Then:
Q (ʃ5 Cι,<2)
P (т5 c1,c2)
β - P7iC i
^r -' - 1
Л----7=---
Beτ√A + 1
τ (r - δ - — "ʌ iζ 1 + -^ τβ + τ√"7^ + 2 In ——+-—
V 7 J 72 L Be^ + 1
Caveat. This solution is valid for (ζ 1,C2) = (O, O),∙ Given the trial solution for Ψ and the dehnition
of a ch.f., it must be the case that q (τ; O, O) = p (τ;O, O) = O, Vτ. I check that the above expressions
for q (τ; ζι,C2) and p (τ; ζ 1,C2) do not contradict the uniform continuity property of Ψ, that is, whether
lim(c,G)'→(o,o)' q( τ; ci,c2) = iim(c,G)'→(o,o)' p( τ; Cι,C2) = o :
Iim
(ζ1,ζ2)'→(0,0)'
q(τ ; c 1 ,c 2)
1
72
' Γ2eτ^2 - lim1∕B (C 1,C2)
β - V β —/=-----------
eτ√^2 + lim1∕B (C 1,C2).
-2 [β - β] = o,
72
lim
(GX2)'→(0,0)'
p(τ ; c 1 ,c 2)
ɪ τβ + τ√β2 + 2ln X±lim1/B (C 1,C2)
7 L eτ√β2 + lim1∕B (C 1,C2).
72 [τβ + τβ - 2 ln eτ"] = O,
because lim(ζ1,ζ2)'→(o,o)' B (C 1, C2)
= ∞, if β > O (it is straightforward to extend the proof
to the subcase β = O^.
Case 2. 7 = O, β > O.
|
q (τ ; c 1,c 2) = |
- 2ξ [e βτ (c 1+ iC 1+ 2 '/C2) - c 1 - iC 1], |
p (τ; c 1, c2) = τ (r - £)iC 1 - 2⅛ [2 (e βτ - 1)iC2+ τ'β 1+ τc2] -
2β
|
α (e ^τ - 1) r 9 -l - 2.,.2 [C ? + ≈C l]. 2β |
Case 3. 7 = β = O.
q (τ; c 1,c2) = -2 [C1 + iC 1] + iC2,
p (τ; c 1,c2) = τ [r - β iC 1 - ɪ [τ (c 1 + iC 1) - 4iC2] .
BlackScholes obtains with 7 = β = α = O and p (τ; C1, C2) = τ [r - £] iC 1.