Computation of First-Order Greeks for Barrier Options Using Chain Rules for Wiener Path Integrals

Construction and sample path properties of Brownian house-moving between two curves

“…2 (1 − t, y) J (b) 3 (s, x, t, y) J (b) 2 (1 − s, x) dy, (3) 2010 Mathematics Subject Classification: Primary 60F17; Secondary 60J25. (n t−s (y − x + 2kη) − n t−s (y + x + 2kη)),…”

“…). Then we are interested in whether the law of Y b under P (3) 0 coincides with that of H 0→b . To answer this question, we must, for example, deal with the density P (3) 0 (Y b (t) ∈ dy), which is given by…”

“…However, the integrand of the above integral contains an infinite series of Bessel functions ( [5], [8], [2]) and hence is not easy to handle. Thus, our next focus is to identify the distribution of Y b under P (3) 0 and compare it with that of H 0→b . In addition, for the BES(δ)-bridge r 0→b δ = {r 0→b δ (t)} t∈[0,1] (b, δ > 0) from 0 to b on [0, 1], we are interested in finding the weak limit of r 0→b δ | K − (b+η) as η ↓ 0.…”

“…In addition, for the BES(δ)-bridge r 0→b δ = {r 0→b δ (t)} t∈[0,1] (b, δ > 0) from 0 to b on [0, 1], we are interested in finding the weak limit of r 0→b δ | K − (b+η) as η ↓ 0. Recently, [3] developed a chain rule for Wiener path integrals between two curves that arise in the computation of firstorder Greeks for barrier options, and demonstrated the effectiveness of this chain rule through numerical examples. In this chain rule, BES(3)-bridge and Brownian meander played an important role.…”

“…By Skorohod's theorem, we may assume that X n and X are defined on the same probability space and X n → X holds almost surely. Then, by the dominated convergence theorem, the first term of the right-hand side of If W [0,T ] = W (1) [0,T ] , W (2) [0,T ] , W (3) [0,T ] is a three-dimensional Brownian motion starting at 0, then P T g < T, R [0,T ] ∈ K − [0,T ] (g), R [0,T ] (T ) ∈ db = E P (g(s), 0, 0) + W [s,T ] ∈ K − [s,T ] (g), (g(s), 0, 0) + W [s,T ] (T ) ∈ db s=T g ; T g < T holds. Therefore, we only have to show that P (g(s), 0, 0) + W [s,T ] ∈ K − [s,T ] (g), (g(s), 0, 0) + W [s,T ] (T ) ∈ db = 0 for s ∈ [0, T ).…”