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

CC BY-SA 4.0
2023-09-18T19:28:06.720Z
2023-09-18T19:28:06.720Z
Michael Greinecker
<p>I am self-studying game theory using Myerson's <em>Game Theory: Analysis of Conflict</em>. I got some trouble understanding his Bayesian conditional-probability system. The Bayesian conditional-probability system in his textbook is based on his 1986 paper, <em>Multistage Games with Communication</em>, 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.</p>
<h2>Background Information</h2>
<p>Here is how Myerson (1986) define the Bayesian conditional probability system:</p>
<blockquote>
<p>A <em>conditional probability system</em> is any function <span class="math-container">$\mu$</span> that specifies a nonnegative number <span class="math-container">$\mu(X|Z)$</span> for every <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> such that <span class="math-container">$X \subseteq \Omega$</span> and <span class="math-container">$\emptyset \neq Z \subseteq \Omega$</span>, and that satisfies the following three properties, for every <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span> such that <span class="math-container">$X \subseteq \Omega$</span>, <span class="math-container">$Y \subseteq \Omega$</span>, and <span class="math-container">$\emptyset \neq Z \subseteq \Omega$</span>:
<span class="math-container">\begin{align}
    &amp; \text{if}\quad X \cap Y = \emptyset \quad \text{then}\quad \mu(X \cup Y|Z) = \mu(X|Z) + \mu(Y|Z);\tag1\\
    &amp; \mu(\Omega|Z) = \mu(Z|Z) = 1;\tag2\\
    &amp; \text{if}\quad X \subseteq Y \subseteq Z \quad \text{then}\quad \mu(X|Z) = \mu(X|Y)\mu(Y|Z).\tag3
\end{align}</span></p>
<p>We let <span class="math-container">$\Delta^*(\Omega)$</span> denote the set of all conditional probability system on <span class="math-container">$\Omega$</span>. Given any probability distribution <span class="math-container">$\eta$</span> in <span class="math-container">$\Delta(\Omega)$</span>, we say that a conditional probability system <span class="math-container">$\mu$</span> in <span class="math-container">$\Delta^*(\Omega)$</span> is an <em>extension</em> of <span class="math-container">$\eta$</span> if and only if
<span class="math-container">\begin{align*}
    \mu(X|\Omega) = \eta(X),\quad \forall X \subseteq \Omega.
\end{align*}</span></p>
<p>One way to construct a conditional probability system on <span class="math-container">$\Omega$</span> is to start with a probability distribution that assigns positive probability to every point in <span class="math-container">$\Omega$</span>. If <span class="math-container">$\mu(Z) &gt; 0$</span> for every nonempty set <span class="math-container">$Z$</span>, 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 <span class="math-container">$\Omega$</span> can be characterized as the limit of conditional probability systems that are constructed in this way.</p>
</blockquote>
<p>Here comes Theorem 1 and its proof:</p>
<blockquote>
<p><strong>Theorem 1</strong><span class="math-container">$\space\space\space\space$</span> <em><span class="math-container">$\mu$</span> is a conditional probability system on <span class="math-container">$\Omega$</span> if and only if there exists a sequence of probability distributions <span class="math-container">$\{\eta^j \}_{j = 1}^{\infty}$</span> such that</em>
<span class="math-container">\begin{align*}
    &amp; \eta^j(\{\omega \}) &gt; 0,\quad \forall j,\quad \forall \omega \in \Omega;\quad \textit{and}\\
    &amp; \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*}</span></p>
<p><strong>Proof</strong><span class="math-container">$\space\space\space\space$</span> Suppose first that there exists a sequence of probability distributions <span class="math-container">$\{\eta^j \}_{j = 1}^{\infty}$</span> such that <span class="math-container">$\eta^j(\{\omega \}) &gt; 0$</span> for every <span class="math-container">$j$</span> and <span class="math-container">$\omega$</span> in <span class="math-container">$\Omega$</span>, and
<span class="math-container">\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*}</span>
Then (1)--(3) can be checked as follows. If <span class="math-container">$X \cap Y = \emptyset$</span>, then
<span class="math-container">\begin{align*}
    \mu(X \cup Y|Z) &amp; = \lim_{j \to \infty} \frac{\eta^j((X \cup Y) \cap Z)}{\eta^j(Z)}\\
    &amp; = \lim_{j \to \infty} \frac{n^j((X \cap Z) \cup (Y \cap Z))}{\eta^j(Z)}\\
    &amp; = \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)}\\
    &amp; = \mu(X|Z) + \mu(Y|Z).
\end{align*}</span>
Moreover,
<span class="math-container">\begin{align*}
    1 &amp; = \lim_{j \to \infty} \frac{\eta^j(Z)}{\eta^j(Z)} = \mu(Z|Z)\\
    &amp; = \lim_{j \to \infty} \frac{\eta^j(\Omega \cap Z)}{\eta^j(Z)} = \mu(\Omega|Z).
\end{align*}</span>
If <span class="math-container">$X \subseteq Y \subseteq Z$</span> and <span class="math-container">$Y \neq \emptyset$</span>, then
<span class="math-container">\begin{align*}
    \mu(X|Z) &amp; = \lim_{j \to \infty} \frac{\eta^j(X \cap Z)}{\eta^j(Z)}\\
    &amp; = \lim_{j \to \infty} \frac{\eta^j(X)}{\eta^j(Z)}\\
    &amp; = \lim_{j \to \infty} \left(\frac{\eta^j(X)}{\eta^j(Y)} \cdot \frac{\eta^j(Y)}{\eta^j(Z)} \right)\\
    &amp; = \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)\\
    &amp; = \mu(X|Y) \mu(Y|Z).
\end{align*}</span>
Thus, <span class="math-container">$\mu$</span> is a conditional probability system.</p>
<p>Conversely, suppose now that <span class="math-container">$\mu$</span> is a conditional probability system. We construct <span class="math-container">$\{\eta^j\}_{j = 1}^{\infty}$</span> as follows. Let <span class="math-container">$W_0 = \Omega$</span> and then inductively define <span class="math-container">$W_h$</span> for <span class="math-container">$h \geq 1$</span> by
<span class="math-container">\begin{align*}
    W_h = \left\{\omega \in W_{h-1}|\mu(\{\omega\}|W_{h-1}) = 0 \right\}.
\end{align*}</span>
Since these sets are strictly decreasing in size and <span class="math-container">$\Omega$</span> is finite, there exists some <span class="math-container">$H$</span> such that <span class="math-container">$W_H \neq \emptyset$</span> and <span class="math-container">$W_{H+1} = \emptyset$</span>. For any <span class="math-container">$X \subseteq \Omega$</span>, let
<span class="math-container">\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*}</span>
Each <span class="math-container">$\eta^j$</span> is a probability distribution on <span class="math-container">$\Omega$</span>, giving positive probability to every point. Given any sets <span class="math-container">$X$</span> and <span class="math-container">$Z$</span> such that <span class="math-container">$Z \neq \emptyset$</span>, let <span class="math-container">$g$</span> be the highest number such that <span class="math-container">$Z \neq \emptyset$</span>, let <span class="math-container">$g$</span> be the highest number such that <span class="math-container">$Z \subseteq W_g$</span>. Then <span class="math-container">$\mu(Z|W_g) &gt; 0$</span> and <span class="math-container">$\mu(Z|W_h) = 0$</span> for every <span class="math-container">$h &lt; g$</span>. Thus, using (3),
<span class="math-container">\begin{align*}
    \mu(X|Z) &amp; = \mu(X \cap Z|Z) = \frac{\mu(X \cap Z|W_g)}{\mu(Z|W_g)}\\
    &amp; = \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*}</span></p>
</blockquote>
<h2>My Question</h2>
<p>I have difficulties understanding his construction of <span class="math-container">$\{\eta^j\}_{j=1}^{\infty}$</span>. Myerson (1986) defines <span class="math-container">$\eta^j(X) = \left(\frac{j - 1}{j - (\frac{1}{j})^H} \right) \sum_{h=0}^{H} \mu(X|W_h) \left(\frac{1}{j} \right)^h$</span>, and then asserts that &quot;Each <span class="math-container">$\eta^j$</span> is a probability distribution on <span class="math-container">$\Omega$</span>, giving positive probability to every point.&quot; So here is a couple of my questions regarding this construction:</p>
<p>(1) I cannot see how this complex equation is been come up with.</p>
<p>(2) When <span class="math-container">$j = 1$</span>, isn't <span class="math-container">$\eta^j(X)$</span> not defined, because the denominator is zero?</p>
<p>(3) How to check that the <span class="math-container">$\eta^j$</span> such defined is indeed a probability distribution on <span class="math-container">$\Omega$</span>? (It does seem to me that, other than the case of <span class="math-container">$j = 1$</span>, <span class="math-container">$\eta^j$</span> gives positive probability to every point.)</p>
<p>(4) Moreover, I am confused about &quot;<span class="math-container">$\mu(Z|W_g) &gt; 0$</span> and <span class="math-container">$\mu(Z|W_h) = 0$</span> for every <span class="math-container">$h &lt; g$</span>&quot;. If <span class="math-container">$h &lt; g$</span>, then <span class="math-container">$W_g \subseteq W_h$</span>. Then, if <span class="math-container">$\mu(Z|W_g) &gt; 0$</span>, why would <span class="math-container">$\mu(Z|W_h) = 0$</span>?</p>
<p>(5) Finally, when he is doing the limit as <span class="math-container">$j$</span> goes to <span class="math-container">$\infty$</span>, did he apply the L'Hôpital's rule to get the final result?</p>
<p>Could someone please help me with these questions? I really appreciate it!</p>

<microeconomics><mathematical-economics><game-theory><decision-theory><probability>
About The Bayesian Conditional-Probability Systems in Myerson's Game Theory: Analysis of Conflict
Beerus
# About The Bayesian Conditional-Probability Systems in Myerson's Game Theory: Analysis of Conflict

(Asked by: username_0 on 2023-09-18)

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!

**Tags:** microeconomics, mathematical-economics, game-theory, decision-theory, probability

---

## 1 answers

### Answer by username_1 on 2023-09-18. Score: 2 | Accepted answer

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 $\mu(Z\vert W\_h)=0$.


---

