Subgame Perfect Nash Equilibrium

In a sequential game, we will study the subgame perfect Nash equilibrium. Subgame perfect Nash equilibrium is a popular refinement to the Nash equilibrium under the se-quential game. It guarantees that all players choose strategies rationally in every possible subgame. A subgame is a part of the original game. In Chinese restaurant game, any game process begins from player i, given all possible actions before player i, could be a subgame.

Definition 15. A subgame in Chinese restaurant game is consisted of two elements: 1) It begins from customer i; 2) The current grouping before customer i is n_i = (n_i,1, ..., n_i,J) with^∑J_j=1ni,j = i− 1.

Definition 16. A Nash equilibrium is a subgame perfect Nash equilibrium if and only if it is a Nash equilibrium for any subgame.

We would like to show the existence of subgame perfect Nash equilibrium in Chinese restaurant game by constructing one. Basically, as a rational customer, customer i should predict the final equilibrium grouping according to his current observation on the choices of previous customers n_i and the system state θ. Then, he may choose the table with highest expected utility according to the prediction. Following from this idea, we derive the best response of customers in a subgame.

We first implement the prediction part through two functions as follows. First, let EG(X^s, N^s) be the function that generates the equilibrium grouping for a table setX^sand number of customers N^s. The equilibrium grouping is generated by the greedy algorithm

shown in previous section with X being replaced by X^s and N being replaced by N^s. Notice thatX^scould be any subset of the total table setX = {1, ..., J}, and N^sis less or equal to N .

Then, let P C(X^s, n^s, N^s), where n^sdenotes the current grouping observed by the cus-tomer, be the algorithm that generates the set of available tables given n^sin the subgame.

The algorithm removes the tables that already occupied by more than the expected number of customers in the equilibrium grouping. This helps the customer remove those unrea-sonable choices and correctly predict the final equilibrium grouping in every subgame.

The basic flow of this algorithm is shown as follows 1) calculate the equilibrium grouping n^e given the table set X^s and number of customers N^s, 2) check if there is any overly occupied table by comparing n^s with n^e. If so, 3) remove these tables fromX^s and the customers occupying these tables from N^s, and go back to 1). Otherwise, the algorithm terminates. The procedures of implementing P C(X^s, n^s, N^s) are described as follows:

1. Initialize: X^o =X^s, N^t= N^s

2. X^t =X^o, n^e = EG(X^t, N^t),X^o ={x|x ∈ X^t, n^e_j ≥ n^sj}, N^t= N^s−^∑x∈X^s\X^on^s_x. 3. IfX^o ̸= X^t, go back to step 2.

4. OutputX^o.

Now, we propose a method to construct a subgame perfect Nash equilibrium. This equilibrium also satisfies (6.5). For each customer i, his strategy in a subgame is

BE_i^se(n_i, θ) = arg max

x∈X^i,cand,ni,x<n^i,candx

U (R_x(θ), n^i,cand_x ), (6.13)

whereX^i,cand = P C(X, ni, N ), N^i,cand = N−^∑x∈X\X^i,candn_i,x, and n^i,cand = EG(X^i,cand, N^i,cand).

The proposed best response BE_i^se∗(n_i, θ) chooses the table with the highest utility ac-cording to the predicted equilibrium grouping n^i,candand candidate table setX^i,cand. The equilibrium grouping n^i,candis obtained by EG(X^i,cand, N^i,cand), where the candidate ta-ble setX^i,cand is derived by P C(X, ni, N ). In Lemma 6, we show that the above strategy results in the equilibrium grouping in any subgame.

Lemma 6. Given the available table setX^s = P C(X, n^s, N ), N^s= N−^∑x∈X\X^sn^s_x, the proposed strategy shown in (6.13) leads to an equilibrium grouping n^∗ = EG(X^s, N^s) overX^s.

Proof. We prove this by contradiction. Let n = (n_j|j ∈ X^s) be the final grouping after all customers choose their tables according to (6.13). Suppose that n̸= n^∗ = EG(X^s, N^s), then there exists some tables j that n_j > n^∗_j. Let table j be the first table that ex-ceeds n^s_j in this sequential subgame. Since n_j > n^∗_j, there are at least n^∗_j + 1 cus-tomers choosing table j. Suppose the n^∗_j + 1-th customer choosing table j is customer i. Let ni = (ni,1, ni,2, ..., ni,J) be the current grouping observed by customer i before he chooses the table. Since customer i is the n^∗_j + 1-th customer choosing table j, we have n_i,j = n^∗_j. Since table j is the first table exceeding n^∗ after customer i's choice, we have n_i,x ≤ n^∗x ∀x ∈ X^s.

According to the definition of P C(·), none of the tables will be removed from candi-dates. Thus,X^i,cand =X^sand N^i,cand = N^s. We have

n^i,cand = EG(X^i,cand, N^i,cand) = EG(X^s, N^s) = n^∗. (6.14)

However, according to (6.13), the customer i should not choose table j since n_i,j = n^∗_j = n^i,cand_j . This contradicts with our assumption that customer i is the n^∗_j+1-th customer choosing table j. Thus, the strategy (6.13) should lead to the equilibrium grouping n^∗ = EG(X^s, N^s).

Note that Lemma 6 also shows that the final grouping of the sequential game should be n^∗ = EG(X, N) if all customers follow the proposed strategy in (6.13). In the following Lemma, we show that P C(X^s, n^s, N^s) removes the tables that are dominated by other tables if all customers follow (6.13).

Lemma 7. Given a subgame with current grouping n^s, if table j ̸∈ X^s = P C(X, n^s, N ), then table j is never the best response of the customer if all other customers follow (6.13).

Proof. Let n^′ = EG(X, N), and n^∗ be the final grouping. We first show that for every table under the final grouping n^∗, there always exists a table providing a less or equal utility

under the grouping n^′. According to Lemma 6, the final grouping n^∗ is an equilibrium grouping overX^sif all customers follow (6.13). Additionally, n^∗_j = n^s_j since no customers will choose table j. Assuming that there exists a table k ∈ X^s with n^′_k < n^∗_k. Since n^∗_j = n^s_j > n^′_j, we have^∑_x∈X\{j}n^∗_x <^∑_x∈X\{j}n^′_x. Therefore,∃k^′ ∈ X^s that n^′_k′ > n^∗_k′. Since n^′ and n^∗ are equilibrium groupings overX^s, similar to (6.10), we have

U (R_k(θ), n^′_k+ 1)≥ U(Rk(θ), n^∗_k)≥ U(Rk^′(θ), n^∗_k′ + 1)≥ U(Rk^′(θ), n^′_k′)≥ U(Rk(θ), n^′_k+ 1)(6.15)

The first and third inequalities are due to n^′_k < n^∗_kand n^′_k′ > n^∗_k′, and the second and fourth ones come from the equilibrium grouping condition in (6.5). The equation is valid only when all equalities hold. Thus, if n^′_k < n^∗_k,∃k^′ ∈ X^sthat U (R_k(θ), n^∗_k) = U (R_k′(θ), n^′_k′), which means that we can always find a table k^′providing the same utility as U (R_k(θ), n^∗_k) under grouping n^′. When n^′_k ≥ n^∗_k, we have U (R_k(θ), n^∗_k) ≥ U(Rk(θ), n^′_k). Therefore,

∀k ∈ X^s,∃k^′ ∈ X^sthat U (R_k(θ), n^∗_k)≥ U(Rk^′(θ), n^′_k′).

Then, we show that table j is dominated by all other tables under n^∗. Since table j is removed by P C(X, n^s, N ), we have n^s_j > n^′_j. Therefore, according the above discussion and the fact that n^′ is an equilibrium grouping, we have∀k ∈ X^s,

U (R_k(θ), n^∗_k)≥ min

k^′∈X^sU (R_k′(θ), n^′_k′)≥ U(Rj(θ), n^′_j+ 1) > U (R_j(θ), n^s_j + 1). (6.16) Since U (Rj(θ), n^s_j+ 1) is the highest utility that can be offered by table j, it is dominated by all other tables inX^sunder the final grouping n^∗. So, table j is never the best response of the customer.

Theorem 16. There always exists a subgame perfect Nash equilibrium with the cor-responding equilibrium grouping n^∗ satisfying (6.5) in a sequential Chinese restaurant game.

Proof. We would like to show that the proposed strategy in (6.13) forms a Nash equilib-rium. Suppose customer i chooses table j in his round according to (6.13). Then, customer i's utility is u_i = U (R_j(θ), n^∗_j) since based on Lemma 6, the equilibrium grouping n^∗will be reached at the end.

Now we show that table j is indeed customer i's best response. Let's assume that customer i is the last customer, i.e, i = N , and chooses another table j^′ ̸= j in his round, then his utility becomes U (R_j′(θ), n^∗_j′ + 1). However, according to (6.5), we have

u^∗_j = U (R_j(θ), n^∗_j)≥ U(Rj^′(θ), n^∗_j′ + 1). (6.17)

Thus, choosing table j is never worse than choosing table j^′ for customer N .

For the case that customer i is not the last customer, we assume that he chooses table j^′instead of table j in his round. Since all customers before customer i follows (6.13), we have n_i,j ≤ n^∗j ∀j ∈ X. Otherwise, n^∗cannot be reached, which contradicts with Lemma 6.

If n_i,j′ < n^∗_j′, we have n_i+1,j′ ≤ n^∗j^′. In addition, we have n_i+1,j = n_i,j ≤ n^∗j ∀j ∈ X \ {j^′}, since other tables are not chosen by customer i. Thus, X^i+1,cand = P C(X, ni+1, N ) and N^i,cand = N . According to Lemma 6, the final grouping should be n^∗ = EG(X, N).

Thus, the new utility of customer i becomes u^′_i = U (R_j′(θ), n^∗_j′). However, according to (6.13), we have

u_i = U (R_j(θ), n^∗_j) = arg max

x∈X,ni,x<n^∗_xU (R_x(θ), n^∗_x)≥ U(Rj^′(θ), n^∗_j′) = u^′_i.(6.18)

Thus, choosing table j^′never gives customer i a higher utility.

If n_i,j′ = n^∗_j′, and the final grouping is n^′ = (n^′₁, n^′₂, ..., n^′_J). Since customer i chooses table j^′ when n_i,j′ = n^∗_j′, we have n^′_j′ ≥ ni+1,j^′ = n_i,j′ + 1 = n^∗_j′+ 1. Thus, we have

u_i = U (R_j(θ), n^∗_j)≥ U(Rj^′(θ), n^∗_j′ + 1)≥ U(Rj^′(θ), n^′_j′) = u^′_i, ∀j^′ ∈ X,(6.19)

where the first inequality comes from the equilibrium grouping condition in (6.5), and the second inequality comes from the fact that U (R, n) is decreasing over n and n^′_j′ ≥ n^∗_j′ + 1. Thus, under both cases, choosing table j^′ is never better than choosing table j.

We conclude that{BEi^se(·)} in (6.13) forms a Nash equilibrium, where the grouping being the equilibrium grouping n^∗.

Finally, we show that the proposed strategy forms a Nash equilibrium in every sub-game. In Lemma 7, we show that if the table j is removed by P C(X, n^s, N ), it is never the best response of all remaining customers. Thus, we only need to consider the re-maining table candidatesX^s = P C(X, n^s, N ) in the subgame. Then, with Lemma 6, we show that for every possible subgame with correspondingX^s, the equilibrium grouping n^∗ = EG(X^s, N^s) will be achieved at the end of the subgame. Moreover, the above proof shows that if the equilibrium grouping n^s will be achieved at the end of the subgame, BE_i^se(·) is the best response function. Therefore, the proposed strategy forms a Nash equilibrium in every subgame, i.e., we have a subgame perfect Nash equilibrium.

In the proof of Theorem 16, we observe that the sequential game structure brings ad-vantages for those customers making decisions early. According to (6.13), customers who make decisions early can choose the table providing the largest utility in the equilibrium.

When the number of customers choosing that table reaches equilibrium number, the sec-ond best table will be chosen until it is full again. For the last customer, he has no choice but to choose the worst one.