## Confusing slice sampler 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 $u_i<\text{logit}(\alpha+\beta x_i)$ 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 $\alpha+\beta x_i > \log(u_i/(1-u_i))$

and the second case the equivalent condition $\alpha+\beta x_i < - \log(u_i/(1-u_i))$

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. $\left\{ (a,b)\,:\ y_i(a+bx_i) > \log\dfrac{u_i}{1-u_i} \right\}$

This set is missing (a) an intersection sign before the curly bracket and (b) a $(1-)^y_i$ instead of the $y_i$. It should be $\displaystyle{\bigcap_{i=1}^n} \left\{ (a,b)\,:\ (-1)^{y_i}(a+bx_i) > \log\dfrac{u_i}{1-u_i} \right\}$

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