Exercise 7.6 (Boundary conditions for lookback option). The look- back option price v(t,x,y) of (7.4.35) must satisfy the boundary conditions (7.4.7)–(7.4.9). As we saw in Subsection 7.4.3, this is equivalent to the function u(t, z) of (7.4.16) given by (7.4.36), u(t, z) = = (1 + 2/7) N (6 + (T, 2)) + e ¯* * N ( − 8_ (7, 2)) TT - 2r 02 e-πr 21-34 N ( - 8 - (7, z ¯¹³)) − z, 0 ≤ t < T, 0 < z ≤ 1, 2r е ТТ - satisfying the boundary conditions (7.4.19)–(7.4.21). This function was shown to satisfy boundary condition (7.4.20) in Exercise 7.5(v). Here we verify by direct computation that the limit of u(t, z) as z ↓0 satisfies (7.4.19) and the limit of u(t, z) as t↑ T (7↓↓0) satisfies (7.4.21). (i) If you have not worked Exercise 7.2, then verify (7.8.11), the second equal- ity in (7.8.14) and (7.8.16). (ii) Use (7.8.11) and the second part of (7.8.14) to show that lim↓0 u(t, z) e- for 0

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Exercise 7.6 (Boundary conditions for lookback option). The look-
back option price v(t,x,y) of (7.4.35) must satisfy the boundary conditions
(7.4.7)–(7.4.9). As we saw in Subsection 7.4.3, this is equivalent to the function
u(t, z) of (7.4.16) given by (7.4.36),
u(t, z) =
=
(1 + 2/7) N (6 + (T, 2)) + e ¯* * N ( − 8_ (7, 2))
TT
-
2r
02 e-πr 21-34 N ( - 8 - (7, z ¯¹³)) − z, 0 ≤ t < T, 0 < z ≤ 1,
2r
е
ТТ
-
satisfying the boundary conditions (7.4.19)–(7.4.21). This function was shown
to satisfy boundary condition (7.4.20) in Exercise 7.5(v). Here we verify by
direct computation that the limit of u(t, z) as z ↓0 satisfies (7.4.19) and the
limit of u(t, z) as t↑ T (7↓↓0) satisfies (7.4.21).
(i) If you have not worked Exercise 7.2, then verify (7.8.11), the second equal-
ity in (7.8.14) and (7.8.16).
(ii) Use (7.8.11) and the second part of (7.8.14) to show that lim↓0 u(t, z)
e- for 0<t<T.
-
(iii) Use (7.8.16) to show that lim₁40 u(t, z) = 1 − z for 0 < z ≤ 1.
=
Transcribed Image Text:Exercise 7.6 (Boundary conditions for lookback option). The look- back option price v(t,x,y) of (7.4.35) must satisfy the boundary conditions (7.4.7)–(7.4.9). As we saw in Subsection 7.4.3, this is equivalent to the function u(t, z) of (7.4.16) given by (7.4.36), u(t, z) = = (1 + 2/7) N (6 + (T, 2)) + e ¯* * N ( − 8_ (7, 2)) TT - 2r 02 e-πr 21-34 N ( - 8 - (7, z ¯¹³)) − z, 0 ≤ t < T, 0 < z ≤ 1, 2r е ТТ - satisfying the boundary conditions (7.4.19)–(7.4.21). This function was shown to satisfy boundary condition (7.4.20) in Exercise 7.5(v). Here we verify by direct computation that the limit of u(t, z) as z ↓0 satisfies (7.4.19) and the limit of u(t, z) as t↑ T (7↓↓0) satisfies (7.4.21). (i) If you have not worked Exercise 7.2, then verify (7.8.11), the second equal- ity in (7.8.14) and (7.8.16). (ii) Use (7.8.11) and the second part of (7.8.14) to show that lim↓0 u(t, z) e- for 0<t<T. - (iii) Use (7.8.16) to show that lim₁40 u(t, z) = 1 − z for 0 < z ≤ 1. =
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