[Game Theory] [Chapter 4] Static Game with Incomplete Information (2)

(Book continued from above)

5. Bayesian game and mixed strategy equilibrium:

Purification theorem (Harsanyi, 1973): A mixed strategy game in a static game with complete information can almost always be interpreted as a pure strategy Bayesian Nash equilibrium of an approximate game with a small amount of incomplete information.
It can be further understood as “The fundamental characteristic of a mixed-strategy Nash equilibrium is not that participants select strategies (i.e., behaviors) in a random manner, but that each participant’s choices for other participants cannot be determined. This uncertainty can be random. It can also be caused by the incompleteness of a small amount of information."
(The vernacular translation is: a mixed strategy Nash equilibrium can be simply divided into several pure strategy Nash equilibrium problems)

[Example] The dispute between husband and wife

has two pure strategy Nash equilibria (fashion, fashion) and (football, football), and a mixed strategy Nash equilibrium, that is, the wife chooses fashion with a probability of 3/4, and chooses fashion with a probability of 1/4. When choosing football, the husband chooses fashion with a probability of 1/3 and football with a probability of 2/3.
The above is the solution to the problem in the first two chapters.
This problem can also be solved using the one-in-one probability method:

[Solution 1]
The husband’s benefits to his wife from watching the fashion show are not completely clear, and the wife’s benefits to her husband’s watching football are not completely clear either. At this point we transform a static game with complete information into a static game with incomplete information.
This incomplete information is reflected in the fact that the husband is not completely clear about his wife's benefits from watching the fashion show, and the wife is not completely clear about her husband's benefits from watching football. But we need to have an understanding of its probability distribution. Here we assume that this distribution is a $[0,x]$ uniform distribution, then the density function is$1/x$ . We use participant 1 to represent the wife and participant 2 to represent the husband.
Use the following formula to calculate how the wife can achieve the best benefit (that is, given the husband’s choice, how should the wife choose, so as to maximize the wife’s own expected utility)
$$\begin{array}{rl}& \underset{t_1\in T_1}{\max }{\int }_0^x\left(s_1,s_2^{\ast }\right)\frac{1}{x}d{t}_2\\ & =\underset{t_1\in T_1}{\max }\{\begin{array}{l}\left[{\int }_0^{t_2^{\ast }}\left(s_1=a_1,s_2^{\ast }=a_1\right)\frac{1}{x}d{t}_2+{\int }_{t_2^{\ast }}^x\left(s_1=a_1,s_2^{\ast }=a_2\right)\frac{1}{x}d{t}_2\right]if{t}_1\ge t_1^{\ast }\\ (When\; the\; wife’s\; desire\; to\; see\; the\; fashion\; show\; is\; very\; high,\; the\; wife\; will\; choose\; the\; fashion\; show\; no\; matter\; what\; the\; husband\; chooses)\\ \left[{\int }_0^{t_2^{\ast }}\left(s_1=a_2,s_2^{\ast }=a_1\right)\frac{1}{x}d{t}_2+{\int }_{t_2^{\ast }}^x\left(s_2=a_2,s_2^{\ast }=a_2\right)\frac{1}{x}d{t}_2\right]if{t}_1<t_1^{\ast }\\ (When\; the\; wife\; doesn’t\; want\; to\; watch\; the\; fashion\; show\; that\; much,\; no\; matter\; what\; the\; husband\; chooses,\; the\; wife\; will\; choose\; football\end{array}\\ & =\underset{t_1\in T_1}{\max }\{\begin{array}{l}\left[{\int }_0^{t_2^{\ast }}\left(2+t_1\right)\frac{1}{x}d{t}_2+{\int }_{t_2^{\ast }}^{x}0\times \frac{1}{x}d{t}_2\right]=\frac{\left(2+t_1\right)t_2^{\ast }}{x}\\ \left[{\int }_0^{t_2^{\ast }}0\times \frac{1}{x}d{t}_2+{\int }_{t_2^{\ast }}^{x}1\times \frac{1}{x}d{t}_2\right]=\frac{x-t_2^{\ast }}{x}\\ \Rightarrow \frac{\left(2+t_1^{\ast }\right)t_2^{\ast }}{x}=\frac{x-t_2^{\ast }}{x}if{t}_1=t_1^{\ast }\\ (Let\; the\; wife’s\; expected\; return\; of\; choosing\; a\; fashion\; show\; or\; football\; be\; equal.\; This\; is\; the\; parameter\; distribution\; corresponding\; to\; the\; wife’s\; best\; return\end{array}\end{array}$$
$$\begin{array}{rl}& \underset{t_2\in T_2}{\max }{\int }_0^x\left(s_1^{\ast },s_2\right)\frac{1}{x}d{t}_1\\ & =\underset{t_2\in T_2}{\max }\{\begin{array}{l}\left[{\int }_0^{t_1^{\ast }}\left(s_1^{\ast }=a_2,s_2=a_1\right)\frac{1}{x}d{t}_1+{\int }_{t_1^{\ast }}^x\left(s_1^{\ast }=a_1,s_2=a_1\right)\frac{1}{x}d{t}_1\right]if{t}_2<t_2^{\ast }\\ \left[{\int }_0^{t_1^{\ast }}\left(s_1^{\ast }=a_2,s_2=a_2\right)\frac{1}{x}d{t}_1+{\int }_{t_1^{\ast }}^x\left(s_1^{\ast }=a_1,s_2=a_2\right)\frac{1}{x}d{t}_1\right]if{t}_2\ge t_2^{\ast }\end{array}\\ & =\underset{t_2\in T_2}{\max }\{\begin{array}{l}\left[{\int }_0^{t_1^{\ast }}0\times \frac{1}{x}d{t}_1+{\int }_{t_1^{\ast }}^{x}1\times \frac{1}{x}d{t}_1\right]=\frac{x-t_1^{\ast }}{x}\\ \left[{\int }_0^{t_1^{\ast }}\left(3+t_2\right)\times \frac{1}{x}d{t}_1+{\int }_{t_2^{\ast }}^{x}0\times \frac{1}{x}d{t}_1\right]=\frac{\left(3+t_2\right)t_1^{\ast }}{x}\\ \Rightarrow \frac{x-t_1^{\ast }}{x}=\frac{\left(3+t_2^{\ast }\right)t_1^{\ast }}{x}if{t}_2=t_2^{\ast }\end{array}\end{array}$$

( 2 + t 1 ∗ ) t 2 ∗ x = x − t 2 ∗ x x − t 1 ∗ x = ( 3 + t 2 ∗ ) t 1 ∗ x } ⇒ { t 1 ∗ = − 3 + 9 + 3 x 2 t 2 ∗ = − 6 + 2 9 + 3 x 3 a ∗ = { s 1 ∗ ( T 11 ) = a 1 , s 1 ∗ ( T 12 ) = a 2 ; s 2 ∗ ( T 21 ) = a 1 , s 2 ∗ ( T 22 ) = a 2 } T 11 = [ t 1 ∗ , x ] , T 12 = [ 0 , t 1 ∗ ] T 21 = [ 0 , t 2 ∗ ] , T 22 = [ t 2 ∗ , x ] \begin{aligned} &\left.\begin{array}{l} \frac{\left(2+t_1^*\right) t_2^*}{x}=\frac{x-t_2^*}{x} \\ \frac{x-t_1^*}{x}=\frac{\left(3+t_2^*\right) t_1^*}{x} \end{array}\right\} \Rightarrow\left\{\begin{array}{l} t_1^*=\frac{-3+\sqrt{9+3 x}}{2} \\ t_2^*=\frac{-6+2 \sqrt{9+3 x}}{3} \end{array}\right.\\ &\begin{aligned} & a^*=\left\{s_1^*\left(T_{11}\right)=a_1, s_1^*\left(T_{12}\right)=a_2 ; s_2^*\left(T_{21}\right)=a_1, s_2^*\left(T_{22}\right)=a_2\right\} \\ & T_{11}=\left[t_1^*, x\right], T_{12}=\left[0, t_1^*\right] T_{21}=\left[0, t_2^*\right], T_{22}=\left[t_2^*, x\right] \end{aligned} \end{aligned} ​x(2+t1∗​)t2∗​​=xx−t2∗​​xx−t1∗​​=x(3+t2∗​)t1∗​​​}⇒{ t1∗​=2−3+9 + 3x _ ​​t2∗​=3−6+29 + 3x _ ​​​​a∗={ s1∗​(T11​)=a1​,s1∗​(T12​)=a2​;s2∗​(T21​)=a1​,s2∗​(T22​)=a2​}T11​=[t1∗​,x],T12​=[0,t1∗​]T21​=[0,t2∗​],T22​=[t2∗​,x]​​
We can also prove the solution results of the mixed probability model from the above conclusion:
$$\begin{array}{rl}& P\left(0\le t_1^{\ast }<\frac{-3+\sqrt{9+3x\_}}{2}\right)=\frac{-3+\sqrt{9+3x\_}}{2x\_}\\ & =\frac{3}{2(3+\sqrt{9+3x\_})}{|}_{x\to 0}=\frac{1}{4}\\ & P\left(0\le t_2^{\ast }<\frac{-6+2\sqrt{9+3x\_}}{3}\right)=\frac{-6+2\sqrt{9+3x\_}}{3x\_}\\ & =\frac{4}{6+2\sqrt{9+3x\_}}{|}_{x\to 0}=\frac{1}{3}\end{array}$$
[Solution 2]
Let’s assume the wife’s strategy: when $t_w>When\; W$ , choose the fashion show, otherwise choose the football game; the husband's strategy is similar to that of the wife: when$t_h>h$ , choose the football game, otherwise choose the fashion show. Under the strategies of both parties above, because they are both$[0,$standard distribution on$x$$]$$(x-w)/x$ , the probability of choosing the football game is$w/x$ ; The probability that the husband chooses the fashion show is$h/x$ , the probability of choosing a football match is$(x—h)/x$ . In order to make the above strategic combination of both parties a Bayesian Nash equilibrium, the parameter$W$ and$h$ must take an appropriate value.
From the wife's perspective, assuming that her husband has adopted the above strategy, her expected benefits from choosing the fashion show and the football game are respectively:
$$\begin{array}{c}\frac{h}{x}\left(2+t_w\right)+\frac{x-h}{x}\times 0=\frac{h}{x}\left(2+t_w\right)\frac{h}{x}\times 0+\frac{x-h}{x}\times 1=\frac{x-h}{x}\\ w=\frac{x}{h}-3\end{array}$$
Similarly, assuming that the wife has adopted the aforementioned critical value strategy, the husband’s expected benefits from choosing the football game and the fashion show are respectively:
$$\begin{array}{rl}& \frac{w}{x}\left(3+t_h\right)+\frac{x-w}{x}\times 0=\frac{w}{x}\left(3+t_h\right)\\ & \frac{w}{x}\times 0+\frac{x-w}{x}\times 1=\frac{x-w}{x}\end{array}$$
When the expected benefit from choosing football is not less than the expected benefit from choosing fashion show, the husband chooses football game, that is:
$t_h>=\frac{x}{w}-4$ , from which we can get:$h=\frac{x}{w}-4$.
$$\{\begin{array}{l}w=\frac{x}{h}-3\\ h=\frac{x}{w}-4\end{array}$$
$$\{\begin{array}{l}w=\frac{-3\pm \sqrt{9+3x\_}}{2}=\frac{-3+\sqrt{9+3x\_}}{2}\\ h=\frac{-6\pm 2\sqrt{9+3x\_}}{3}=\frac{-6+2\sqrt{9+3x\_}}{3}\end{array}$$
When the parameters W and h are determined by the above two equations, it is easy to prove that the aforementioned Lin Jue strategies of both parties constitute a Bayesian Nash equilibrium. According to the above values ​​of W and h, the probability of the wife choosing the fashion show is:
$$\frac{x-w}{x}=1-\frac{w}{x}=1-\frac{-3+\sqrt{9+3x\_}}{2x\_}$$
The probability that the husband chooses the football game is:
$$\frac{x-h}{x}=1-\frac{h}{x}=1-\frac{-6+2\sqrt{9+3x\_}}{3x\_}$$
When x approaches 0, that is, when incomplete information is close to disappearing or insignificant, the above two probabilities tend to 3/4 and 2/3 respectively, and this is also the random choice of the mixed strategy equilibrium of the complete information husband and wife game. The probability. Therefore, the mixed strategy equilibrium can be regarded as a Bayesian Nash equilibrium of a static Bayesian game with a very small amount of incomplete information, which proves that Harsanyi's conclusion is valid.

[Example] Nonlinear Pricing
Suppose there is a monopoly manufacturer with constant marginal cost $c$ produces a certain product. When the manufacturer sells$q\ge When\; 0$ quantity of goods are given to the consumer, the quantity TTis obtained from the consumer.$T$ ’s currency; while consumers consume$q$ quantity of goods and payTTThe utility obtained when the cost of$T$$u(q,T,i)=\theta v(q)-T.$ _ Here$\theta V\left(q\right)$ is the total consumer surplus,$V(.)$ is a definite value aboutqqFunction of$q$$V(0)=0,V^{\prime }>0，V^{\prime \prime }<0$;$\theta$ is the type information of the consumer, and its value may be$i_1$orθ $i_2$,$i_2>i_1>0$ . Assume that θ 1 \theta_1in the total population$i_1$The proportion of type consumers is $p_1$，$i_2$The proportion of type consumers is $p_2$, and $p_1+p_2=1$ . Proof: The profit maximizing behavior of the enterprise (strategic optimality) means that the following conditions are true:
(1)$i_2{V}^{\prime }\left(q_2\right)=c$ ;
(2)$i_1{V}^{\prime }\left(q_1\right)=\frac{c}{1-\frac{p_2\left(i_2-i_1\right)}{p_1{i}_1}}$

[Solution]
First understand the meaning of the question: $V\left(q\right)$ is a parameter about the quantity of goods purchasedqqThe first-order differential of the function q is greater than zero and the second-order differential is less than zero. This means that as the number of goods purchased by a consumer gradually increases, the marginal utility that the consumer thinks the product can bring to him gradually decreases. This is a rate of change$.$ A function similar to the logarithmic function.
Secondly,$V\left(q\right)$ is the utility attribute of a commodity itself, but because the types of consumers are different, the utility it brings to consumers is also different. For example, pepper itself has its inherent value, but for people who like spicy food The utility it brings will be higher than the utility it brings to people who don’t like spicy food. This is affected by the type of consumer.
Let us prove below: the commodity utility corresponding to the two participant types: for$i_1$, when its consumption quantity is $q_2$When, its marginal utility equals marginal cost. For $i_2$, when its consumption quantity is $q_1$When , the marginal utility is greater than the marginal cost. For those who particularly want the product, the utility is equal to the marginal cost. For those who do not want the product very much, the marginal utility is greater than the marginal cost.
Proof:
This game process offers a price $T\left(q\right)$ , the consumer either accepts or rejects. Therefore, the company will have two sets of plans:
$(q_1,T_1)$ sold to$i_1$Consumer of type, $(q_2,T_2)$ sold to$i_2$type of consumer.
The company's expected return is: $E{u}_0=p_1(T_1-c{q}_1)+p_2(T_2-c{q}_2)$
The above are all conditions offered by the seller. As for whether the buyer will accept such conditions, there is also a participation constraint, which is that the buyer evaluates whether this condition is in line with its own interests: participation constraint (IR, individual - rationality): (
IR
$$\begin{gathered} \left(I{R}_1\right)i_{1}V(q_1)-T_1\ge 0 \\ \left(I{R}_2\right)i_{2}V\left(q_2\right)-T_2\ge 0 \end{gathered}$$
Another problem is that the company develops different plans for different groups of people. For$i_1$For this type of people, the formula is $(q_1,T_1)$ , for$i_2$For this type of people, the formula is
$(q_2,T_2)$ . Therefore, it is necessary to ensure$i_1$at $(q_1,T_1)$ criteria, consumption will get the maximum benefit, rather than what they finally found in$(q_2,T_2)$ criteria, consumption will get better benefits. If this is the case, it means that the company's consumption plan is wrongly formulated, so there is the following incentive compatibility:
incentive compatibility (C, incentive - compatibility) constraint:
$$\begin{gathered} (I{C}_1)i_{1}V(q_1)-T_1\ge i_{1}V\left(q_2\right)-T_2 \\ \left(I{C}_2\right)i_{2}V\left(q_2\right)-T_2\ge i_{2}V\left(q_1\right)-T_1 \end{gathered}$$
Therefore, when solving game theory, it is actually an optimization problem under multiple constraints, so it also has corresponding optimization algorithms to solve this kind of problem.
If $I{R}_1$和$I{C}_2$For example, $\left(I{R}_1\right)i_{1}V\left(q_1\right)-T_1\ge 0\to i_{1}V\left(q_1\right)\ge T_1$
$$\begin{array}{rl}{i}_{2}V\left(q_2\right)-T_2\ge i_{2}V\left(q_1\right)-T_1& \Rightarrow i_{2}V\left(q_2\right)-T_2\ge i_{2}V\left(q_1\right)-i_{1}V\left(q_1\right)\\ & =\left(i_2-i_1\right)V\left(q_1\right)\ge 0\\ V\left(q_1\ne 0\right)>0,i_2-i_1>0\Rightarrow & i_{2}V\left(q_2\right)-T_2>0\end{array}$$
Therefore, when the manufacturer's strategy is optimal, $I{R}_1$must be established under the equal sign, otherwise for any small $\epsilon$有：
$\left(I{C}_1\right)i_{1}V\left(q_1\right)-\left(T_1+e\right)\ge i_{1}V\left(q_2\right)-\left(T_2+\epsilon \right)$
$$\left(I_2\right)i_{2}V\left(q_2\right)-\left(T_2+e\right)\ge i_{2}V\left(q_1\right)-\left(T_1+\epsilon \right)$$
This does not comply with the manufacturer's optimization strategy, so$i_{1}V(q_1)=T_1$
So there is:
$$\begin{array}{rl}& i_{2}V\left(q_2\right)-T_2=i_{2}V\left(q_1\right)-T_1=i_{2}V\left(q_1\right)-i_{1}V\left(q_1\right)\\ & \Rightarrow T_2=i_{2}V\left(q_2\right)-i_{2}V\left(q_1\right)+i_{1}V\left(q_1\right)\\ & E{u}_0=p_1\left(T_1-c{q}_1\right)+p_2\left(T_2-c{q}_2\right)\\ & E{u}_0=p_1\left(i_{1}V\left(q_1\right)-c{q}_1\right)+p_2\left[i_{2}V\left(q_2\right)-i_{2}V\left(q_1\right)+i_{1}V\left(q_1\right)-c{q}_2\right)]\\ & =\left[p_1{i}_1-p_2\left(i_2-i_1\right)\right]V\left(q_1\right)-p_{1}c{q}_1+p_2\left[i_{2}V\left(q_2\right)-c{q}_2\right]\end{array}$$
Find the conditions for each section separately:
$i_2{V}^{\prime }(q_2)=c$
$i_1{V}^{\prime }\left(q_1\right)=\frac{c}{1-\frac{p_2(i_2-i_1)}{p_1{i}_1}}$