Madras exercise 4.24: Confusion about function $A:[0,\infty] \rightarrow [0,1]$ such that $A(z) = z A(1/z)$ for all $z \in [0,\infty]$.

Question

My question is from an exercise in "Lectures on Monte-Carlo Methods" by Madras:

My questions:

1. How do I interpret the function $A(z)$ when $z = 0$ or $+\infty$? When $z = 0$, the condition $A(z) = z A(1/z)$ becomes an issue: What is $0 \cdot A(1/0)$? And what is $\infty \cdot A(1/\infty)$? Should I think of $A(\infty)$ as $\lim\limits_{z \to \infty} A(z)$?

2. How do I interpret $t_{ij}$ when $q_{ij} = 0$? In the case where $q_{ij} = 0$ and $q_{ji} \neq 0$, I am inclined to say that $t_{ij} = \infty$, and that $A(t_{ij}) = A(+\infty) = \lim\limits_{z \to +\infty} A(z)$. But I don't see how to even justify that this limit exists...

3. And how to make sense of the case where $q_{ij} = q_{ji} = 0?$ (Should I assume this doesn't occur?)

I am rather baffled by this problem. Any help or insights would be greatly appreciated!

Answer 1

Q1 For all $z\in (0,\infty)$, we have the given relationship and

$$ A(z) \in [0,1].$$

We can define $A$ at $0$ by limiting:

$$ A(0):=\lim_{z\rightarrow 0} A(z) = \lim_{z\rightarrow 0} zA\left(\frac{1}{z}\right) =0, $$

as $A$ is bounded above and below.

We also note that:

$$ A(z)= \min \left\{ 1,A\left(z\right) \right\}. $$

For $\infty$ we then have:

$$ A(\infty):=\lim_{z\rightarrow \infty} A(z) = \lim_{ z\rightarrow \infty} \min \left\{ 1,A\left(z\right) \right\} \in [0,1]. $$

I'm not sure if more can be said for $A(\infty)$. Functions of type (iii) that respect the given relationship are of form:

$$ A_b(z) = \min \left\{1,\frac{z^{\frac{b+1}{2}}}{1+z^b}\right\}. $$

We see that $A_0(z) = \min \{1,\sqrt{z}$}, so $A_0(\infty) = 1$. Also, $A_1(\infty) = 1$. On the other hand $A_2(z) = \min \{1, z^{3/2}(1+z^2)^{-1}\}$, so $A_2(\infty) = 0$.

Note: We also have $A(1/z)\in [0,1]$, and hence $$ A(z)=zA\left(\frac{1}{z}\right)\leq \min\{1,z\}. $$ Right hand side of the inequality is the basic function that satisfies the given relationship and dominates all the others.

Note 2: For (iii), we have equivalences:

$$ z^a(1+z^{b})^{-1} = z^{1-a}(1+z^{-b})^{-1} $$

$$ z^a(1+z^{b})^{-1} = z^{1-a+b}(1+z^{b})^{-1} $$

$$ z^a = z^{1-a+b} $$

$$ a = 1-a+b. $$

Q2 and Q3

This source says that acceptance distribution needs to be deliberately defined for the cases you are interested in. In particular, definitions (1), which in your notation translate to:

For $i\not= j$ and $q_{ij}\not= 0$, use $A(t_{ij})= \min\{1, A(t_{ij})\}$.

For $i= j$ or $i\not=j$ and $q_{ij}= 0$, define $A(t_{ij})=0$.

For $i=j$, this is in line with the way the probability transition matrix, $P$, is then defined.

For $i\not=j$ and $q_{ij}= 0$, there is no need to figure out what $t_{ij}$ means, one just sets the acceptance entry to $0$ for it.