USC Math. Finance April 22, Path-dependent Option Valuation under Jump-diffusion Processes. Alan L. Lewis

Transcription

1 USC Math. Finance April 22, 23 Path-dependent Option Valuation under Jump-diffusion Processes Alan L. Lewis These overheads to be posted at (Publications) Topics Why jump-diffusion models? Review of the vanilla Euro-style Valuation A Standard Machine for Path-dependent Options: a. Perpetual American b. Knock-outs (Down-and-out call, etc.) c. One-touch Options (Rebate term) 1

2 Why Jump-Diffusion Models for Options? I. Benchmark model (exponential Brownian motion): Attractive features: limited liability stock prices uncorrelated, level independent returns simple formulas (methods) for option prices (euro,amer) Weak points: Actual stock price distributions have wider tails Lacks volatility clustering (auto-corr. of absolute returns) Lacks stock price jumps Poor fit to real-world option prices (smile/skew) II. Jump-diffusion processes (exponential Lévy processes) - Stationary, independent increment processes - Continuous-time analog of Random Walk - Brownian motion plus Poisson-driven jump process Attractive features: all the attractive benchmark features + large flexible class of models, each with a few parameters wide return tails common (exponential decay, moments) Good fits to expiring options (fear of jumps/crashes?) Weak points: Lacks volatility clustering (auto-corr. of absolute returns) Brownian motion Large number of small jumps 2

3 Stock price Evolution and Examples St = S exp( Xt), (Assumption: this is under the martingale pricing measure Q ) where Xt = ct + σbt + Xt; Jump probability(t) X t nt () = i= 1 λ t y~( p y) (Jump distribution) y i Examples of Jump distributions: (A.1) Merton s 1976 jump-diffusion model with log-normally distributed jumps: p( y) = 1 exp ( y µ J ) / 2δ 2 2πδ 2 2 (A.2) Degenerate Case: Point-jump: p( y) = δ ( y µ ) J Typical SPX Smile Fit: λ 3., µ 25,. δ 1. J 3

4 Figure 1 SPX Options: Implied Volatility vs. Strike on Aug. 16, 22 H1 monthto ExpirationL ConstantVolatility +Jumps Ø 25 2 SPX

5 Vanilla European-style options Solutions in Fourier space are simple Ingredients: 1. The generalized Fourier transform of the payoff function: For the call option: 1 iz izx x + K gz ˆ( ) = e ( e K) dx =, Im z < 1 2 ( z + iz) 2. The characteristic function of the Lévy process: ϕt( z) = e izx pt( x) dx = E [ exp( izxt )] = exp ( TΨ( z) ) where Ψ ( z) is the characteristic exponent. For the Point Jump model: ψ( z) = izω z σ λ {exp( iµ J z) 1} (Entire) 2 3. Finally, the Call Option price is given by : rt iz S Tψ ( z) C( S, K, T) e K e = dz π K, 2 z( z+ i) Im z< 1 The integration is along a line parallel to the real z-axis. 5

6 European-style options (cont.) The solution is very easy to derive and obvious : First, we need the inversion formula for the payoff function: iν + izx g( x) = 1 e gˆ ( zdz ) π, x = log S 2 T, z Payoff strip iν Then, by martingale pricing: rτ iν + rτ i ( ) E[ (log T) ] e z C S = e g S = E ( ST) gˆ ( z) dz 2π iν rτ i e ν + iz izx T = E ( S ) e gˆ ( z) d π z 2 iν rt iν + e iz Tψ ( z = ( S ) e )ˆ g( z) π dz 2 iν, Ok to exchange the integrations (sufficient conditions) if: 1. gx ( ) is Fourier integrable in some Payoff strip bounded for x <. 2. ψ ( z) is regular in some strip S X : α < Im z < β 3. ν = Im z lies in the intersection of these two strips S g and 6

7 A Standard Machine: the Down-and-out Call (or Down-and-out anything ) How we will do it. 1. Write the payoff in terms of its Fourier Transform 2. Write the barrier condition using a representation for 1 { Y>}. 3. Bring an expectation inside an integral. 4. Find a Fluctuation Identity to do the expectation. 5. Done with General Formula! This general procedure works for all the problems I listed at the beginning and probably lots of others. It saves having to learn a lot of the heavy machinery of The Boyarchenko/Levendorskii approach. 6. Then, for your particular model: Try to do as many integrals as possible analytically; (Residue Calculus). 7. Do the remaining integrals numerically. 7

8 Down-and-out Call (or Down-and-out anything ) (cont). 1. Write the payoff in terms of its Fourier Transform Minimum Process: S min T = t T St CDOC ( S, K, H, T) e rt E ( ST K) + = 1 ST > H or, with x = log S, h = log H, and NT = min t T Xt rt x XT f ( x, T) = e E ( e K) + + DOC 1 N > h x T rt = e E dz exp { iz( x+ X )} gˆ ( z) 1 T 2π Im z< 1 T N > h x 8

9 Down-and-out Call (or Down-and-out anything ) (cont) 2. Write the barrier condition using a representation for 1 { Y>}. 1 Y> = Imξ< dξ exp( iξy) 2πξ i, where Y = x h+ N T. This produces: rt fdo( x, T) = e d E dz ξ gˆ( z)exp ( izx+ iξ( x h) ) exp( izxt + iξnt 2π 2πiξ Im z< 1 Imξ< ) 3. Bring an expectation inside an integral. We need a formula for E[ exp( izxt iξ NT )] +. 9

10 4. Find a Fluctuation Identity to do the expectation Some Fluctuation Identities: 1. Factorization identities: (Spitzer, Rogozin, others) where q = φq q+ ψ( z) + q ( z) φ ( z), φ φ + qt q ( ) = [ exp( τ ( q) )] = izm t z E izm q e E e dt qt iznt q ( z) = E[ exp( iznτ ( q) )] = q e E e dt τ ( q) is an independent (exponentially distributed) random stopping time. Not computationally effective. But, this one is: φ q ( ξ) = exp dη ξlog [ q + ψ( η) ] ( 2πi ) η( ξ η) { } + + (Im ξ) < Imη< σ 1

11 4. Find a Fluctuation Identity to do the expectation Some Fluctuation Identities: 2. Here s the one we really need: + T T q q qt q e E[ exp ( izx + iξn )] dt = φ ( z) φ ( ξ + z) By Laplace Inversion: dq qt + E[ exp ( izxt + iξnt )] = e φq ( z) φq ( ξ + z) 2 πiq Req> r 11

12 5. Done with General Formula! Here it is: dq qt fdo ( x, T) = e FDO ( x, q ), 2 πi Req> r where for the call option payoff, it reads F DOC C ( x, q) rt Ke = q dξ dz exp { iz( h k) + iξ( x h) } + φq ( z) φq ( ξ) 2πi 2π ( z ξ) z( z+ i) C 1 2 Integration contours: C : Imλ 1 < Imξ < 1, C2 : Im α 1 ( q) < Im z< 1 and Imξ < Im z 12

13 6. Then, for your particular model: Try to do as many integrals as possible analytically; (Residue Calculus). 7. Do the remaining integrals numerically. Example 1: Suppose your model has No negative jumps. (This means the Barrier is crossed continuously). When there are no negative jumps, the Laplace transform of the down-and-out call option is given by: ( x > h) F ( xq, ) = F ( xq, ) exp { iγ ( q)( x h) } F ( hq, ), DOC NNJ E E where FE ( xqis, ) the Laplace transform of the European (no barrier) call option: rt FE ( xq, ) = dz Ke exp { izx ( k) } π zz ( + i)( q+ ψ( z)) 2. Im α1 ( q) < Im z< 1 (This one could be proved directly from the PIDE). 13

14 6. Then, for your particular model: Try to do as many integrals as possible analytically; (Residue Calculus). 7. Do the remaining integrals numerically. Example 2: Suppose your model has a Negative Point Jump. Then, you have to investigate the roots (zeros) of q + ψ( z) =, where 1 q+ ψ( z) = q izω + z 2 σ 2 λ ( exp( iµ J z) ) 2 1 F It turns out there are (almost certainly) an infinity of these in the complex z-plane. Then, I have a (conjectured) result using these roots: DOC x rt N ie γ exp { i( γi α)( x )} ( x, q) = 1 h + lim α( α+ i) ψ'( α) Nγ ψ'( γi ) i = 1 α is the single lower half-plane root. (I m α < ) γ are the infinity of upper half-plane roots. (Im γ > ) i i 14

15 A plot of f '( z)/ f( z ) where f ( z) = r+ ψ( z) for the Merton jump-diffusion 15

16 Figure 2. The location of some roots of r+ ψ( z ) = for the point jump model. 16