H , we have E [ ∫ 1 0 V$_{z}$ ( Y$_{s}$, Z$_{s}$, s ) dB 2 s ] = 0. Therefore Moreover, due to the properties of V we have The above inequalities become equalities if and only if the following conditions hold: Δ θ = 0 for equation (3.15); [ θ $^{c}$]$_{1}$ = 0 for equation (3.16); H ( Y ∗ 1 , 1) = f ( Z$_{1}$ ) for equation (3.17). Therefore, for any function V satisfying equations (3.12), (3.14) and the final condition given by V ( y, z, 1) ≥ V ( y $^{∗}$( z ) , z, 1) = 0 for every z and any y = y $^{∗}$( z ) (where y $^{∗}$( z ) is the solution of H ( y $^{∗}$( z ) , 1) = f ( z )), we have that This expression holds with equality if and only if θ is continuous and condition (3.10) is satisfied. Hence, if θ is such that these conditions are satisfied, then it is optimal. (Necessity) Consider the continuous martingale given by This martingale is well defined since H is an admissible pricing rule. Consider θ$_{t}$ = ∫ t 0 X$_{s}$ − Y$_{s}$ 1 − s ds . In this case, we can solve the stochastic differential equation for 15 X$_{t}$ = G ( Z$_{t}$, t ) = E [ h − $^{1}$( f ( Z$_{1}$ )) |F I t ] . E [ X θ 1 ] ≤ V (0 , Z$_{0}$, 0) . ∑ s ≤ 1 (Δ V ( Y$_{s}$, Z$_{s}$, s ) − V$_{y}$ ( Y$_{s}$, Z$_{s}$, s )Δ Y$_{s}$ ) ≤ 0 , (3.15) − ∫ 1 0 V$_{yy}$ ( Y$_{s}$, Z$_{s}$, s ) 2 d [ θ $^{c}$]$_{s}$ ≤ 0 , (3.16) − V (1 , Y$_{1}$, Z$_{1}$ ) ≤ − V (1 , y $^{∗}$( Z$_{1}$ ) , Z$_{1}$ ) . (3.17) E [ X θ 1 ] = E [ V (0 , Z$_{0}$, 0) − V ( Y$_{1}$, Z$_{1}$, 1) − 1 2 ∫ 1 0 V$_{yy}$ ( Y$_{s}$$_{−}$, Z$_{s}$, s ) d [ θ $^{c}$]$_{s}$ + ∑ s ≤ 1 [Δ V ( Y$_{s}$, Z$_{s}$, s ) − V$_{y}$ ( Y$_{s}$, Z$_{s}$, s )Δ Y$_{s}$ ]   .