Appendix D.1.1. Case 1a: b_{X} ≥ 0, a_{Y} + b_{Y} > 0
In this section, we consider a relaxed version of the class of coordination game as in
Section 4.3. We prove theorems presented in
Section 4.3, showing that these results can in fact be extended to the case where
${a}_{Y}+{b}_{Y}>0$, instead of requiring
${a}_{Y}>0$ and
${b}_{Y}>0$.
First,
${a}_{Y}+{b}_{Y}>0$,
${y}^{II}$ is an increasing function of
x, meaning
This implies that both players tend to agree to each other. Intuitively, if ${a}_{Y}\ge {b}_{Y}$, then both players agree that the first action is the better one. For this case, we can show that, no matter what ${T}_{y}$ is, the principal branch lies on $x\in \left(\frac{1}{2},1\right)$. In fact, this can be extended to the case whenever ${T}_{y}>{T}_{I}$, which is the first part of Theorem 1.
Proof of Part 1 of Theorem 1. We can find that, for ${T}_{y}>{T}_{I}$, ${y}^{II}(1/2,{T}_{Y})>\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$ for any ${T}_{y}$ according to Proposition A1. Since ${y}^{II}$ is monotonically increasing with x, ${y}^{II}>\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$ for $x>1/2$. This means that ${T}_{X}^{II}>0$ for any $x\in (1/2,1)$. Additionally, it is easy to see that ${lim}_{x\to {1}^{-}}{T}_{X}^{II}=0$. As a result, $(0.5,1)$ contains the principal branch. ☐
For Case 1a with ${a}_{Y}\ge {b}_{Y}$, on the principal branch, the lower the ${T}_{x}$, the closer x is to 1. We are able to show these monotonicity characteristics in Proposition A2, and they can be used to justify the stability owing to Lemma 1.
Proposition A2. In Case 1a, if ${a}_{Y}\ge {b}_{Y}$, then $\frac{\partial {T}_{X}^{II}}{\partial x}<0$ for $x\in \left(\frac{1}{2},1\right)$.
Proof. It suffices to show that
$L(x,{T}_{y})>\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$ for
$x\in \left(\frac{1}{2},1\right)$. Note that, according to Proposition A1, if
${a}_{Y}\ge {b}_{Y}$,
Since
${y}^{II}(x,{T}_{y})$ is monotonically increasing when
${a}_{Y}+{b}_{Y}>0$,
${y}^{II}(x,{T}_{y})>\frac{1}{2}$ for
$x\in \left(\frac{1}{2},1\right)$. As a result,
$1-2{y}^{II}<0$; hence, we can see that, for
$x\in \left(\frac{1}{2},1\right)$,
Consequently, for $x\in \left(\frac{1}{2},1\right)$, $L(x,{T}_{y})>\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$; hence, $\frac{\partial {T}_{X}^{II}}{\partial x}<0$ according to Lemma A2.
Proof of Part 1 of Theorem 3. According to Lemma 1, Proposition A2 implies that all $x\in (0.5,1)$ is on the principal branch. This directly leads us to Part 1 of Theorem 3. ☐
Next, if we look into the region $x\in (0,1/2)$, we can find that, in this region, QREs appears only when ${T}_{x}$ and ${T}_{y}$ are low. This observation can be formalized in the proposition below. We can see that this proposition directly proves Parts 2 and 3 of Theorem 2, as well as Part 2 of Theorem 3.
Proposition A3. Consider Case 1a. Let ${x}_{1}=min\left\{\frac{1}{2},\frac{-{T}_{y}ln\left(\frac{{a}_{X}}{{b}_{X}}\right)+{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right\}$ and ${T}_{B}=\frac{{b}_{Y}}{ln({a}_{X}/{b}_{X})}$. The following statements are true for $x\in (0,1/2)$:
- 1.
If ${T}_{y}>{T}_{B}$, then ${T}_{X}^{II}<0$.
- 2.
If ${T}_{y}<{T}_{B}$, then ${T}_{X}^{II}>0$ if and only if $x\in (0,{x}_{1})$.
- 3.
$\frac{\partial L}{\partial x}>0$ for $x\in (0,{x}_{1})$.
- 4.
If ${T}_{y}<{T}_{I}$, then $\frac{\partial {T}_{X}^{II}}{\partial x}>0$.
- 5.
If ${T}_{y}>{T}_{I}$, then there is a nonnegative critical temperature ${T}_{C}\left({T}_{y}\right)$ such that ${T}_{X}^{II}(x,{T}_{Y})\le {T}_{C}\left({T}_{y}\right)$ for $x\in (0,1/2)$. If ${T}_{Y}<{T}_{B}$, then ${T}_{C}\left({T}_{y}\right)$ is given as ${T}_{X}^{II}\left({x}_{L}\right)$, where ${x}_{L}\in (0,{x}_{1})$ is the unique solution to $L(x,{T}_{y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$.
Proof. For the first and second part, consider any
$x\in (0,1/2)$ and we can see that
Note that for ${T}_{y}>\frac{{b}_{Y}}{ln({a}_{X}/{b}_{X})}={T}_{B}$, we have ${x}_{1}<0$; hence, ${T}_{X}<0$.
From the above derivation, for all
$x\in (0,1/2)$ such that
${T}_{X}^{II}(x,{T}_{y})>0$,
${y}^{II}<1/2$ since
$\frac{{b}_{X}}{{a}_{X}+{b}_{X}}<1/2$. Then
Further, when ${T}_{y}<{T}_{I}$, ${y}^{II}(1/2,{T}_{y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. This implies that, for $x\in (0,1/2)$, ${y}^{II}(x,{T}_{y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. Since $\frac{\partial L}{\partial x}>0$, and L is continuous, $L(x,{T}_{y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$ for $x\in (0,1/2)$. This implies the fourth part of the proposition.
Next, if we look at the derivative of
${T}_{X}^{II}$,
We can see that any critical point in $x\in (0,1/2)$ must satisfy $L(x,{T}_{y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. When ${T}_{y}>{T}_{I}$, ${x}_{1}<1/2$, and $L({x}_{1},{T}_{y})>{y}^{II}({x}_{1},{T}_{y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. If ${T}_{y}<\frac{{b}_{Y}}{ln({a}_{X}/{b}_{X})}$, then ${lim}_{x\to 0+}{T}_{X}={y}^{II}(0,{T}_{Y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. Hence, there is exactly one critical point for ${T}_{X}$ for $x\in (0,{x}_{1})$, which is a local maximum for ${T}_{X}$. If ${T}_{y}>\frac{{b}_{Y}}{ln({a}_{X}/{b}_{X})}$, then we can see that ${T}_{X}$ is always negative, in which case the critical temperature is zero. ☐
The results in Proposition A3 not only apply for the case ${a}_{Y}\ge {b}_{Y}$ but also general cases about the characteristics on $(0,1/2)$. According to this proposition, we can conclude the following for the case ${a}_{Y}\ge {b}_{Y}$, as well as the case ${a}_{Y}<{b}_{Y}$ when ${T}_{y}>{T}_{I}$:
The temperature ${T}_{B}=\frac{{b}_{Y}}{ln({a}_{X}/{b}_{X})}$ determines whether there is a branch appears in $x\in (0,1/2)$.
There is some critical temperature ${T}_{C}$. If we raise ${T}_{x}$ above ${T}_{C}$, then the system is always on the principal branch.
The critical temperature ${T}_{C}$ is given as the solution to the equality $L(x,{T}_{Y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$.
When there is a positive critical temperature, though it has no closed form solution, we can perform a binary search to look for $x\in (0,{x}_{1})$ that satisfies $L(x,{T}_{y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$.
Another result we are able to obtain from Proposition A3 is that the principal branch for Case 1a when ${T}_{y}<{T}_{I}$ lies on $(0,1/2)$.
Proof of Part 2 of Theorem 1. First, we note that ${T}_{y}<{T}_{I}$ is meaningful only when ${b}_{Y}>{a}_{Y}$, for which case we always have ${T}_{I}<{T}_{B}$. From Proposition A3, we can see that for ${T}_{Y}^{II}<{T}_{I}$, we have ${x}_{1}=1/2$; hence, ${T}_{X}^{II}>0$ for $x\in (0,1/2)$. From Proposition A1, we already have ${lim}_{x\to {\frac{1}{2}}^{-}}{T}_{X}^{II}=\infty $. Additionally, it is easy to see that ${lim}_{x\to {0}^{+}}{T}_{X}^{II}=0$. As a result, since ${T}_{X}^{II}$ is continuously differentiable over $(0,0.5)$, for any ${T}_{x}>0$, there exists $x\in (0,0.5)$ such that ${T}_{X}^{II}(x,{T}_{y})={T}_{x}$. ☐
What remains to be shown is the characteristics on the side
$(1/2,1)$ when
${b}_{Y}>{a}_{Y}$. In
Figure 4 and
Figure 5, for low
${T}_{y}$, the branch on the side
$(1/2,1)$ demonstrated a similar behavior as what we have shown in Proposition A3 for the side
$(0,1/2)$. However, for a high
${T}_{y}$, while we still can find that
$(0,1/2)$ contains the principal branch, the principal branch is not continuous. These observations are formalized in the following proposition. From this proposition, the proof of Part 4 of Theorem 2 directly follows.
Proposition A4. Consider Case 1a with ${b}_{Y}>{a}_{Y}$. Let ${x}_{2}=max\left\{\frac{1}{2},\frac{-{T}_{Y}ln\left(\frac{{a}_{X}}{{b}_{X}}\right)+{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right\}$ and ${T}_{A}=max\left\{0,\frac{-{a}_{Y}}{ln({a}_{X}/{b}_{X})}\right\}$. The following statements are true for $x\in (1/2,1)$.
If ${T}_{y}<{T}_{A}$, then ${T}_{X}^{II}<0$.
If ${T}_{y}>{T}_{A}$, then ${T}_{X}^{II}>0$ if and only if $x\in ({x}_{2},1)$.
For $x\in \left[\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}},1\right)$, we have $\frac{\partial L}{\partial x}>0$.
If ${T}_{y}\in ({T}_{A},{T}_{I})$, then there is a positive critical temperature ${T}_{C}\left({T}_{y}\right)$ such that ${T}_{X}^{II}(x,{T}_{y})\le {T}_{C}\left({T}_{y}\right)$ for $x\in (1/2,1)$, given as ${T}_{C}\left({T}_{y}\right)={T}_{X}^{II}\left({x}_{L}\right)$, where ${x}_{L}\in (1/2,1)$ is the unique solution of $L(x,{T}_{y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$.
Proof. For the first part and the second part, consider
$x\in (1/2,1)$, and we can find that
Note that, for ${T}_{y}>{T}_{I}$, ${x}_{2}=1/2$. Additionally, if ${T}_{y}<{T}_{A}$, then ${T}_{X}^{II}<0$ for all $x\in (1/2,1)$.
For the third part,
${y}^{II}\ge \frac{1}{2}$ for all
$x\ge \frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}}$ and
$\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}}>\frac{1}{2}$. Thus,
For the fourth part, we can find that any critical point of
$L(x,{T}_{Y})$ in
$(0,1)$ must be either
$x=\frac{1}{2}$ or satisfies the following equation:
Consider
$G(x,{T}_{y})=(1-2x)+x(1-x)(1-2{y}^{II})\frac{{a}_{Y}+{b}_{Y}}{{T}_{y}}$. For
${b}_{Y}>{a}_{Y}$,
${y}^{II}(1/2,{T}_{y})$ is strictly less than
$1/2$. Additionally,
$\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}}>1/2$. Now,
$G(1/2,{T}_{y})>0$ and
$G(\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}},{T}_{y})<0$. Next, we can see that
$G(x,{T}_{y})$ is monotonically decreasing with respect to
x for
$x\in \left(\frac{1}{2},\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right)$ by looking at its derivative:
As a result, there is some ${x}^{*}\in \left(\frac{1}{2},\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right)$ such that $G({x}^{*},{T}_{y})=0$. This implies that $L(x,{T}_{y})$ has exactly one critical point for $x\in \left(\frac{1}{2},\frac{{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right)$. Additionally, if $G(x,{T}_{y})>0$, $\frac{\partial L}{\partial x}<0$; if $G(x,{T}_{y})<0$, then $\frac{\partial L}{\partial x}>0$. Therefore, ${x}^{*}$ is a local minimum for L.
From the above arguments, we can conclude that the shape of $L(x,{T}_{y})$ for ${T}_{y}<{T}_{I}$ is as follows:
There is a local maximum at $x=1/2$, where $L(1/2,{T}_{y})=y(1/2,{T}_{y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$.
L is decreasing on the interval $\left(\frac{1}{2},{x}^{*}\right)$, where ${x}^{*}$ is the unique solution to Equation (A7).
L is increasing on the interval $({x}^{*},1)$. If ${T}_{y}>{T}_{A}$, then ${lim}_{x\to {1}^{-}}L(x,{T}_{y})=y(1,{T}_{y})>\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$.
Finally, we can claim that there is a unique solution to $L(x,{T}_{Y})=\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$, and such a point gives a local maximum to ${T}_{X}^{II}$. ☐
The above proposition suggests that, for
${T}_{y}\in ({T}_{A},{T}_{I})$, we are able to use binary search to find the critical temperature. For
${T}_{y}>{T}_{I}$, unfortunately, with a similar argument of Proposition A4, we can find that there are potentially at most two critical points for
${T}_{X}^{II}$ on
$(1/2,1)$, as shown in
Figure 5, which may induce an unstable segment between two stable segments. This also proves Part 3 of Theorem 3.
Now, we have enough materials to prove the remaining statements in
Section 4.3.
Proof of Parts 1, 5, and 6 of Theorem 2, Part 4 of Theorem 3. For ${T}_{y}>{T}_{I}$, by Proposition A3, we can conclude that, for $x\in (0,{x}_{L})$, we have $\frac{\partial {T}_{X}^{II}}{\partial x}>0$, for which the QREs are stable by Lemma 1. With similar arguments, we can conclude that the QREs on $x\in ({x}_{L},{x}_{1})$ are unstable. Additionally, given ${T}_{x}$, the stable QRE ${x}_{a}\in (0,{x}_{L})$ and the unstable ${x}_{b}\in ({x}_{L},{x}_{1})$ that satisfies ${T}_{X}^{II}({x}_{a},{T}_{y})={T}_{X}^{II}({x}_{b},{T}_{y})={T}_{x}$ appear in pairs. For ${T}_{y}<{T}_{I}$, with the same technique and by Proposition A4, we can claim that the QREs in $x\in ({x}_{2},{x}_{L})$ are unstable, while the QREs in $x\in ({x}_{L},1)$ are stable. This proves the first part of Theorem 2 and Part 4 of Theorem 3.
Parts 5 and 6 of Theorem 2 are corollaries of Part 5 of Proposition A3 and Part 4 of Proposition A4. ☐
Appendix D.1.2. Case 1b: b_{X} > 0, a_{Y} + b_{Y} < 0
In this case, both players have different preferences. For the game within this class, there is only one Nash equilibrium (either pure or mixed). We presented examples in
Figure A1 and
Figure A2. We can see that, in these figures, there is only one QRE given
${T}_{x}$ and
${T}_{y}$. We show in the following two propositions that this observation is true for all instances.
Proposition A5. Consider Case 1b. Let ${x}_{3}=max\left\{0,\frac{-{T}_{y}ln({a}_{X}/{b}_{X})+{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right\}$. If ${T}_{y}<{T}_{I}$, then the following statements are true:
- 1.
${T}_{X}^{II}(x,{T}_{y})<0$ for $x\in (1/2,1)$.
- 2.
${T}_{X}^{II}(x,{T}_{y})>0$ for $x\in \left({x}_{3},\frac{1}{2}\right)$.
- 3.
$\frac{\partial {T}_{X}^{II}(x,{T}_{y})}{\partial x}>0$ for $x\in \left({x}_{3},\frac{1}{2}\right)$.
- 4.
$\left({x}_{3},\frac{1}{2}\right)$ contains the principal branch.
Proof. Note that, if
${T}_{y}<{T}_{I}$,
${x}_{3}<1/2$. Additionally, according to Proposition A2,
${y}^{II}(1/2,{T}_{y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. Since
${y}^{II}$ is continuous and monotonically decreasing with
x,
${y}^{II}<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$ for
$x>1/2$. Therefore, the numerator of Equation (
6) is always positive for
$x\in (1/2,1)$, which makes
${T}_{X}^{II}$ negative. This proves the first part of the proposition.
For the second part, observe that, for $x\in (0,1/2)$, ${T}_{X}^{II}>0$ if and only if ${y}^{II}<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$. This is equivalent to $x>\frac{-{T}_{y}ln({a}_{X}/{b}_{X})+{b}_{Y}}{{a}_{Y}+{b}_{Y}}$.
For the third part, note that, for $x\in (0,1/2)$, $x(1-x)ln(1/x-1)\frac{\partial {y}^{II}}{\partial x}<0$. This implies $L(x,{T}_{y})<{y}^{II}(x,{T}_{y})<\frac{{b}_{X}}{{a}_{X}+{b}_{X}}$ for $x\in ({x}_{3},1/2)$, from which we can conclude that $\frac{\partial {T}_{X}^{II}(x,{T}_{y})}{\partial x}>0$.
Finally, we note that if ${x}_{3}>0$, then ${T}_{X}^{II}({x}_{3},{T}_{y})=0$. If ${x}_{3}=0$, we have ${lim}_{x\to {0}^{+}}{T}_{X}^{II}=0$. As a result, we can conclude that $({x}_{3},1/2)$ contains the principal branch. ☐
With the similar arguments, we are able to show the following proposition for ${T}_{y}>{T}_{I}$:
Proposition A6. Consider Case 1b. Let ${x}_{3}=min\left\{1,\frac{-{T}_{y}ln({a}_{X}/{b}_{X})+{b}_{Y}}{{a}_{Y}+{b}_{Y}}\right\}$. If ${T}_{y}>{T}_{I}$, then the following statements are true:
- 1.
${T}_{X}^{II}(x,{T}_{y})<0$ for $x\in (0,1/2)$.
- 2.
${T}_{X}^{II}(x,{T}_{y})>0$ for $x\in \left(\frac{1}{2},{x}_{3}\right)$.
- 3.
$\frac{\partial {T}_{X}^{II}(x,{T}_{y})}{\partial x}<0$ for $x\in \left(\frac{1}{2},{x}_{3}\right)$.
- 4.
$\left(\frac{1}{2},{x}_{3}\right)$ contains the principal branch.