# About The Bayesian Conditional-Probability Systems in Myerson's Game Theory: Analysis of Conflict

I am self-studying game theory using Myerson's Game Theory: Analysis of Conflict. I got some trouble understanding his Bayesian conditional-probability system. The Bayesian conditional-probability system in his textbook is based on his 1986 paper, Multistage Games with Communication, and so my question will be based on his paper. Specifically, I cannot understand part of his proof of Theorem 1. I will first provide some background information, and then I will describe my question.

## Background Information

Here is how Myerson (1986) define the Bayesian conditional probability system:

A conditional probability system is any function $$\mu$$ that specifies a nonnegative number $$\mu(X|Z)$$ for every $$X$$ and $$Z$$ such that $$X \subseteq \Omega$$ and $$\emptyset \neq Z \subseteq \Omega$$, and that satisfies the following three properties, for every $$X$$, $$Y$$, and $$Z$$ such that $$X \subseteq \Omega$$, $$Y \subseteq \Omega$$, and $$\emptyset \neq Z \subseteq \Omega$$: \begin{align} & \text{if}\quad X \cap Y = \emptyset \quad \text{then}\quad \mu(X \cup Y|Z) = \mu(X|Z) + \mu(Y|Z);\tag1\\ & \mu(\Omega|Z) = \mu(Z|Z) = 1;\tag2\\ & \text{if}\quad X \subseteq Y \subseteq Z \quad \text{then}\quad \mu(X|Z) = \mu(X|Y)\mu(Y|Z).\tag3 \end{align}

We let $$\Delta^*(\Omega)$$ denote the set of all conditional probability system on $$\Omega$$. Given any probability distribution $$\eta$$ in $$\Delta(\Omega)$$, we say that a conditional probability system $$\mu$$ in $$\Delta^*(\Omega)$$ is an extension of $$\eta$$ if and only if \begin{align*} \mu(X|\Omega) = \eta(X),\quad \forall X \subseteq \Omega. \end{align*}

One way to construct a conditional probability system on $$\Omega$$ is to start with a probability distribution that assigns positive probability to every point in $$\Omega$$. If $$\mu(Z) > 0$$ for every nonempty set $$Z$$, then the conditional probabilities that are defined by equation (1) do satisfy (2)--(4), as is straightforward to check. In fact, every conditional probability system on $$\Omega$$ can be characterized as the limit of conditional probability systems that are constructed in this way.

Here comes Theorem 1 and its proof:

Theorem 1$$\space\space\space\space$$ $$\mu$$ is a conditional probability system on $$\Omega$$ if and only if there exists a sequence of probability distributions $$\{\eta^j \}_{j = 1}^{\infty}$$ such that \begin{align*} & \eta^j(\{\omega \}) > 0,\quad \forall j,\quad \forall \omega \in \Omega;\quad \textit{and}\\ & \mu(X|Z) = \lim_{j \to \infty} \frac{\eta^j(X \cap Z)}{\eta^j(Z)},\quad \forall X, \quad \forall Z \neq \emptyset. \end{align*}

Proof$$\space\space\space\space$$ Suppose first that there exists a sequence of probability distributions $$\{\eta^j \}_{j = 1}^{\infty}$$ such that $$\eta^j(\{\omega \}) > 0$$ for every $$j$$ and $$\omega$$ in $$\Omega$$, and \begin{align*} \mu(X|Z) = \lim_{j \to \infty} \frac{\eta^j(X \cap Z)}{\eta^j(Z)},\quad \forall X,\quad \forall Z \neq \emptyset. \end{align*} Then (1)--(3) can be checked as follows. If $$X \cap Y = \emptyset$$, then \begin{align*} \mu(X \cup Y|Z) & = \lim_{j \to \infty} \frac{\eta^j((X \cup Y) \cap Z)}{\eta^j(Z)}\\ & = \lim_{j \to \infty} \frac{n^j((X \cap Z) \cup (Y \cap Z))}{\eta^j(Z)}\\ & = \lim_{j \to \infty} \frac{n^j(X \cap Z)}{\eta^j(Z)} + \lim_{j \to \infty} \frac{n^j(Y \cap Z)}{n^j(Z)}\\ & = \mu(X|Z) + \mu(Y|Z). \end{align*} Moreover, \begin{align*} 1 & = \lim_{j \to \infty} \frac{\eta^j(Z)}{\eta^j(Z)} = \mu(Z|Z)\\ & = \lim_{j \to \infty} \frac{\eta^j(\Omega \cap Z)}{\eta^j(Z)} = \mu(\Omega|Z). \end{align*} If $$X \subseteq Y \subseteq Z$$ and $$Y \neq \emptyset$$, then \begin{align*} \mu(X|Z) & = \lim_{j \to \infty} \frac{\eta^j(X \cap Z)}{\eta^j(Z)}\\ & = \lim_{j \to \infty} \frac{\eta^j(X)}{\eta^j(Z)}\\ & = \lim_{j \to \infty} \left(\frac{\eta^j(X)}{\eta^j(Y)} \cdot \frac{\eta^j(Y)}{\eta^j(Z)} \right)\\ & = \left(\lim_{j \to \infty} \frac{\eta^j(X \cap Y)}{\eta^j(Y)} \right) \cdot \left(\lim_{j \to \infty} \frac{\eta^j(Y \cap Z)}{\eta^j(Z)} \right)\\ & = \mu(X|Y) \mu(Y|Z). \end{align*} Thus, $$\mu$$ is a conditional probability system.

Conversely, suppose now that $$\mu$$ is a conditional probability system. We construct $$\{\eta^j\}_{j = 1}^{\infty}$$ as follows. Let $$W_0 = \Omega$$ and then inductively define $$W_h$$ for $$h \geq 1$$ by \begin{align*} W_h = \left\{\omega \in W_{h-1}|\mu(\{\omega\}|W_{h-1}) = 0 \right\}. \end{align*} Since these sets are strictly decreasing in size and $$\Omega$$ is finite, there exists some $$H$$ such that $$W_H \neq \emptyset$$ and $$W_{H+1} = \emptyset$$. For any $$X \subseteq \Omega$$, let \begin{align*} \eta^j(X) = \left(\frac{j - 1}{j - (\frac{1}{j})^H} \right) \sum_{h=0}^{H} \mu(X|W_h) \left(\frac{1}{j} \right)^h. \end{align*} Each $$\eta^j$$ is a probability distribution on $$\Omega$$, giving positive probability to every point. Given any sets $$X$$ and $$Z$$ such that $$Z \neq \emptyset$$, let $$g$$ be the highest number such that $$Z \neq \emptyset$$, let $$g$$ be the highest number such that $$Z \subseteq W_g$$. Then $$\mu(Z|W_g) > 0$$ and $$\mu(Z|W_h) = 0$$ for every $$h < g$$. Thus, using (3), \begin{align*} \mu(X|Z) & = \mu(X \cap Z|Z) = \frac{\mu(X \cap Z|W_g)}{\mu(Z|W_g)}\\ & = \lim_{j \to \infty} \frac{\sum_{h=g}^{H} \mu(X \cap Z|W_h) \left(\frac{1}{j} \right)^h}{\sum_{h=g}^{H} \mu(Z|W_h) \left(\frac{1}{j} \right)^h} = \lim_{j \to \infty}\frac{\eta^j(X \cap Z)}{\eta^j(Z)}.\\ \end{align*}

## My Question

I have difficulties understanding his construction of $$\{\eta^j\}_{j=1}^{\infty}$$. Myerson (1986) defines $$\eta^j(X) = \left(\frac{j - 1}{j - (\frac{1}{j})^H} \right) \sum_{h=0}^{H} \mu(X|W_h) \left(\frac{1}{j} \right)^h$$, and then asserts that "Each $$\eta^j$$ is a probability distribution on $$\Omega$$, giving positive probability to every point." So here is a couple of my questions regarding this construction:

(1) I cannot see how this complex equation is been come up with.

(2) When $$j = 1$$, isn't $$\eta^j(X)$$ not defined, because the denominator is zero?

(3) How to check that the $$\eta^j$$ such defined is indeed a probability distribution on $$\Omega$$? (It does seem to me that, other than the case of $$j = 1$$, $$\eta^j$$ gives positive probability to every point.)

(4) Moreover, I am confused about "$$\mu(Z|W_g) > 0$$ and $$\mu(Z|W_h) = 0$$ for every $$h < g$$". If $$h < g$$, then $$W_g \subseteq W_h$$. Then, if $$\mu(Z|W_g) > 0$$, why would $$\mu(Z|W_h) = 0$$?

(5) Finally, when he is doing the limit as $$j$$ goes to $$\infty$$, did he apply the L'Hôpital's rule to get the final result?

The point of conditional probability systems is to have probabilities even defined conditional on events that have probability zero. A normal probability distribution corresponds to $$\mu(\cdot\vert\Omega)$$.
If $$\mu(X\vert\Omega)=0$$ and $$\mu(Y\vert \Omega)>0$$, then $$\mu(X\vert Y)=0$$.
Indeed, $$\mu(Y\vert Y)=1$$ implies that $$\mu(X\vert Y)=\mu(X\cap Y\vert > Y)$$. Since $$\mu(X\cap Y\vert\Omega)\leq\mu(X\vert\Omega)=0$$ an $$\mu(X\cap Y\vert\Omega)=\mu(X\cap Y\vert Y)\mu(Y\vert\Omega)=0$$, we must have $$\mu(X\cap Y\vert Y)=\mu(X\vert Y)=0$$.
Consequently, we only get something new if we condition on events that have probability zero. The largest set of initial probability zero is $$W_1$$. Repeating, the logic, if $$\mu(X\vert W_1)=0$$ and $$\mu(Y\vert W_1)>0$$, then $$\mu(X\vert W_1)=0$$. So, intuitively, $$W_0$$ is infinitely more probable than $$W_1$$, $$W_1$$ is infinitely more probable than $$W_2$$, and so on. To represent this in terms of the limits, he wants to have a sequence $$(\alpha_0^j,\alpha_1^j,\ldots,\alpha_H^j)$$ of strictly positive weights that sum to one, such that $$\eta^j(\cdot)=\alpha_0^j \mu(\cdot\vert W_0)+\alpha_1^j\mu(\cdot\vert W_1)+\cdots+\mu(\cdot\vert W_H)$$ with $$\lim_j \alpha_h^j/\alpha_{h+1}^j=\infty$$. This is the case here, since $$\lim_{j\to\infty} \frac{\big(\frac{1}{j}\big)^h}{\big(\frac{1}{j}\big)^{h+1}}=\lim_{j\to\infty} j=\infty.$$ The complicated expression everything is multiplied with \begin{align*} \left(\frac{j - 1}{j - (\frac{1}{j})^H} \right) \end{align*} is only there so that the sum of the weights is one and $$\eta^j$$ is a probability distribution as a convex combination of probability distributions. The case $$j=1$$ is indeed not well-defined, but since only the limit of the sequence matters, finitely many undefined terms pose no real problem.
Lastly, if $$Z\subseteq W_g$$ and $$h, then $$\mu(W_g\vert W_h)=0$$ by construction. A smaller set cannot get a larger conditional probability, so $$\mu(Z\vert W_h)=0$$.