Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. Most embarrassingly, Liaosa Xu from Virginia Tech sent the following email almost a month ago and I forgot to reply:

I have a question regarding your example 7.11 in your book Introducing Monte Carlo Methods with R.  To further decompose the uniform simulation by sampling a and b step by step, how you determine the upper bound for sampling of a? I don’t know why, for all y(i)=0, we need a+bx(i)>- log(u(i)/(1-u(i))).  It seems that for y(i)=0, we get 0>log(u(i)/(1-u(i))).  Thanks a lot for your clarification.

There is nothing wrong with our resolution of the logit simulation problem but I acknowledge the way we wrote it is most confusing! Especially when switching from $(alpha,beta)$ to $(a,b)$ in the middle of the example….

Starting with the likelihood/posterior $L(alpha, beta | mathbf{y}) propto prod_{i=1}^n left(dfrac{e^{ alpha +beta x_i }}{1 + e^{ alpha +beta x_i }}right)^{y_i}left(dfrac{1}{1 + e^{ alpha +beta x_i }}right)^{1-y_i}$

we use slice sampling to replace each logistic expression with an indicator involving a uniform auxiliary variable $U_i sim mathcal{U}left( 0,dfrac{e^{ y_i(alpha +beta x_i) }}{1 + e^{ alpha +beta x_i }} right)$

[which is the first formula at the top of page 220.] Now, when considering the joint distribution of $(alpha,beta,u_1,...,u_n)$,

we only get a product of indicators. Either indicators that $u_i or of $u_i<1-text{logit}(alpha+beta x_i)$,

depending on whether yi=1 or yi=0. The first case produces the equivalent condition

This is how we derive both uniform distributions in $alpha$ and \$beta\$.

What is both a typo and potentially confusing is the second formula in page 220, where we mention the uniform over the set.

instead of the $y_i$. It should be