How to prove that a strategy profile is a Proper Equilibrium?

Question

Consider the example given in Wikipedia's article on Proper Equilibrium, which is as follows:

Further, let us denote by $H$ the action "hide heads up"; by $h$ the action "guess heads up", by $T$ the action "hide tails up" and by $t$ the action "guess tails up". Also, let $g$ denote the action "grab a penny". Finally, $\sigma_i(X)$ denotes the probability with which Player $i$ chooses pure strategy $X$.

Then, it is easy to see that the set of strategy profiles given by $\sigma_1(H) \in [0,1]$ and $\sigma_2(g)=1$ constitutes the set of Nash Equilibria of this game. Further, it can also be shown that any such pair constitutes a Trembling Hand Perfect Equilibrium. To see so, let us define the $\epsilon$-perturbed game in which Player $i$ plays $H$ with probability $\sigma_i(H)\in[0,1]$ and plays $T$ with probability $\sigma_i(T)=1-\sigma_i(H)$; while Player $2$ plays $g$ with probability $1-\epsilon_2^h-\epsilon_2^t$, plays $t$ with probability $\epsilon_2^t$ and plays $h$ with probability $\epsilon_2^h$. Hence, the payoffs of the $\epsilon$-perturbed game are as follows:

1. For Player 1, $u_1(H) = \epsilon_2^t-1$ and $u_1(T) = \epsilon_2^h-1$.

2. For Player 2, $u_2(h) = \sigma_1(H)$, $u_2(t) = \sigma_1(T)$ and $u_2(g) = 1$.

Therefore, it follows that:

1. For Player 1, $u_1(H) = u_1(T) \Leftrightarrow \epsilon_2^t-1 = \epsilon_2^h-1$. Of course, it is possible to define a sequence $\{\epsilon_2^t,\epsilon_2^h\}_{k\geqslant 1}$ such that $u_1(H) = u_1(T)$ holds true for any $k \geqslant 1$ and $\{\epsilon_2^t,\epsilon_2^h\}_{k\geqslant 1} \longrightarrow 0$ as $k \longrightarrow \infty$. For example, the sequence $\{\epsilon_2^t,\epsilon_2^h\}_{k\geqslant 1} \equiv \{1/k,1/k\}_{k\geq 1}$ satisfies the requirements.

2. For Player 2, $u_2(g) > u_2(h)$ and $u_2(g) > u_2(t)$ for any sequence $\{\epsilon_1^h\}_{k\geqslant 1}$, for instance $\{\epsilon_1^h\}_{k\geqslant 1}\equiv\{1/k\}$. Hence, in any $\epsilon$-perturbed game, $\sigma_2(g)=1$.

I think that this suffices to show that any Nash Equilibrium strategy profile of this game is also Trembling Hand Perfect. However, I don't see how I can further refine the argument to show that only one of these strategy profiles is actually Proper. Following the Wikipedia's article on Proper Equilibrium, the only Proper Equilibrium strategy profile of this game is:

$\sigma^{*} = \Big\{ \sigma_1^{*}(H)=\frac{1}{2}$ and $\sigma_2^{*}(g)=1\Big\}$

Intuitively, it makes sense. However, I can't come up with a way of showing that $\sigma^{*}$ is the unique Proper Equilibrium. Therefore, my questions are:

1. How can I formally show that $\sigma^{*}$ is the unique Proper Equilibrium of this game?

2. Can you improve my argument to show that any Nash Equilibrium is also Trembling Hand Perfect of this game?

Thank you all very much for your time.

Answer 1

I never planned to answer this question myself, but it turns out that months have passed without an answer and I now know how to go about solving this problem. Since this is the Wikipedia example, I think that answering this question may be useful for other people. Let's address both questions.

1. I have made some edits to my original question. Now, I think that the reasoning posted above is a sufficiently good argument to prove that all Nash Equilibria of that game are also Trembling Perfect.

2. The additional requirement when moving from Trembling Hand Perfection to Properness is that, given some NE strategy profile and some accordingly $\epsilon$-perturbed game, costlier mistakes are less likely. To see how this principle formally translates in our game, let us assume that $\sigma_2(g)=1$ and, without loss of generality, let $\sigma_1(H)>\sigma_1(T)$ (by symmetry, an analogous argument applies if $\sigma_1(H)<\sigma_1(T)$). Then, $\{\epsilon_2^t,\epsilon_2^h\}_{k\geqslant1}$ must satisfy $\{\epsilon_2^t,\epsilon_2^h\}_{k\geqslant1}\to 0$ as $k\to\infty$ and $\epsilon_2^t<\epsilon_2^h$ for all $k\geqslant 1$. However, $u_1(H)<u_1(T)$ for any sequence of mistakes satisfying $\epsilon_2^t<\epsilon_2^h$, thus implying that $\sigma_1(H)<\sigma_1(T)$. This contradiction shows that no strategy profile involving $\sigma_1(H)\neq\sigma_1(T)$ can be a proper Equilibrium. Because the set of Proper Equilibrium strategy profiles is non-empty for finite games and is also a (potentially proper) subset of Trembling Hand Perfect Equilibrium, the proof is done.

I hope this helps someone else!