I'm working on a game theory model of imperfect information, where players observe certain attributes via noisy signals. Specifically, Player 1 has the opportunity to choose any value $\eta$ from the interval $[0, h]$ to transfer to Player 2. $h$, the maximum he may transfer, is drawn from the uniform distribution $U(0,1)$. Player 1 knows $h$, but Player 2 only knows its distribution, and rather than observe it directly, receives a noisy signal $S_h \sim U(h - \epsilon, h + \epsilon)$ about its value. Last, Player 2 also receives a noisy signal $S_\eta \sim U(\eta - \epsilon, \eta + \epsilon)$ about Player 1's choice.

Both players also know:

  • $a$ is some constant known to both players. $0 \leq a \leq 1$. $a$ defines the threshold that Player 2 will use to decide how to behave. For example, if $a = .75$, then Player 2 will be cooperative if Player 1 chooses $\eta$ of at least 75% of $h$.
  • $\epsilon$ is some number known to both players. $0 < \epsilon \leq 1$.

How can I find the CDF:

$$P(\eta - ah < x \mid S_\eta, S_h, a, \epsilon)$$

This is the probability from Player 2's point of view that Player 1 chose an $\eta$ above the threshold ($ah$) given Player 2's signals $S_\eta$ and $S_h$. Evaluated at 0, this will give $P(\eta < ah \mid S_\eta, S_h, a, \epsilon)$.

I assume you may need a prior about $\eta$ to solve this. Since Player 1 may choose any $\eta \in [0, h]$, I suppose the best to use is just the Uniform distribution $U(0, S_h)$, but I'm really not sure. Since Player 2 doesn't know $h$, it's hard to have a reliable prior distribution. I'm not sure what's normally done in these circumstances.


I've posted about this before, but in that case trying to solve the CDF from Player 1's point of view. In case it's helpful, here is what I found:

$$ P(S_{\eta} - aS_{h} < x \mid \eta, h, a, \epsilon) = \begin{cases} 0, & \text{ if } x < \eta - \epsilon - ah - a\epsilon \\ \dfrac{1}{8a\epsilon^2}(x - \eta + \epsilon + ah + a\epsilon)^2, & \text{ if } x \in [\eta - \epsilon - ah - a\epsilon, \eta - \epsilon - ah + a\epsilon] \\ \dfrac{a}{2} + \dfrac{1}{2\epsilon}(x-\eta + \epsilon + ah - a\epsilon), & \text{ if } x \in [\eta - \epsilon - ah + a\epsilon, \eta + \epsilon - ah - a\epsilon] \\ 1 - \dfrac{1}{8a\epsilon^2}(\eta + \epsilon - ah + a\epsilon - x)^2, & \text{ if } x \in [\eta + \epsilon - ah - a\epsilon, \eta + \epsilon - ah + a\epsilon] \\ 1, & \text{ if } x > \eta + \epsilon - ah + a\epsilon \\ \end{cases} $$

This distribution is trapezoidal, hence the title of the question.

  • $\begingroup$ I think that because $a h$ and $a S_h$ are both uncertain with respect to each other by $a \epsilon$ and because $\eta$ and $S_{\eta}$ are similarly uncertain by $\epsilon$ that the answers are actually the same, just replace $h$ with $S_h$ and $\eta$ with $S_{\eta}$ $\endgroup$ – MikeP Mar 4 '16 at 20:28
  • $\begingroup$ Do you know how I might be able to test that with a simulation? The only way I could come up with is to sample randomly from $U(0,1)$ to get an $h$, and then sample randomly from $U(0,h)$ to get an $\eta$. Then find $S_\eta$ and $S_h$ by drawing from $U(\eta-\epsilon, \eta+\epsilon)$ and $U(h-\epsilon, h+\epsilon)$ respectively, and keep track of how many times $\eta < ah$ for each unique $S_h, S_\eta$ pair. The problem is that I think you would have to do this 10^12 times or more to get anything useful. $\endgroup$ – sundance Mar 5 '16 at 4:31
  • $\begingroup$ I can't really think of any way other that what you suggest. However, it might be possible to design the experiment a little less randomly (maintaining the distributions) and get clear results with fewer runs. Let me play with some things in Matlab and I'll try to validate my conjecture :) $\endgroup$ – MikeP Mar 7 '16 at 13:14

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