Is the following claim true?

$\textbf{Claim:}$ The finer a partition $P$ it becomes, the more informative is.

Let me give an example of a partition on some space where the Lebesgue measure is defined properly.

Suppose that there is a finite set $\Omega$ and $\omega\in\Omega$ states the state of the world. There is prior $\mu_0(\omega)$ and $\mu$ is the posterior belief about the state of the world. Also there is a random variable $X$ that is indepedent of $\Omega$ and uniformly distributed on $[0,1]$ and denote as $x$ a realization of $X$. A signal is a partition of $\Omega\times [0,1]$ denoted as $\pi$ such that $\pi\subset S$ where $S$ is the non-emtpy set that contains all Lebesgue measurable subsets of $\Omega\times [0,1]$. An element $s\in S$ is a signal realization.

With each $\pi$ is assocaited an $S-$valued random variable that $s\in\pi$ when $(\omega, x)\in s$. Let $\Lambda$ be the Lebesgue measure defined as $\Lambda(\{x|(\omega, x)\in s \})=\mathbb{P}(s|\omega)$ which is the conditional probability of $s$ given $\omega$ and $\mathbb{P}(s)=\sum_{\omega\in\Omega}\Lambda(\{x|(\omega, x)\in s \})=\sum_{\omega\in\Omega}\mathbb{P}(s|\omega)$ which is the unconditional probability of $s$.

Assuming that we have two different partitions $\pi_1$ and $\pi_2$ defined as above, by combining both of them to their common refinement $\pi^*=\pi_1 \vee \pi_2$ in the sense that the signal $\pi^*$ yields the same information as observing both $\pi_1$ and $\pi_2$, this mean that $\pi^*$ is more informative of $\pi_1$ and $\pi_2$ alobe because it consists of all the points of both of them (correct me if I am mistaken). Does this also mean that $\pi^*$ (partition or signal) is a finer than $\pi_1$ or $\pi_2$ alone as well?

$\textbf{Note:}$ The operator $\vee$ stands for the join of two different paritions, which seems like some kind of ``adding" the different signals (partitions).


2 Answers 2


Some of the sentences in the answer you wrote don't make sense to me: e.g. ''$g$ stands for the signal that the good state occurs with probability Q almost surely'' doesn't make any sense: isn't $g$ a potential signal? How does it stand for the probability $Q$ almost surely? What does that even mean, almost surely with respect to what? Further, talking about paritions doesn't really require us to invoke any probability at all. Nevertheless, much of what you write makes sense. In any case, it'd perhaps be simpler to just try to clearly understand partitions, without all of the baggage of the particular setting attached.

Take any set $\mathcal{S}$. Then a partition $ \pi$ of $\mathcal{S}$ consists of disjoint nonempty subsets of $\mathcal{S}$ that cover it, i.e., $\bigcup_{\mathcal{T} \in \pi} \mathcal{T} = \mathcal{S},$ and if $\mathcal{T,U} \in \pi, \mathcal{T}\cap \mathcal{U} = \emptyset$. A partition $\sigma$ is finer than a partition $\pi$ if for any $\mathcal{U} \in \sigma,\exists \mathcal{T} \in \pi: \mathcal{U} \subseteq \mathcal{T}.$ In other words, if you can interpret $\sigma$ as a union of partitions over the parts of $\pi$ itself, then $\sigma$ is finer. If not, then it is not meaningful to say that $\sigma$ is finer: e.g. for $\mathcal{S} = \{1,2,3,4\},$ if $\pi_1 = \{ \{1,2\}, \{3,4,\}\}$ and $\pi_2 = \{ \{1\}, \{2,3,4\}\}$, then $\pi_2$ is not finer that $\pi_1$ (since it doesn't separate $2$ from $(3,4)$), but also $\pi_1$ is not finer than $\pi_2$, because it doesn't separate $1$ from $2$. Of course, $\pi_3 = \{ \{1\}, \{2\}, \{3\},\{4\}\}$ is finer than both $\pi_1$ and $\pi_2$.

Partitions are intimately associated with an experimental interpretation (which is where their use in economics, engineering &c arises). Suppose that there is some underlying signal $s$ that you care about, which takes value in the set $\mathcal{S}$ (e.g. $\mathcal{S}$ is $[0,\infty)$, and $s$ is the weight of someone in pounds). Further suppose that you can perform some experiment, and at its conclusion, you get to observe $f(s)$ for some function $f:\mathcal{S} \to \mathcal{O}$ for some set of possible observable values $\mathcal{O}$ (for example, you ask the person to step on some scales, that tell you the weight to the nearest pound if the weight is $< 400$, and otherwise tells you that the weight exceeds $400.$ Here $\mathcal{O} = \{0,1,\cdots, 399\},$ and $f(s) = i$ if $s \in [i - 0.5, i + 0.5)$). Now, this function $f$ induces a partition on $\mathcal{S}$, namely $\pi_f = \{ f^{-1}(i) : i \in \mathcal{O}\}$ (in our previous example, this would be $\pi_f = \{ [0,0.5), [0.5, 1.5), \cdots, [399.5,400), [400, \infty)\}$). The point is that after the conclusion of the experiment, you can certainly say that $s \in f^{-1}(o)$ for the output $o$, and so the partition is capturing how the experiment localises the signal in the signal space.

The join of two partition $\pi_1, \pi_2$ is the crudest partition that is finer than both $\pi_1$ and $\pi_2$. This is the same as the partition that contains $\mathcal{T}_1 \cap \mathcal{T}_2$ for any $\mathcal{T}_1 \in \pi_1, \mathcal{T}_2 \in \pi_2,$ but you should prove this (I haven't seen the notation $\pi_1 \cap \pi_2$ before, but this is natural, interpreting $\cap$ in the Minkowski sense). The experimental interpretation is as follows: say you have two experiments $f, g$, and observe $f(s) = i, g(s) = j$. Individually these tell you that $s \in f^{-1}(i),$ and $s\in g^{-1}(j),$ but logically you can infer that $s \in f^{-1}(i) \cap g^{-1}(j),$ and you have localised $s$ to a smaller set than you could with either just $f$ or just $g$. The corresponding partition is $\pi_{(f,g)} = \{ (f^{-1}(i) \cap g^{-1}(j) ) : i \in \mathcal{O}_g, j \in \mathcal{O}_g\}$, which happens to be exactly $\pi_f \vee \pi_g.$ Note here that $(f,g)$ is more informative than $f$ (or $g$), since trivially, anything you could say about $s$ after viewing just $f(s)$, you could also say after viewing both $f(s)$ and $g(s)$.

This brings us to finer=more informative. One natural definition of an experiment $f$ being more informative than an experiment $g$ is if anything you could conclude about $s$ after observing $g(s)$ is also a conclusion after observing $f(s)$. This indeed is the same as saying that the parition $\pi_f$ is finer than the parition $\pi_g$. Indeed, if $\pi$ is finer than $\rho$, and you conclude that $s \in \mathcal{T}_\rho$ for some $\mathcal{T}_\rho \in \rho,$ and that $s \in \mathcal{T}_\pi$ for some $\mathcal{T}_\pi \in \pi,$ then clearly it holds $\mathcal{T}_\pi \subseteq \mathcal{T}_\rho$, and $\mathcal{T}_\pi \not\subset \mathcal{T}'_\rho$ for any $\mathcal{T}'_\rho \in \rho$ (why?), and so you can exactly infer what you could conclude from $\rho$. So, in our weight example, if I gave you scales that told you the weight to the nearest half-pound, you could predict what the original nearest-pound scales would have said, and this is more informative. You should work out the corresponding partition, and verify that it is finer than the nearest-pound partition.

It's worth noting that this definition of informativeness doesn't completely capture everything, since if $\pi$ and $\rho$ are incomparable in that neither is finer than the other, this definition says nothing about which is more informative. Nevertheless, if $\pi = \{ \{1,2\}, \{3,4,5\}\}\}$ and $\rho = \{ \{1\} ,\{2,3\}, \{4\}, \{5\}\}$, then it seems that $\rho$ is more informative than $\pi$. Other notions of informativeness can capture such properties, if they are of interest.

Now after all this talk of partitions, you start talking about sigma algebrae. These are qualitatively different types of objects, and in particular no partition is a sigma-algebra, because the latter must contain the empty set, and the former must not. There certainly is an associated notion of informativeness attached to sigma-algebrae, which you need if you start talking about randomised experiments, but that is its own can of worms that requires a solid understanding of sigma-algebrae themselves. Just roughly, though, sigma algebrae capture something like a set of possible questions you could answer about the signal given the output of a randomised experiment (the subtlety is that for continuous sets such as $[0,1]$, this can't be all questions, due to technicalities involving probability). Then a sigma-algebra is "finer" if the set of questions you can answer is richer, again in an inclusion sense. The join operator carries a similar experimental interpretation, but the join of two sigma-algebrae is a much more complicated object).

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    $\begingroup$ Thank you for your response. I will make some changes in my answer and I will let you know...I need a second view and I think your response helped me to some point understand some things $\endgroup$ Nov 7, 2023 at 19:03

Suppose that $\Omega =\{0,1\}$ where $0$ denotes the good and $1$ denotes the bad state of the world. If $\pi_1=\{g, b\}$ where $g$ stands for the signal that the good state occurs with probability $\mathbb{Q}$ almost surely and $b$ stands for the signal that the bad state occurs with probability $\mathbb{Q}$ almost surely as well such taht $g=\left((0,[0,0.6])\cup(1, [0, 0.25])\right)$ and $b=\left((0,[0.6,1])\cup(1, [0.25, 1])\right)$ or else

$$\pi_1=\{g, b\}=\{\underbrace{\left((0,[0,0.6])\cup(1, [0, 0.25])\right)}_{g},\underbrace{ \left((0,[0.6,1])\cup(1, [0.25, 1])\right)\}}_{b}$$

and also there is $\pi_2$ such that

$$\pi_2=\{g^{'}, b^{'}\}=\{\underbrace{\left((0,[0,0.65])\cup(1, [0, 0.2])\right)}_{g^{'}},\underbrace{ \left((0,[0.65,1])\cup(1, [0.2, 1])\right)\}}_{b^{'}}$$


$$\pi_1\vee\pi_2=\{\underbrace{(0,[0,0.6])\cup(1,[0,0.2])}_{(g,g^{'})},\underbrace{(0,[0.6,0.65])}_{(b,g^{'})},\underbrace{(1,[0.2,0.25]))}_{(g,b^{'})},\underbrace{(0,[0.65,1])\cup(1,[0.25,1])}_{(b,b^{'})},\}$$ in other words


If we define the information environment as $\Pi := \Omega\times [0,1]$ and x is a realization of $X$ which is a random variable that is uniformly distributed on $[0,1]$ then any interval has probability proportional to its length. This means that if $x=0.4$ is a realization of X, then for $\omega=0$

$$\Lambda(\{x=0.4|(0, 0.4)\in \pi_1\vee\pi_2\})=\Lambda(\{x=0.4|(0, 0.4)\in s=(g,g^{'})\})=\Lambda(\{x=0.4|\underbrace{(0, 0.4)\in (0,[0,0.6])\cup(1,[0,0.2])}_{(0,0.4)\in (0,[0,0.6]),\quad\text{since $\omega=0$ and $0.4\in [0,0.6]$ }} \})=\mathbb{P}(s=(g,g^{'})|\omega=0)=\mathbb{P}([0,0.4]\cup[0.4,0.6]|\omega=0)=|0.4-0|+|0.6-0.4|=0.6$$

while for $\omega=1$

$$\Lambda(\{x=0.4|(1, 0.4)\in \pi_1\vee\pi_2\})=\Lambda(\{x=0.4|(1, 0.4)\in s=(b,b^{'})\})=\Lambda(\{x=0.4|\underbrace{(1, 0.4)\in (0,[0.65,1])\cup(1,[0.25,1])}_{(1, 0.4)\in (1,[0.25,1]),\quad\text{since $\omega=1$ and $0.4\in [0.25,1]$}}\})=\mathbb{P}(s=(b,b^{'})|\omega=1)=\mathbb{P}([0.25,0.4]\cup[0.4,1]|\omega=1)=|0.4-0.25|+|1-0.4|=0.75$$

The above construction based on my definition gives me a clear view about what all these partitions of the information environment $\Pi=\Omega\times[0,1]$ is. In a chapter of Handbook of game theory with economic applications of Geanakopolos common knowledge survey it is quoted that if $P$ and $Q$ are two different partitions of the same space $\Pi$ then $P\vee Q= P\cap Q$. Maybe this does not answer exaclty to my claim, but a finer partition like $\pi^*$ is more informative that $\pi_1$ and $\pi_2$ alone and because it is the intersection of $\pi_1$ and $\pi_2$, this is close to the definition of a sigma algebra where someone conseders a smaller collection of privileged subsets of $\Pi=\Omega\times [0,1]$ which are reflected in $\pi^*$...

Am I wrong? Is there anything that I am missing? Please any additional comment or help could be very important for my understanding. Thank you in advance


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