have only one limit price. Therefore, the partitioning of E should work in exactly the same way. There is no outstanding order or limit price associated with the fifth Arrow-Debreu security. Thus, we can define only one subset of the set E : E 1 $_{5}$. We introduce new notation as (17). The idea is as follows. Having n$_{1}$ distinct limit prices associated with the first security yields 2 n$_{1}$ + 1 distinct subsets of E . Choose one subset out of these 2 n$_{1}$ + 1 subsets. Similarly, choose one of 2 n$_{2}$ + 1 subsets of E that we generate from the limit prices associated with the second security. Repeat this process for the remaining Arrow-Debreu securities. Once we have one subset for each type of Arrow-Debreu security, we can obtain the intersection of those K subsets, which is shown in (17). There are a total of Π K k $_{=1}$(2 n$_{k}$ + 1) ways to choose a combination of subsets. Now consider the optimization problem (9). Imagine replacing the constraint N ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 with a more restrictive one (17). The part that causes problem (9) to be non-convex is (18). However, once the feasible set of the state price vector ξ is restricted to a smaller set E ( ℓ$_{1}$, ℓ$_{2}$, ..., ℓ$_{K}$ ), (18) can be replaced with x$_{j}$ = 0 or x$_{j}$ ∈ [0 , Q$_{j}$ ] or x$_{j}$ = Q$_{j}$ for ∀ j . x$_{j}$ = 0 if ( A $^{T}$ξ ) j > B$_{j}$b$_{j}$ x$_{j}$ ∈ [0 , Q$_{j}$ ] if ( A $^{T}$ξ ) j = B$_{j}$b$_{j}$ x$_{j}$ = Q$_{j}$ if ( A $^{T}$ξ ) j < B$_{j}$b$_{j}$ for ∀ j ∈ { 1 , ..., J } (18) 18 E ( ℓ$_{1}$, ℓ$_{2}$, ..., ℓ$_{K}$ ) = E ℓ$_{1}$ 1 ∩ E ℓ$_{2}$ 2 ∩ ... ∩ E ℓ$_{K}$ K where 1 ≤ ℓ$_{1}$ ≤ 2 n$_{1}$ + 1 , ..., 1 ≤ ℓ$_{K}$ ≤ 2 n$_{K}$ + 1 (17) E 1 5 = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 } (16) E 1 2 = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 , 0 ≤ ξ$_{2}$ ≤ 0 . 18 } E 2 = ξ ∈ R $^{5}$| 5 i =1 ξ$_{i}$ = 1 , ξ ≥ 0 ,ξ$_{2}$ = 0 . 18 } E 3 2 = { ξ ∈ R $^{5}$| 5 ∑ i =1 ξ$_{i}$ = 1 , ξ ≥ 0 , 0 . 18 ≤ ξ$_{2}$ ≤ 1 } (15)