from normal distributions. These normal distributions have mean zero and variances that are given by the rate of evolution and the branch length of the tree, as stated in equation 3.1. Note that we are modeling changes on these branches, so even if $\bar{z}(0) \neq 0$ the values for changes on branches are drawn from a distribution with a mean of zero. In the case of the tree in Figure 3.1, x1 ∼ N(0, σ2t1). Similarly, x2 ∼ N(0, σ2t2) and x3 ∼ N(0, σ2t3). If I set σ2 = 1 for the purposes of this example, I might obtain x1 = −1.6, x2 = 0.1, and x3 = −0.3. These values represent the evolutionary changes that occur along branches in the simulation. To calculate trait values for species, we add: xa = θ + x1 + x2 = 0 − 1.6 + 0.1 = −1.5, and xb = θ + x1 + x3 = 0 − 1.6 + −0.3 = −1.9.