A decision maker (DM) has to choose action $y\in \mathcal{Y}$ possibly without being fully aware of the state of the world. $\mathcal{Y}$ is a finite set.

The state of the world is a random variable $V$ with support $\mathcal{V}$. $\mathcal{V}$ is not a finite set. For instance, $\mathcal{V}=[a,b]$ or $\mathcal{V}=\mathbb{R}$.

When the DM chooses action $y\in \mathcal{Y}$ and the state of the world is $v\in \mathcal{V}$, she receives the payoff $u(y,v)$.

Let $P_V\in \Delta(\mathcal{V})$ be the DM's prior about the state of the world $V$ (probability density function).

The DM also processes some signal $T$ with support $\mathcal{T}$ and distribution $P_{T|V}$ conditional on $V$ to refine his prior and get a posterior on $V$, denoted by $P_{V|T}$, via the Bayes rule.

Let $S\equiv \{\mathcal{T}, P_{T|V}\}$ be called "information structure".

A strategy for the DM is $P_{Y|T}$. Such a strategy is optimal if it maximises his expected payoff, where the expectation is computed using the posterior, $P_{V|T}$.


I would like to represent the information structure that gives complete information ("complete info structure") to the DM. When $\mathcal{V}$ is finite, this is easy: we can set $\mathcal{T}=\mathcal{V}$ and $P_{T|V}(t|v)=1$ if $t=v$ and 0 otherwise. When $\mathcal{V}$ is not finite, however, I'm struggling to find an appropriate probability density function.

Note One way to represent the complete info structure is to set $|\mathcal{V}|=1$. This is not the way I am looking for. I don't want to "manipulate" $\mathcal{V}$ to obtain the desired information structure. Let us assume that $\mathcal{V}$ is fixed (e.g., $\mathcal{V}=\mathbb{R}$) and find a density $P_{T|V}$ giving complete info to the DM.


2 Answers 2


Not every probability distribution comes from a density. In particular, point masses have no density (with respect to Lebesgue measure). Here, being perfectly informed requires that the conditional distribution is concentrated on the true value and, therefore, not describable by a density function. If you want to make these things rigorous, you will have to invest in learning some measure theory.


There is a canonical way to represent information structures popularized in the literature on Bayesian persuasion and information design. You identify signals with the distribution over the posteriors they induce. Here is how this works: First, assume that all spaces under question are Polish (separable and completely metrizable) and endowed with their Borel $\sigma$-algebra. If $X$ is a Polish space, we let $\Delta(X)$ be the space of Borel probability measures on it, endowed with the topology of weak convergence and the corresponding Borel $\sigma$-algebra. One can show that this is the smallest $\sigma$-algebra such that the function $\mu\mapsto\mu(E)$ for every Borel set $E\subseteq X$ is measurable.

Now, to fill in some details in the answer by user42421, a signal is a measurable function $\sigma:\mathcal{V}\to\Delta(\mathcal{T})$. There is a probability measure $\rho$ on $\mathcal{V}\times\mathcal{T}$ defined by $$\rho(E)=\int\int 1_E(v,t)~\mathrm d\sigma_v(t)~\mathrm dP_V(v)$$ for every Borel set $E\subseteq\mathcal{V}\times\mathcal{T}$. One can actually reverse this process of integration: If $\rho$ is a probability measure on $\mathcal{V}\times\mathcal{T}$ whose $\mathcal{V}$-marginal is $P_V$, that is, $\rho(F\times\mathcal{T})=P_V(F)$ for every Borel set $F\subseteq\mathcal{T}$, one can find a measurable function $\sigma$ such that $\rho$ can be given as above. One calls then $\sigma$ a disintegration of $\rho$. Such disintegrations are unique, except for a $P_V$-null set. One can use disintegrations also to define conditional beliefs given signal realizations. If $\rho_\mathcal{T}$ is the $\mathcal{T}$-marginal of $\rho$, then there exists a measurable function $\beta:\mathcal{T}\to\Delta(\mathcal{V})$ such that $$\rho(E)=\int 1_E(v,t)~\mathrm d\beta_t(v)~\mathrm d\rho_\mathcal{T}(t)$$ for every Borel set $E\subseteq\mathcal{V}\times\mathcal{T}$. Of course, the belief $\beta(t)$ for an individual signal realization $t$ is not very meaningful since $\beta$ is only unique up to $\rho_\mathcal{T}$-null sets. Now, we also get a distribution over posteriors (the values of $\beta$), namely the element $\tau$ of $\Delta(\Delta(\mathcal{V}))$ such that $\tau=\rho_\mathcal{T}\circ\beta^{-1}$. A convenient property of $\tau$ is that for every Borel set $F\subseteq V$, one has $$\int \mu(F)~\mathrm d\tau(\mu)=P_V(F),$$ the average posterior equals the prior. It turns out that this is the only restriction on a set of posteriors induced by a signal. If $\tau\in \Delta(\Delta(\mathcal{V}))$ satisfies the condition $$\int \mu(F)~\mathrm d\tau(\mu)=P_V(F),$$ for every Borel set $F$, then there exists a space $T$ and a signal $\sigma:\mathcal{V}\to\Delta(\mathcal{T})$ such that $\tau$ is the induced distribution over posteriors. Indeed, one can let $T=\Delta(\mathcal{V})$. There exists a probability measure $\rho$ on $\mathcal{V}\times\mathcal{T}=\mathcal{V}\times \Delta(\mathcal{V})$ defined by $$\rho(E)=\int\int 1_E(v,t)~\mathrm d\mu(v)~\mathrm d \tau(\mu)$$ for every measurable set $E\subseteq\mathcal{V}\times\mathcal{T}$, The condition that $\tau$ averages to $P_V$ guarantees that the $\mathcal{V}$-marginal of $\rho$ is $P_V$. One then takes a disintegration $\sigma:\mathcal{V}\to\Delta(\mathcal{V})$ of $\rho$ as the desired signal. It has the convenient property that one can take the belief induced by $\sigma(v)$ to be $\sigma(v)$. Here is the intuition behind it. Suppose you cannot learn the value of the signal but only learn what you would believe if you would have learned the value. Your updated belief should be then this belief, so taking posteriors as signals works.

Now, what about perfect information? Well, then the posteriors should be point-masses, Dirac-measures. Indeed, you can take $\sigma$ to be given by $\sigma(v)=\delta_v$ and $\tau=P_V\circ \sigma^{-1}$.

  • $\begingroup$ Yes, I would like to make things rigorous. I've studied measure theory, but I cannot see how to formalise this here. Any rigorous help appreciated. $\endgroup$
    – Star
    Commented Oct 5, 2022 at 8:33
  • $\begingroup$ The most helpful version of conditioning for this kind of problems is given by regular conditional probabilities: en.wikipedia.org/wiki/Regular_conditional_probability $\endgroup$ Commented Oct 5, 2022 at 18:15
  • $\begingroup$ Would it be possible to formalise an answer? I've tried to apply the Wiki article you linked but I'm not sure how to proceed. $\endgroup$
    – Star
    Commented Jan 24, 2023 at 18:20
  • $\begingroup$ @user3285148 I've added the basics of the theory. $\endgroup$ Commented Jan 24, 2023 at 22:21
  • $\begingroup$ Thanks. Just some clarifications on the last line "You can take $\sigma$ to be given by $\sigma_v=\delta_v$". Take $t\in \mathcal{T}$. This means that $$\sigma_v(t)=\begin{cases} 1 & \text{if $t=v$} \\ 0 & \text{otherwise}\end{cases}$$ Now, I don't understand what is the meaning of $t=v$ in the above expression given that $\mathcal{V}$ is not finite. I am back to the issue of null sets. $\endgroup$
    – Star
    Commented Jan 26, 2023 at 18:37

Its basically the same function, $P(t|v)=1$ if $t=v$ and zero otherwise.

Let $V$ be a state space . An information structure if a map $q:V\to \Delta(T)$. $P$ and $q$ induce a distribution over $V\times T$, call it $Q$. $P_{V|T}$ is the Bayesian update of $Q$ on $T$. This is how I typically see these type of information structures modeled. But I hope you can see that it is equivalent to how you introduced it.

The complete information structure would be defined as follows: Let $\tau:V\to T$ be an injective function. Let $t_v=\tau(v)$, so it $v\neq v'$ then $t_v\neq t_{v'}$. Let $q(v)=\delta_{t_v}(\cdot)$ where $\delta_t$ is the Dirac measure on $t$


$\delta_t(t')=1$ if $t=t'$ and zero otherwise, that is$\delta_t(\cdot)$ puts all of its mass on $t$.

With this $q$, $P_{V|T}(v|t)=1$ if $v=t_v$ and zero otherwise.

Its just the same concept as before (1) $T$ is a relabeling of the states in $V$, so $t_v$ corresponds to $v$. (2) Have $q$ simply announce state by putting all of the weight on the relabeled state, $q(v)(t_v)=1$.


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