ARCH 2014.1 Proceedings July 31-August 3, 2013 Option Pricing - PDF document

Article from: ARCH 2014.1 Proceedings July 31-August 3, 2013

Option Pricing Without Tears: Valuing Equity-Linked Death Benefits Elias S. W. Shiu Department of Statistics & Actuarial Science The University of Iowa U.S.A. Joint work with Hans U. Gerber & Hailiang Yang

Let T x denote the time-until-death random variable for a life aged x. Let S(t) be the time-t price of a stock or mutual fund. Consider death benefits that depend on the value of S(T x ), i.e., consider b(S(T x )) for some function b(.). Examples: b(s) = Max(s, K) b(s) = (s – K) + Problem : Evaluate −δ T E[e b(S(T ))] x x where the expectation is taken with respect to an appropriate probability distribution and δ is a continuously compounded interest rate.

−δ T E[e b(S(T ))] x x −δ = T | E[ E[ e b(S(T )) T ] ] x x x ∞ = ∫ −δ t | E[ e b(S(t)) T =t f ] (t) dt x T x 0 ∞ = ∫ −δ t E[ e b(S(t)) ] f (t ) dt T x 0 if T x is independent of {S(t)}.

So we want to calculate ∞ ∫ −δ t E[e b(S(t))]f (t)dt. T x 0 = ∑ If f (t) c f (t) , τ T j x j j then ∞ ∫ −δ t E[e b(S(t))]f (t)dt T x 0 ∞ ∑ ∫ −δ t = c E[e b(S(t))]f (t)dt τ j j j 0 ∑ − τ δ τ j = c E[e b (S( ) ]. ) j j j

The time-until-death density function can be approximated by linear combinations of exponential density functions ∑ ∑ −λ t ≈ × × λ j f (t) c f (t) = c e . τ T j j j x j j j Thus, our valuation problem becomes finding E[e −δτ b(S( τ ))], where τ is an exponential random variable independent of {S(t)}. It turns out to be an elementary calculus exercise for geometric Brownian motion {S(t)}.

Let S(t) = S(0)e µ t + σ Z(t) , t ≥ 0 , where {Z(t)} is a standard Brownian motion. Let τ be an independent exponential random variable with mean 1/ λ . Then, E[e −δτ b(S( τ ), Max{S(t); 0 ≤ t ≤ τ })]   ∞ λ m 2 ∫ ∫ −α − β−α = x m x ( )m   b S(0)e ( , S(0)e ) e d x e dm σ 2   −∞ 0 where α < 0 and β > 0 are the solutions of ½ σ 2 x 2 + µ x − (λ + δ ) = 0.

E[e −δτ b(S( τ ), Max{S(t); 0 ≤ t ≤ τ })]   ∞ λ m 2 ∫ ∫ −α − β−α = x m x ( )m   b S(0)e ( , S(0)e ) e d x e d m. σ 2   −∞ 0 Examples: b(s, u) = (s – K) + call option b(s, u) = (K – s) + put option b(s, u) = u high water mark payoff

Barrier Options Assume S(0) < B, a barrier. b(s, u) = I(u < B)× π (s) Up-and-out option b(s, u) = I(u ≥ B)× π (s) Up-and-in option Useful for incorporating lapses or surrenders.

Assume S(t) = S(0)e X(t) , t ≥ 0, where N (t) N (t) ν ω ∑ ∑ = µ + σ + − X(t) t Z(t) J K j k = = j 1 k 1 m ∑ − = v x > f (x) A v e , x 0 i J i i = i 1 n ∑ − = > w x f (x) B w e , x 0 i K i i = i 1 m n ∑ ∑ = = A 1, B 1 i i = = i 1 i 1

S(t) = S(0)e X(t) , t ≥ 0 Running maximum M(t) := Max{X(u); 0 ≤ u ≤ t } Running minimum m(t) := Min{X(u); 0 ≤ u ≤ t } Because {X(u)} is a Levy process, (i) M( τ ) and [X( τ) − M (τ )] are independent random variables, (ii) [X( τ) − Μ(τ )] has the same distribution as m( τ ). (i) is hard to prove; (ii) is easy.

In fact, (ii) is true for each fixed t. X(t) – M(t) = X(t) – Max{X(s ); 0 ≤ s ≤ t} = X(t) + Min{ − X(s); 0 ≤ s ≤ t} = Min{X(t) − X(s); 0 ≤ s ≤ t} = Min{X(t − s); 0 ≤ s ≤ t} in distribution = Min{X(s ); 0 ≤ s ≤ t} = m(t)

Running maximum M(t) := Max{X(u); 0 ≤ u ≤ t} Running minimum m(t) := Min{X(u); 0 ≤ u ≤ t} (i) M( τ ) and [X( τ) − M (τ )] are independent r.v.’s. (ii) [X( τ) − M (τ )] and m( τ ) have the same distribution. Then, E[e zX( τ ) ] = E[e z[X( τ ) −Μ(τ)+ M( τ )] ] = E[e z[X( τ ) − M( τ )] ] × E[e zM( τ ) ] = E[e zm( τ ) ] × E[e zM( τ ) ], which is a version of Wiener-Hopf factorization.

Assume N (t) N (t) ν ω ∑ ∑ = µ + σ + − X(t) t Z(t) J K j k = = j 1 k 1 where m ∑ − = > v x f (x) A v e , x 0 i J i i = i 1 n ∑ − = > w x f (x) B w e , x 0 i K i i = i 1 Then, E[e zX(t) ] = e t Ψ (z) for each t ≥ 0 , with m n z z ∑ ∑ Ψ = µ + σ + ν − ω 2 2 (z) z ½ z A B i i − + v z w z = = i i i 1 i 1

m n z z ∑ ∑ Ψ = µ + σ + ν − ω 2 2 (z) z ½ z A B i i − + v z w z = = i i i 1 i 1 can be extended by analytic continuation . The moment-generating function of X( τ ) is E[e zX( τ ) ] = E[E[e zX( τ ) | τ ]] = E[e Ψ (z) τ ] λ = λ − Ψ . (z) The zeros of the RHS are the poles of Ψ (z). The poles of the RHS are the zeros of λ − Ψ (z).

Label the parameters (the poles of Ψ (z)) such that v 1 < v 2 < … < v m w 1 < w 2 < … < w n If the weights A’s and B’s are positive, then − ∞< α n+1 < − w n <...<− w 1 <α 1 <0< β 1 < v 1 <...< v m <β m+1 < ∞

Label the parameters (the poles of Ψ (z)) such that v 1 < v 2 < … < v m w 1 < w 2 < … < w n If the weights A’s and B’s are positive, then − ∞< α n+1 < − w n <...<− w 1 <α 1 <0< β 1 < v 1 <...< v m <β m+1 < ∞ E[e zX( τ ) ] = E[e zm( τ ) ] ×E[e zM( τ ) ]. Wiener-Hopf: For z > 0, 0 < E[e zm( τ ) ] ≤ 1. No positive zeros or poles. For z < 0, 0 < E[e zM( τ ) ] ≤ 1. No negative zeros or poles.

− ∞< α n+1 < − w n <...<− w 1 <α 1 <0< β 1 < v 1 <...< v m <β m+1 < ∞    + n n 1 1 ∏ ∏ τ ∝  +   zm( ) E[e ] (z w )    j − α z     = = j j 1 j 1    + m m 1 1 ∏ ∏ τ ∝ − zM( )    E[e ] (z v )    j − β z    = = j j 1 j 1    + − α + n n 1 z w ∏ ∏ τ j j =    zm( ) E[e ]    − α w z    = = j j j 1 j 1    − β + m m 1 v z ∏ ∏ τ j j =  zM( )   E[e ]    β − v z    = = j j j 1 j 1

τ z M( ) E[e ]    − β + m m 1 v z ∏ ∏ j j =       β − v z    = = j j j 1 j 1    −β β + + β m m 1 m 1 v ∑ ∏ ∏ j k j =    k .    β −β β − v z    = = = ≠ j j k k k 1 j 1 j 1,j k + m 1 ∑ −β x = > Thus, f ( ) x b e , x 0, k τ M( ) k = k 1    + −β β m m 1 v ∏ ∏ j k j    = β where b k k    β −β v    = = ≠ j j k j 1 j 1,j k

τ z m( ) E[e ]    + −α + n n 1 z w ∏ ∏ j j =       − α w z    = = j j j 1 j 1    α + −α + + −α n 1 n n 1 w ∑ ∏ ∏ k j j =    k    α − α − α w z    = = = ≠ j k j k k 1 j 1 j 1 , j k + n 1 ∑ −α = x < Thus, f ( ) x a e , x 0, k τ m( ) k = k 1    α + −α + n n 1 w ∏ ∏ k j j = −α    where a ( )    k k α − α w    = = ≠ j k j j 1 j 1, j k

For y ≥ max(x, 0), f X( τ), M (τ ) (x, y) = f M( τ), X (τ)−Μ(τ ) (y, x − y) × 1 = f M( τ) (y) f X (τ)−Μ(τ ) (x − y) = f M( τ) (y) f m (τ ) (x − y)    +  + m 1 n 1 ∑ ∑ −α − −β (x y) = y     j b e a e k  k j    = = k 1 j 1 + + m 1n 1 ∑∑ −α − β −α x ( )y = × j k j a b e e j k = = k 1 j 1

Further Work 1. Use SOA CAE research grant to hire graduate students to estimate the parameters and to program the formulas. 2. Binomial tree version.

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