# References for this game

I would like to know how the following game is known in the literature and, possibly, to have references for related papers.

Description of the game: Fix a space $X$ and two Borel probability measures $\mu$ and $\nu$ over $X$. There are two players, $A$ and $B$. They both know $\mu$ and $\nu$. Player $A$ chooses between $\mu$ and $\nu$. Player $B$ can not observe Player $A$'s choice. Say $\mu$ is chosen by $A$. Then an element $x\in X$ is randomly chosen in accordance with $\mu$. Now player $B$, looking at $x$, must guess the choice of $A$, i.e., Player $B$ must say "you piked $\mu$", or "you picked $\nu$". Player $B$ wins if their guess is correct. Player $A$ wins otherwise.

How to formalize the game:

1. The strategies for Player $A$ can be formalized as (randomized) choices over the two element set, i.e., as elements in $[0,1]$.

2. A strategy for Player $B$ can be formalized as a map $\sigma: X\rightarrow [0,1]$: if $x$ is the outcome, then guess $\mu$ with probability $\sigma(x)$ and $\nu$ with probability $1-\sigma(x)$.

3. Since Player $B$ can not observe Player $A$'s choice, the game can be consider as played concurrently.

I believe the game has an optimal equilibrium and its value is a function of $\displaystyle \bigsqcup_{B\ Borel} | \mu(B) - \nu(B) |$, i.e., of the total variation distance between $\mu$ and $\nu$.

Thank you in advance for any information.

Matteo

• Looks like Bayesian statistics. – Gerald Edgar Apr 1 '12 at 18:53

I interpret your game as follows. $$A$$ chooses $$\alpha\in[0,1]$$, $$B$$ chooses some measurable $$\sigma:X\to[0,1]$$, resulting in a probability $$u(\alpha,\sigma)=\int [\alpha(1-\sigma)\ d\mu + (1-\alpha)\sigma\ d\nu]$$ of $$B$$ guessing incorrectly. Here, the players move simultaneously, $$A$$ wants to maximize $$u$$ and $$B$$ wants to minimize $$u$$.

It's possible to explicitly compute a value for this game in general, but let's consider a special case (which is, in a sense, not very special) that cleans up notation, in order to specifically address the question of how the value of the game relates to total variation.

Specifically, for a given nondecreasing $$f:[0,1]\to[0,2]$$ with $$\int_0^1 f=1$$, suppose $$X=[0,1]$$, measure $$\mu$$ has Lebesgue density $$f$$, and $$\nu$$ has Lebesgue density $$2-f$$. Then, consider $$\alpha^*:=1-\tfrac12 f(\tfrac12)$$ and $$\sigma^*:=\mathbf1_{[\tfrac12,1]}$$. One can verify that this is a Nash equilibrium (a.k.a. saddle point): $$\alpha^*$$ maximizes $$u(\cdot,\sigma^*)$$ and $$\sigma^*$$ minimizes $$u(\alpha^*,\cdot)$$, and the value it produces is exactly $$v = \int_0^{\tfrac12}f.$$ Meanwhile, the total variation between $$\mu$$ and $$\nu$$ can be computed as $$TV = 2\int_0^1 \max\{f-1, 0\}.$$

So the question, restricted to this special case, reduces to: When looking at nondecreasing functions $$f:[0,1]\to[0,2]$$, can one express $$\int_0^{\tfrac12}f$$ as a function of $$2\int_0^1 \max\{f-1, 0\}$$?

If it so happens that $$f(\tfrac12)=1$$, then $$\tfrac12 TV=\int_0^1 \max\{f-1, 0\}=\int_{\tfrac12}^1 (f-1)=\left(\int_0^1 f \right) - \tfrac12 - \int_0^{\tfrac12} f = \tfrac12-v,$$ implying $$v=\tfrac{1-TV}2$$.

But, if $$\int_0^1 \max\{f-1, 0\}\neq\int_{\tfrac12}^1 (f-1)$$, then the above algebra shows that $$v\neq\tfrac{1-TV}2$$. Finally, it's easy to find two continuous, strictly increasing functions $$f,\hat f:[0,1]\to[0,2]$$ such that $$\int_0^1 \max\{f-1, 0\}=\int_0^1 \max\{\hat f-1, 0\}$$ but $$f(\tfrac12)=1\neq \hat f(\tfrac12)$$.

So in summary, the value cannot generally be expressed as a function of total variation. However, the original post reflected a good intuition. If we restrict to the case that there is "enough symmetry" between $$\mu$$ and $$\nu$$ that player $$A$$ could choose a 50/50 mixture in equilibrium, then the value is a function of total variation.