Systemic risk models for disjoint and overlapping groups with equilibrium strategies

A.1 Comparison between trivial grouping and multi-groups

We first look at the trivial grouping, i.e., m = h = 1 and all k ∈ I 1 = { 1 , 2 , … , N } . The group parameter and group vectors are

Then, following (2.17), the systemic risk allocation of individual 𝑖 is

The total systemic risk allocation for the system is

Then, for the multi-group case, assuming m = 1 , … , h ( h ≥ 2 ) and for k m ∈ I m ≠ ∅ , the systemic risk allocation of individual k m is

The total risk allocation is

We need to compare the total risk of trivial grouping (A.1) and the total risk of nontrivial grouping (A.2). For simplicity, we take h = 2 in the nontrivial grouping case and compare.

Assume an 𝑁-individual system is divided into two subgroups with sizes N 1 = | I 1 | , N 2 = | I 2 | , respectively. Given all risk factors, define

Note that S = S 1 + S 2 . Therefore,

Then we compare the last term in (A.1), (A.2),

We can conclude the trivial grouping has a smaller total systemic risk allocation for the 𝑁-player system compared with two-group case. It can be extended to the general grouping case and shows the advantage of trivial grouping or fewer groups in terms of the total risk. And it is consistent with the monotonicity property proved in [9].

A.2 Proof of Claim 2.1

Under the assumption β m = | I m | ⁢ 1 α , equation (2.17) becomes

First, when ρ ≠ 1 , the function is monotonically decreasing in | I m | . So the risk allocation for every individual/individual is maximized when | I m | = n , i.e., all individuals/individuals are in the same group.

If we assume individuals are separated in several groups which makes it a Nash equilibrium for all. For some individual 𝑖 in the second largest group, moving to the largest group makes it achieve a smaller risk allocation, which is better. If there are only two equal size groups, i.e., “ n 2 - n 2 ”, one individual 𝑖 in group 1 joining the other one gives “ ( n 2 - 1 ) - ( n 2 + 1 ) ” and E Q X 1 ⁢ [ Y X i ] > E Q X 2 ⁢ [ Y X i ] . So “ ( n 2 - 1 ) - ( n 2 + 1 ) ” cannot be a Nash equilibrium neither.

In conclusion, nontrivial grouping strategy cannot be a Nash equilibrium and only | I m | = n is a Nash equilibrium under the case that all standard deviation and utility parameters are the same, and the correlation coefficient is ρ ∈ [ - 1 , 1 ) . When ρ = 1 , the risk for every individual is constant and grouping has no effect.

A.3 Proof of Claim 2.2

First we look at the grouping “ { 1 , 2 } - { 3 , 4 } ” and find the equivalent condition to make it a Nash equilibrium. We compare risks for individual 1 under two cases, according to (2.17),

For individual 1, to make the risk allocation under “ { 1 , 2 } - { 3 , 4 } ” at most that under “ { 2 } - { 1 , 3 , 4 } ”,

For individuals 2, 3, 4, we repeat a similar discussion, and the condition is the same, ρ ≤ - 3 13 . So, in conclusion, if ρ ≤ - 3 13 , grouping “ { 1 , 2 } - { 3 , 4 } ” is a Nash equilibrium.

In the grouping “ { 1 , 3 } - { 2 , 4 } ”, we follow the same discussion about comparing risk allocations for all individuals. For example, for individual 1,

For individual 1, to make the risk allocation under “ { 1 , 3 } - { 2 , 4 } ” at most that under “ { 3 } - { 1 , 2 , 4 } ”,

For individuals 2, 3, 4, we repeat a similar discussion, and the condition is the same, ρ ≥ 3 8 . As a result, grouping “ { 1 , 3 } - { 2 , 4 } ” is a Nash equilibrium if ρ ≥ 3 8 . It is also the equivalent condition for grouping “ { 1 , 4 } - { 2 , 3 } ” to be a Nash equilibrium

A.4 Proof of Claim 2.3

Similar to the proof of Claim 2.2, we look at the grouping “ { 1 , 2 } - { 3 , 4 , 5 } ” and find the equivalent condition to make it a Nash equilibrium. Under the assumptions, we first compare risks for individual 1 under two cases, according to (2.17),

For individual 1, to make the risk allocation under “ { 1 , 2 } - { 3 , 4 , 5 } ” at most that under “ { 2 } - { 1 , 3 , 4 , 5 } ”,

For individual 2, this is the same condition to have a smaller risk allocation. We then compare risks for individual 3 under two cases,

To make the risk allocation for individual 3 under “ { 1 , 2 } - { 3 , 4 , 5 } ” at most that under “ { 1 , 2 , 3 } - { 4 , 5 } ”, we get the equivalent condition ρ ≤ 0 . This is true for individuals 4, 5 as well. In conclusion, when ρ ≤ - 2 7 , “ { 1 , 2 } - { 3 , 4 , 5 } ” is a Nash equilibrium for all individuals.

In the grouping “ { 1 , 3 } - { 2 , 4 , 5 } ”, we follow the same discussion about comparing risk allocations for all individuals. Individual 1 will stay with individual 3 instead of joining the other group if ρ = 1 . Individual 2 will stay if ρ ≥ 0 . Individual 3 will stay with individual 1 if ρ ≥ 2 5 , and individuals 4, 5 will not move for any 𝜌. As a result, grouping “ { 1 , 3 } - { 2 , 4 , 5 } ” is not a Nash equilibrium for any ρ ∈ ( - 1 , 1 ) .

For the grouping “ { 1 , 2 , 3 } - { 4 , 5 } ”, after the discussion about the condition for every individual not moving, we find a contradiction which proves, for any value of 𝜌, this grouping cannot be a Nash equilibrium.

A.5 Proof of Theorem 3.1

We can rewrite the systemic risk measure 𝜌 as

where ( I 1 , … , I h ) are the group index sets given. For any group 𝑚, we define the smallest element as m 0 ∈ I m and fix another element m * ∈ I m and m * ≠ m 0 . (When there is only one element in the group, m * = m 0 , and the discussion will be similar.) In the following proof, we assume | I m | ≥ 2 for all m = 1 , … , h . We also assume 𝑌 is defined on a finite space, Y k , m ∈ { y 1 k , m , … , y M k , m } for all k , m and y j k , m ∈ R . Then we have

We show the results with the Lagrange method with the function defined by

We compute partial derivatives of ℒ with respect to all variables and get all equivalent conditions in the following.

(1) Given 𝑚, for j = 1 , … , M , k ∈ I m and k ≠ m * , we have ∂ ⁡ L ∂ ⁡ y j k , m = 0 if and only if, for every fixed 𝑗,

or, equivalently, we have

This implies

Then, for k ≠ m * ,

and by (A.5) and (A.6), we obtain (details are shown below)

where β m = ∑ k ∈ I m 1 α k , S m = ∑ k ∈ I m w k , m ⁢ X k .

(2) For m = 1 , … , h , the derivative with respect to d m , ∂ ⁡ L ∂ ⁡ d m = 0 if and only if

By (A.5),

i.e.,

(3) The derivative with respect to 𝜆, ∂ ⁡ L ∂ ⁡ λ = 0 if and only if

We then compute d m by inserting (A.10) and (A.7) in (A.9) for k = m 0 ∈ I m , and m 0 ≠ m * ,

So

Then, back to (A.6) and (A.7), for k , m 0 , m * ∈ I m and k ≠ m 0 ≠ m * ,

In addition, the systemic risk measure is given by

A.6 Proof of Proposition 3.1

First we prove the marginal risk allocation for individual 𝑖 in group 𝑗. By Theorem 3.1, for i ∈ I j ,