the fact that H ( y, t ) satisfies equation (3.7), we have that Therefore, by Theorem 26.6 of [8], and Theorem II.6.29 of [14], we have (notice that Z$_{s}$ and B 2 t are continuous) On the other hand, consider a function where y $^{∗}$( z ) is the solution of H ( y $^{∗}$( z ) , 1) = f ( z ). Let This function is well defined (it is easy to check that E [ | J ( B $_{1}$, Z$_{1}$ ) | ] < ∞ ) and satisfies the partial differential equation with terminal condition V ( y, z, 1) = J ( y, z ). Therefore V ( y, z, 1) ≥ V ( y $^{∗}$( z ) , z, 1) = 0 for any fixed z and any y = y $^{∗}$( z ). Moreover, since H ( y, 1) is a nondecreasing continuous function of 13 V$_{t}$ ( y, z, t ) + 1 2 V$_{yy}$ ( y, z, t ) + σ 2 $_{z}$( t ) 2 V$_{zz}$ ( y, z, t ) = 0 (3.12) V ( y, z, t ) = E [ J ( y + B 2 1 − B 2 $_{t}$, z + ∫ 1 t σ$_{z}$ ( s ) dB 1 s )] . (3.11) J ( y, z ) = ∫ y $^{∗}$( z ) y ( f ( z ) − H ( x, 1)) dx, [ θ, F ( Z, · )]$_{1}$ = ∫ 1 0 F$_{z}$ ( Z$_{s}$, s ) d [ θ $^{c}$, Z ]$_{s}$ [ θ, H ( Y, · )]$_{1}$ = 1 0 H$_{y}$ ( Y$_{s}$$_{−}$, s ) d [ θ $^{c}$]$_{s}$ + ∫ 1 0 H$_{y}$ ( Y$_{s}$$_{−}$, s ) d [ θ $^{c}$, B $^{2}$] s + ∑ s ≤ 1 Δ H ( Y$_{s}$, s )Δ θ$_{s}$. H ( Y$_{t}$, t ) = H (0 , 0) + ∫ t 0 H$_{y}$ ( Y$_{s}$$_{−}$, s ) dY c s + 1 2 ∫ t 0 H$_{yy}$ ( Y$_{s}$$_{−}$, s ) d 〈 θ $^{c}$〉 s + ∫ t 0 H$_{yy}$ ( Y$_{s}$$_{−}$, s ) d 〈 θ $^{c}$, B $^{2}$〉 s + ∑ s ≤ t Δ H ( Y$_{s}$, s ) .