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?

Could someone please help me with these questions? I really appreciate it!


1 Answer 1


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 $h<g$, then $\mu(W_g\vert W_h)=0$ by construction. A smaller set cannot get a larger conditional probability, so $\mu(Z\vert W_h)=0$.


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