Electronic Companion A Two-Sie Laplace Inversion Algorithm with Computable Error Bouns an Its Applications in Financial Engineering Ning Cai, S. G. Kou, Zongjian Liu HKUST an Columbia University Appenix A. Proof of Proposition 6.1 Proof. (i) First, F Xt (x) = P (X t x) C 1 because X t has a continuous istribution uner P. Secon, P (X t x) is obviously increasing. Then applying Lemma 3.1, we know F Xt (x) satisfies Assumption 1. Moreover, the ROAC of F Xt (x) is (, u ) because for any (, u ), e x F Xt (x)x = ( ) e x P (X t x) x = E e x x = Ee X t X t < +. Therefore, e x F Xt (x) is of boune variation on R for any (, u ). (ii) EuC(k) can be rewritten as EuC(k) =S E [ e Xt rt ] I {St e k e rt e k P ( S t e k) =S P ( St e k) e rt e k P ( S t e k) := f 1 (k) f (k). (3) where the secon equality hols because of the change of measure with P P F t = e X t rt. Because S t has a continuous istribution uner both P an P, then f 1 (k) an f (k) are in C 1. Besies, P ( S t e k) an P ( S t e k) are both increasing in k. Therefore, applying Lemma 3.1 yiels that both f 1 (k) an f (k) satisfy Assumption 1. Note that for any ( 1, l 1), e k f (k)k =e rt e (+1)k P ( S t e k) k =e rt E [ log S t e (+1)k k ] = e rt S +1 Ee (+1)Xt + 1 < +, an for any (, l 1), e k f 1 (k)k =S = S+1 Ẽe Xt e k ( P St e k) [ ] k = S Ẽ e k k log S t = e rt S +1 Ee (+1)Xt < +. It follows that both e k f 1 (k) an e k f (k) are of boune variation on R for any (, l 1), an so is e k EuC(k). A-1
Appenix B. Proof of Proposition 6. Proof. Replacing s in (17) by + iω with ( ˆM, Ĝ) yiels L fxt ( + iω) = { ( ) e µt iµtω exp tĉ Γ( Ŷ ) ˆM Ŷ + ĜŶ [ exp {tĉ Γ( Ŷ ) ( ˆM ] + + iω)ŷ + (Ĝ iω)ŷ ( ) = exp { µt tĉ Γ( Ŷ ) ˆM Ŷ + ĜŶ [ exp {tĉ ] Γ( Ŷ ) eŷ [ln Z 1 +i arg(z 1 )] + eŷ [ln Z +i arg(z )] ( ) = exp { µt tĉ Γ( Ŷ ) ˆM Ŷ + ĜŶ [ ] exp {tĉ Γ( Ŷ ) Z 1 Ŷ cos(ŷ arg(z 1)) + Z Ŷ cos(ŷ arg(z )) { ( ) = exp µt tĉ Γ( Ŷ ) ˆM Ŷ + ĜŶ Q(Ŷ,, ω), (33) where ln( ) is the real natural logarithm function, arg(z j ) ( π, π) for j = 1 an, Z 1 Z 1 (, ω) := ˆM + + iω, Z Z (, ω) := Ĝ iω, { ] Q(Ŷ,, ω) := exp tĉ [ Z Γ( Ŷ ) 1 Ŷ cos(ŷ arg(z 1)) + Z Ŷ cos(ŷ arg(z )) { = exp tĉ Γ( Ŷ ) cos(ŷ π/) ω Ŷ [f 1 ( ω ) + f ( ω )], an f j ( ω ) := Z j Ŷ ω Ŷ Note that f 1 ( ω ) can be rewritten as f 1 ( ω ) = cos(ŷ arg(z j)) cos(ŷ π/), for j = 1 an. [ 1 + ( ˆM + ) ω Then for any ω >, some algebra yiels where R( ω ) := ( Defining z = arctan ω ˆM+ ] Ŷ / ( )) cos (Ŷ arctan ω ˆM+ cos(ŷ π/). f 1( ω ) = ( ˆM [ + )Ŷ cos(ŷ 1 + ( ˆM ] Ŷ / 1 + ) π/) ω ω R( ω ), (34) ( ( )) ( ( )) ω ω cos Ŷ arctan + sin Ŷ arctan, ˆM + ˆM + ˆM + ω ) (, π/), we obtain R( ω ) = 1 1 ] cos(ŷ z) + sin(ŷ z) = [cos(ŷ tan(z) sin(z) z) cos(z) + sin(ŷ z) sin(z) = 1 cos((ŷ 1)z). sin(z) Since z (, π/), we have R( ω ) > for any Ŷ (, 1) (1, ). Then by (34), we obtain f 1( ω ) < if Y (, 1), an f 1( ω ) > if Ŷ (1, ). Similarly, we can prove f ( ω ) < if Ŷ (, 1), an f ( ω ) > if Ŷ (1, ). It follows that if Ŷ (, 1), f 1( ω ) an f ( ω ) are both ecreasing functions of ω in (, + ). This implies that f 1 ( ω ) + f ( ω ) lim ω + [f 1 ( ω ) + f ( ω )] = for any ω >. Hence for any ω > ω :=, we have Q(Ŷ {tĉ,, ω) exp Γ( Ŷ ) cos(ŷ π/) ω Ŷ, which along with (33) completes the proof for the case of Ŷ (, 1). A-
If Ŷ (1, ), f 1( ω ) an f ( ω ) are both increasing functions of ω in (, + ). Then for any ω > ω := max{ ˆM +, Ĝ tan((π/ + ϵ) /Ŷ ) > with ϵ (, π(ŷ 1)/), we have f j( ω ) cos(π/+ϵ) for cos(ŷ π/) j = 1 an. As a result, for any ω > ω, we have Q(Ŷ {tĉ,, ω) exp Γ( Ŷ ) cos(π/ + ϵ) ω Ŷ, which along with (33) completes the proof of (18). Appenix C. Proof of Proposition 6.3 Proof. Substituting + iω for s in (19) yiels that for any ω > ω :=, L fxt ( + iω) = exp t ( iω) m p l η l µt( + iω) + λt p u η l + + iω + q q j θ j θ j=1 j iω 1 { ( ) = exp t µ exp { t ω m exp λt p l η l (η l + ) p u (η l + ) + ω + q q j θ j (θ j ) (θ j=1 j ) + ω 1 { ( ) exp t µ exp { t ω m exp λt p l η l (η l + ) p u (η l + ) + ω + q q j θ j (θ j ) (θ j=1 j ) + ω 1 { ( ) exp t µ exp { t m ω exp λt p l η l p u η l + + q q j θ j θ j=1 j 1 = exp t m p l η l µ + λ p u η l + + q q j θ j θ j 1 { t exp ω, j=1 which completes the proof. Appenix D. Proof of Propositions 7.1 7.3 We shall efer the proof of Proposition 7.1 to the en because it uses the results of Propositions 7.3. Proof of Proposition 7.. For any ω > ω, we know from Lemma D. in Cai an Shi [5] that β 1,s+r η Y 3 ω θ < Y 3 an β,s+r z ω µ. (35) It follows that η Y 3 < β 1,s+r < Y 3 + η an β,s+r z ω µ ω = µ. (36) where the first inequality hols also because ω Y ( + r) Y 4 = Y3 ηθ. From the secon inequality of (36) we further obtain β,s+r 1 β,s+r 1 ω µ ω ω 1 > + µ ω 1 >, (37) A-3
thanks to the fact that ω > ω Y ( + r) Y 1 = (η + µ ) an η > 1. Moreover, note that Re(z ω ) µ > Re(z µ ω ω ) = µ Y1 µ = η + ( 1) µ > η, This implies that Re(z ω ) η > Re(z ω ) η > µ obtain that for any ω > ω, an hence z ω η z ω η. Then by (35) we β,s+r β 1,s+r z ω η z ω β,s+r β 1,s+r η z ω η z ω β,s+r β 1,s+r η z ω η µ Y 3 >. where the last inequality hols because z ω η µ Y 3 Im(z µ ω ) η = Re(z ω ) µ η >. Accoring to (35), (36) an the equality above, we have that for any ω > ω, I 1 (ω) := β,s+r η β,s+r β 1,s+r 1 + β 1,s+r η β,s+r β 1,s+r Y 3 ω 1 + θ z ω η µ = z ω η µ Y 3 z ω η µ I (ω) := β,s+r β,s+r β 1,s+r 1 + β 1,s+r β,s+r β 1,s+r 1 + Y 3 + η z ω η µ Y3 Y 3 = z µ ω η + η z ω η µ Y3, (38), (39) Besies, efine h(x) := xe bx for x with b >. It is easy to see that h(x) attains its maximum at x = 1 b, an ecreases in x when x > 1 b. Thus by (36) we have that for any ω > ω 1 b I 3 (ω) := ω ( S M ) β,s+r ω e b ω ( ) µ S M h ( 1 b ) ( ) µ S = 1 M be 1 ( M S ) µ. (4) Note that from (1), we can obtain ( ) Re(β1,s+r ) S ( β 1,s+r η )I (ω) L f (s) M M ( β 1,s+r 1 )η(s + r) + M β 1,s+r I 1 (ω)i 3 (ω) η(s + r)( β,s+r 1 ) ω + rm s(s + r). Substituting (35) (4) into the RHS of the inequality above conclues the proof. Proof of Proposition 7.3. Define f(t ) := f(t ) + M >. Then L f (s) = L f (s) + M s for any Re(s) >. Moreover, for any > an ω R, we have L f ( + iω) = e (+iω)t f(t )T e (+iω)t f(t ) T = L f () = L f () + M. Applying Proposition 7. yiels L f () + M L f ( + iω) L(ω; ) :=, for all ω ω ; ζ() ω, for all ω > ω. (41) Then by the Bromwich contour integral we obtain e T f(t ) 1 π L f ( + iω) ω 1 π L(ω; ) ω = δ(). A-4
[ ] Proof of Proposition 7.1. Define a function u(t ) := E max {M, max t T S t 1 {T. Then u(t ) is positive an increasing in T when T, an f(t ) e rt u(t ) M when T. To show e T f(t ) is of boune variation on R for any >, it suffices to prove that for any >, the function g u (T ) := e (+r)t u(t ) is of boune variation on R + for any >. To this en, we shall first compute the total variation of g u (T ) on the interval [, A] for any A > (enote by A (g u)), an then show that lim sup {A> A (g u) < +. Recall that A (g u ) := lim 1 i= g u (x i+1 ) g u (x i ), (4) where : = x < x 1 < < x n = A, is any partition of the interval [, A] with the norm := sup i n 1 x i+1 x i. For any > an < T 1 < T, we have g u (T ) g u (T 1 ) e (+r)t u(t 1 ) g u (T 1 )) + g u (T ) e (+r)t u(t 1 ) = [e (+r)t 1 e (+r)t ]u(t 1 ) + [u(t ) u(t 1 )]e (+r)t (T T 1 )( + r)e (+r)t1 u(t 1 ) + [u(t ) u(t 1 )]e (+r)t, where the last inequality is obtaine base on the mean value theorem. It follows that where 1 i= 1 S1 := ( + r) (x i+1 x i )e (+r)x i u(x i ) an S := i= Letting yiels As for S, note that lim S 1 = ( + r) S = ( + r) 1 i= A g u (x i+1 ) g u (x i ) S 1 + S (43) e (+r)t u(t )T = ( + r) e T f(t ) T + M( + r) [u(x i+1 ) u(x i )]e (+r)x i+1. n 1 i= A [e T f(t ) + e T M]T u(x i )[e (+r)xi e (+r)xi+1 ] u() + u(a)e (+r)a 1 ( + r) g u (x i )(x i x i 1 ) + (e A 1)M + f(a)e A i= 1 ( + r) g u (x i )(x i x i 1 ) + δ(), i= < +. (44) where the first an secon inequalities hol ue to the mean value theorem an Proposition 7.3, A-5
respectively. Then letting, we have lim S ( + r) ( + r) A Substituting (43), (44) an (45) into (4) yiels e (+r)t u(t )T + δ() e T f(t ) T + M ( + r) + δ() < +. (45) A (g u ) ( + r) e T f(t ) T + M ( + r) + δ() < +, for all A >. Since the boun in the inequality above is inepenent of A, it follows that g u (T ) is of boune variation on R +. The proof is complete. A-6