Licensed Unlicensed Requires Authentication Published by De Gruyter March 7, 2015

Fourier inversion formulas for multiple-asset option pricing

Bruno Feunou and Ernest Tafolong
From the journal Studies in Nonlinear Dynamics & Econometrics
https://doi.org/10.1515/snde-2014-0034

Abstract

Plain vanilla options have a single underlying asset and a single condition on the payoff at the expiration date. For this class of options, a well known result of Duffie, Pan, and Singleton (Duffie, D., J. Pan, and K. Singleton. 2000. “Transform Analysis and Asset Pricing for Affine Jump-Diffusions.” Econometrica 68: 1343–1376. http://dx.doi.org/10.1111/1468-0262.00164.) shows how to invert the characteristic function to obtain a closed-form formula for their prices. However, multiple-asset and multiple-condition derivatives such as rainbow options cannot be priced within this framework. This paper provides an analytical solution for options whose payoffs depends on two or more conditions. We take the advantage of the inversion of the Fourier transform, resorting to neither Black and Scholes’s framework, nor the affine models’s settings. Numerical experiments based on the aforementioned class of derivatives are provided to illustrate the usefulness of the proposed approach.

Keywords: derivatives pricing; Fourier-Stieltjes transform; multiple triggers payoff
JEL classifications:: G12
Corresponding author: Bruno Feunou, Bank of Canada, 234, Wellington Street, Ottawa, ON, K1A 0G9, Canada, Phone: +(613) 782-8302, Fax: +(613) 782-7713, e-mail:

Acknowledgments

We thank an anonymous referee and Bruce Mizrach (the editor) for comments and suggestions that improved the article substantially. We also thank Guillaume Nolin and Maren Hansen, Glen Keenleyside and Foutse Khomh for comments and suggestions. This research was initiated while Tafolong was working at Desjardins Group. A previous version circulated under the title: “Pricing Multiple Triggers Contingent Claims.” The views expressed in this paper are those of the authors. No responsibility for them should be attributed to the Bank of Canada or the National Bank of Canada. Remaining errors are ours.

Appendix

A Proof of Proposition 2

For 0<τ1, τ2<+∞ and y1, y2∈ℝ, let us define the expression 𝕀 by the following relation:

(19)I≡∫−τ1τ1∫−τ2τ2eiv2y2iv2[eiv1y1ψ(a−iv1b1−iv2b2, x, t, T)−e−iv1y1ψ(a+iv1b1−iv2b2, x, t, T)iv1]−e−iv2y2iv2[eiv1y1ψ(a−iv1b1+iv2b2, x, t, T)−e−iv1y1ψ(a+iv1b1+iv2b2, x, t, T)iv1]dv1dv2. (19)

Because ψ(a+iv1b1+iv2b2, x, t, T)=∫ℝ2eiv1z1+iv2z2Ga,b1,b2(dz1, dz2; x, T), the relation (19) can be expressed as

I=∫−τ1τ1∫−τ2τ2∫ℝ2eiv1y1+iv2y2−iv1z1−iv2z2−e−iv1y1+iv2y2+iv1z1−iv2z2(iv1)(iv2)−eiv1y1−iv2y2−iv1z1+iv2z2+e−iv1y1−iv2y2+iv1z1+iv2z2(iv1)(iv2)Ga,b1,b2(dz1, dz2; x, T)dv1dv2.

Since we disregard the exact order in which the integral has been composed, integrand permutations conserve the value of the integral, and 𝕀 can be expressed as follows:

(20)I=∫ℝ2∫−τ1τ1∫−τ2τ2eiv1y1+iv2y2−iv1z1−iv2z2−e−iv1y1+iv2y2+iv1z1−iv2z2(iv1)(iv2)−eiv1y1−iv2y2−iv1z1+iv2z2+e−iv1y1−iv2y2+iv1z1+iv2z2(iv1)(iv2)dv1dv2Ga,b1,b2(dz1, dz2; x, T). (20)

To proceed with the determination of 𝕀, we make use of the following relation. For τ1>0, τ2>0,

(21)∫−τ1τ1∫−τ2τ2eiv1y1+iv2y2−iv1z1−iv2z2−e−iv1y1+iv2y2+iv1z1−iv2z2−eiv1y1−iv2y2−iv1z1+iv2z2+e−iv1y1−iv2y2+iv1z1+iv2z2(iv1)(iv2)dv1dv2=∫−τ1τ1∫−τ2τ2eiv1y1+iv2y2−iv1z1−iv2z2+e−iv1y1−iv2y2+iv1z1+iv2z2iv1iv2−e−iv1y1+iv2y2+iv1z1−iv2z2+eiv1y1−iv2y2−iv1z1+iv2z2iv1iv2dv1dv2. (21)

Using the usual trigonometric identity, (21) becomes (22) as follows

(22)∫−τ1τ1∫−τ2τ2eiv1y1+iv2y2−iv1z1−iv2z2−e−iv1y1+iv2y2+iv1z1−iv2z2−eiv1y1−iv2y2−iv1z1+iv2z2+e−iv1y1−iv2y2+iv1z1+iv2z2(iv1)(iv2)dv1dv2=∫−τ1τ1∫−τ2τ2−2cos(v1(y1−z1)+v2(y2−z2))−2cos(v1(y1−z1)−v2(y2−z2))v1v2dv1dv2=∫−τ1τ1∫−τ2τ22[cos(v1(y1−z1)−v2(y2−z2))−cos(v1(y1−z1)+v2(y2−z2))v1v2]dv1dv2=∫−τ1τ1∫−τ2τ22[cos(v1(y1−z1))cos(v2(y2−z2))+sin(v1(y1−z1))sin(v2(y2−z2))v1v2]−2[cos(v1(y1−z1))cos(v2(y2−z2))−sin(v1(y1−z1))sin(v2(y2−z2))v1v2]dv1dv2=∫−τ1τ1∫−τ2τ24sin(v1(y1−z1))sin(v2(y2−z2))v1v2dv1dv2=4∫−τ1τ1sin(v1(y1−z1))v1dv1∫−τ2τ2sin(v2(y2−z2))v2dv2. (22)

Furthermore, the following result holds for every τ>0:

(23)∫−ττe−iv(z−y)−eiv(z−y)ivdv=∫−ττ−2sin(v(z−y))vdv=−2sgn(z−y)∫−ττ−sin(v|z−y|)vdv. (23)

And limτ→∞∫ττsin(v|z−y|)vdv=π.

Pooling together the relation (23) with the bounded convergence theorem and using the fact that lim(y1,y2)→(∞,∞)Ga,b1,b2(y1, y2; x, T)=ψ(a, x, 0, T), 𝕀 as defined in (19), when letting τ1, τ2, →∞, this brings about

(24)lim(τ1,τ2)→(∞,∞)I4π2=∫ℝ2sgn(z1−y1)sgn(z2−y2)Ga,b1,b2(dz1, dz2; x, T)=  ∫y1+∞∫y2+∞Ga,b1,b2(dz1;dz2)  ︸I−  ∫y1+∞∫−∞y2Ga,b1,b2(dz1; dz2)  ︸II−  ∫y2+∞∫−∞y1Ga,b1,b2(dz1;dz2)  ︸III+  ∫−∞y1∫−∞y2Ga,b1,b2(dz1; dz2)  ︸IV. (24)

We determine each of the expressions above in order to compute 𝕀. Firstly, we proceed by computing the expression I

I=∫y1+∞∫y2+∞Ga,b1,b2(dz1; dz2)=∫y1∞[Ga,b1,b2(dz1; +∞)−Ga,b1,b2(dz1; y2)]=∫y1∞Ga,b1,b2(dz1; +∞)−∫y1∞Ga,b1,b2(dz1; y2)=Ga,b1,b2(+∞; +∞)−Ga,b1,b2(y1; +∞)−Ga,b1,b2(+∞; y2)+Ga,b1,b2(y1; y2).

We know that, whenever one of y1 or y2 is +∞, we recover Duffie, Pan, and Singleton (2000) one-condition framework; then, from their Proposition 2, we have

Ga,b1,b2(y1; +∞)=ψ(a, x, t, T)2+14π∫−∞+∞eivy1ψχ(a−ivb1, x, t, T)−e−ivy1ψ(a+ivb1, x, t, T)ivdvGa,b1,b2(+∞; y2)=ψ(a, x, t, T)2+14π∫−∞+∞eivy2ψχ(a−ivb2, x, t, T)−e−ivy2ψ(a+ivb2, x, t, T)ivdv.

Plugging these relations into I yields

(25)I=Ga,b1,b2(y1; y2)−14π∫−∞+∞eivy1ψ(a−ivb1, x, t, T)−eivy1ψ(a+ivb1, x, t, T)ivdv−14π∫−∞+∞eivy2ψ(a−ivb2, x, t, T)−eivy2ψ(a+ivb2, x, t, T)ivdv. (25)

The second expression, II, is derived as follows

II=∫y1+∞∫−∞y2∂2Ga,b1,b2(z1; z2)=∫y1+∞dGa,b1,b2(z1; y2−)=Ga,b1,b2(+∞; y2−)−Ga,b1,b2(y1; y2−),

where Ga,b1,b2(−y1, y2, x, T)=limz1→y1,z1<y1;z2→y2,z2>y2Ga,b1,b2(z1, z2, x, T) and Ga,b1,b2(−y1, −y2, x, T)=limz1→y1,z1<y1;z2→y2,z2<y2Ga,b1,b2(z1, z2, x, T).

From Proposition 2 of Duffie, Pan, and Singleton (2000) we also have

Ga,b1,b2(+∞;y2−)=ψ(a, x, t, T)2+14π∫−∞+∞eivy2ψ(a−ivb2, x, t, T)−e−ivy2ψ(a+ivb2, x, t, T)ivdv and

Ga,b1,b2(y1; y2−)=Ga,b1,b2(y1; y2). The expression II then becomes

(26)II=ψ(a, x, t, T)2+14π∫−∞+∞eivy2ψ(a−ivb2, x, t, T)−e−ivy2ψ(a+ivb2, x, t, T)ivdv−Ga,b1,b2(y1; y2). (26)

Likewise, III is computed as

(27)III=∫y2+∞∫−∞y1Ga,b1,b2(dz1;dz2)=∫y2+∞dGa,b1,b2(y1−; z2)=Ga,b1,b2(y1−; +∞)−Ga,b1,b2(y1−; y2)=ψ(a, x, t, T)2+14π∫−∞+∞eivy1ψ(a−ivb1, x, t, T)−e−ivy1ψ(a+ivb1, x, t, T)ivdv−Ga,b1,b2(y1; y2). (27)

Finally, the last term, IV, is expressed as

(28)IV=∫−∞y1∫−∞y2Ga,b1,b2(dz1;dz2)=∫−∞y1dGa,b1,b2(z1;y2−)=Ga,b1,b2(y1−;y2−)=Ga,b1,b2(y1; y2). (28)

In conclusion, we can express the following limit:

(29)14π2lim(τ1,τ2)→(∞,∞)I=4Ga,b1,b2(y1; y2)−ψ(a, x, t, T)−12π∫−∞+∞eivy2ψ(a−ivb2, x, t, T)−eivy2ψ(a+ivb2, x, t, T)ivdv−12π∫−∞+∞eivy1ψ(a−ivb1, x, t, T)−e−ivy1ψ(a+ivb1, x, t, T)ivdv. (29)

B Proof of Lemma 1

The proof of Lemma 1 relies on the following combinatory identity:

(30)∏j∈A(rj+sj)=∑B⊆A(∏j∈Brj)(∏j∈A\Bsj). (30)

Substituting rj→−eivj(yj−zj)2i, sj→e−ivj(yj−zj)2i and using the fact that sin(vj(yj−zj))=eivj(yj−zj)−e−ivj(yj−zj)2i, we have

−2sin(vj(yj−zj))vj=e−ivj(yj−zj)−eivj(yj−zj)ivj and (30) becomes

(31)(−2)|A|∏j∈Asin(vj(yj−zj))vj=∑B⊆A(−1)|B|eiv(B)(y(B)−z(B))−iv(A\B)(y(A\B)−z(A\B))∏j∈A(ivj). (31)

We therefore show equation (11) of Lemma 1 by combinatory identity.

Substituting QA by its expression (11) in 𝕀A leads to

IA=∫ℝ|A|∫ℝ|A|(−2)|A|∏j=1|A|sin(vj(yj−zj))vj∏j∈AdvjGa,b(dz(A), x, T)=(−2)|A|∫ℝ|A|∏j∈A[∫sin(vj(yj−zj))vjdvj]Ga,b(dz(A), x, T)=(−2π)|A|∫ℝ|A|∏j∈Asgn(zj−yj)Ga,b(dz(A), x, T).

C Proof of Lemma 2

Henceforth, to simplify our notations, we drop the multiple occurrences of Xt, T, χ from the expression of Ga,b(z, Xt, T). By definition, we have

IA=∫ℝ|A|∑k=0|A|(−1)k∑Ak⊆Aeiv(Ak)y(Ak)−iv(A\Ak)y(A\Ak)ψ(a−iv(Ak)b(Ak)+iv(A\Ak)b(A\Ak))∏j∈A(ivj)∏j∈Advj.

Using the fact that

ψ(a−iv(Ak)b(Ak)+iv(A\Ak)b(A\Ak))=∫ℝ|A|e−iv(Ak)z(Ak)+iv(A\Ak)z(A\Ak)Ga,b(dz(A), Xt, T),

we have

IA=∫ℝ|A|∑k=0|A|(−1)k∑Ak⊆Aeiv(Ak)y(Ak)−iv(A\Ak)y(A\Ak)∫ℝ|A|e−iv(Ak)z(Ak)+iv(A\Ak)z(A\Ak)Ga,b(dz(A))∏j∈A(ivj)∏j∈Advj=∫ℝ|A|∫ℝ|A|∑k=0|A|(−1)k∑Ak⊆Aeiv(Ak)(y(Ak)−z(Ak))−iv(A\Ak)(y(A\Ak)−z(A\Ak))∏j∈A(ivj)∏j∈AdvjGa,b(dz(A))=∫ℝ|A|∫ℝ|A|QA∏j∈AdvjGa,b(dz(A), Xt, T),

where

QA≡∑k=0|A|(−1)k∑Ak⊆Aeiv(Ak)(y(Ak)−z(Ak))−iv(A\Ak)(y(A\Ak)−z(A\Ak))∏j∈A(ivj).

Replacing QA by its expression in 𝕀A leads to

IA=∫ℝ|A|∫ℝ|A|(−2)|A|∏j=1|A|sin(vj(yj−zj))vj∏j∈AdvjGa,b(dz(A), Xt, T)=(−2)|A|∫ℝ|A|∏j∈A[∫sin(vj(yj−zj))vjdvj]Ga,b(dz(A), Xt, T)=(−2π)|A|∫ℝ|A|∏j∈Asgn(zj−yj)Ga,b(dz(A), Xt, T).

We are now ready to demonstrate the lemma. The goal is to show that

IA=(−2π)|A|∑k=0|A|(−2)k∑Ak⊆AGa,b(y(Ak)).

In other words, we endeavor to prove that

(32)JA≡∫ℝ|A|∏j∈Asgn(zj−yj)Ga,b(dz(A))=∑k=0|A|(−2)k∑Ak⊆AGa,b(y(Ak)). (32)

To do so, we proceed by induction. Suppose that the relation holds for each subset of E whose member’s cardinality is less than or equal to |A|. Let us show that the relation is also fulfilled for the subset with |A|+1 elements. Denoting A+1≡A∪{y∣A∣+1}, we have

JA+1=∫ℝ|A|+1∏j∈A+1sgn(zj−yj)Ga,b(dz(A+1))=∫ℝsgn(z|A|+1−y|A|+1)d[∫ℝ|A|∏j∈Asgn(zj−yj)Ga,b(dz(A); z|A|+1)].

By induction, we can apply (32) to ∫ℝ|A|∏j∈Asgn(zj−yj)Ga,b(dz(A); z|A|+1). This implies that

(33)∫ℝ|A|∏j∈Asgn(zj−yj)Ga,b(dz(A); z|A|+1)=∑k=0|A|(−2)k∑Ak⊆AGa,b(y(Ak); z|A|+1), (33)

thus

JA+1=∫ℝsgn(z|A|+1−y|A|+1)[∑k=0|A|(−2)k∑Ak⊆AGa,b(y(Ak); dz|A|+1)],

hence

JA+1=∑k=0|A|(−2)k∑Ak⊆A∫ℝsgn(z|A|+1−y|A|+1)Ga,b(y(Ak); dz|A|+1)=∑k=0|A|(−2)k∑Ak⊆A[−∫−∞y|A|+1Ga,b(y(Ak); dz|A|+1)+∫y|A|+1∞Ga,b(y(Ak); dz|A|+1)]=∑k=0|A|(−2)k∑Ak⊆A[−2Ga,b(y(Ak); y|A|+1)+Ga,b(y(Ak); −∞)+Ga,b(y(Ak); ∞)].

By the definition of Ga,b(·), we have

Ga,b(y(Ak); −∞)=0,   Ga,b(y(Ak); ∞)=Ga,b(y(Ak)),

and therefore

JA+1=∑k=0|A|(−2)k∑Ak⊆A[−2Ga,b(y(Ak); y|A|+1)+Ga,b(y(Ak))]=∑k=0|A|(−2)k+1∑Ak⊆AGa,b(y(Ak); y|A|+1)+JA=(−2)|A|+1Ga,b(y(A+1))+∑k=0|A|−1(−2)k+1∑Ak⊆AGa,b(y(Ak); y|A|+1)+JA.

Replacing ∑Ak⊆AGa,b(y(Ak); y|A|+1) by

∑Ak⊆AGa,b(y(Ak);y|A|+1)=[∑Ak⊆AGa,b(y(Ak); y|A|+1)+∑Ak+1⊆AGa,b(y(Ak+1))−∑Ak+1⊆AGa,b(y(Ak+1))]

implies that

JA+1=(−2)|A|+1Ga,b(y(A+1))+∑k=0|A|−1(−2)k+1[∑Ak⊆AGa,b(y(Ak); y|A|+1)+∑Ak+1⊆AGa,b(y(Ak+1))−∑Ak+1⊆AGa,b(y(Ak+1))]+JA.

Hence

JA+1=(−2)|A|+1Ga,b(y(A+1))+∑k=0|A|−1(−2)k+1[∑Ak⊆AGa,b(y(Ak); y|A|+1)+∑Ak⊆A+1Ga,b(y(Ak))]−∑k=0|A|−1(−2)k+1[∑Ak+1⊆AGa,b(y(Ak+1))]+JA.

By noting that

∑Ak⊆AGa,b(y(Ak); y|A|+1)+∑Ak⊆A+1Ga,b(y(Ak))=∑Aj⊆A+1Ga,b(y(Aj)),

we can rewrite 𝕁A+1 as the following

JA+1=(−2)|A|+1Ga,b(y(A+1))+∑k=0|A|−1(−2)k+1∑Ak+1⊆A+1Ga,b(y(Ak+1))−∑k=0|A|−1(−2)k+1[∑Ak+1⊆AGa,b(y(Ak+1))]+JA.

Rearranging the summation leads to

JA+1=(−2)|A|+1Ga,b(y(A+1))+∑j=1|A|(−2)j∑Aj⊆A+1Ga,b(y(Aj))−∑k=0|A|−1(−2)k+1[∑Ak+1⊆AGa,b(y(Ak+1))]+JA,

which implies

JA+1=∑j=1|A|+1(−2)j∑Aj⊆A+1Ga,b(y(Aj))+JA−∑j=1|A|(−2)j[∑Aj⊆AGa,b(y(Aj))].

Using the induction assumption up to 𝕁A, we have

JA−∑j=1|A|(−2)j[∑Aj⊆AGa,b(y(Aj))]=Ga,b(y(∅))=ψ(a, x, t, T).

Thus

JA+1=∑j=1|A|+1(−2)j∑Aj⊆A+1Ga,b(y(Aj))+ψ(a, x, t, T).

In other words

JA+1=∑j=0|A|+1(−2)j∑Aj⊆A+1Ga,b(y(Aj)).

This ends the proof of Lemma 2.

D Proof of Proposition 3

The proof of Proposition 3 relies on Lemma 2 and an application of the Möbius inversion for the Boolean algebra of subsets of a finite set [see Hazewinkel (2002) and Rota (1964)].

Define A⊆E,

(34)G(A)=2|A|Ga,b(y(A)),  F(A)=(−2π)−|A|IA, (34)

where 𝕀A, A⊆E is given by (10).

Lemma 2 implies that

(35)F(A)=∑B⊆A(−1)|A|−|B|G(B). (35)

The Möbius inversion says that (35) is equivalent to

(36)G(A)=∑B⊆AF(B)  ∀B⊆A, (36)

and the proof of Proposition 3 is completed with A=E in (36).

E Some ARV model features

Below we provide some ARV model properties.

E.1 The density

The joint conditional density of returns (Rt) and realized variance-covariance matrix (RVt) is

ft−1(Rt, RVt)=ft−1(Rt|RVt)ft−1(RVt)

ft−1(Rt|RVt)=(2π) − n2|RVt| − 12exp(−12z′1,tz1,t)

where

z1,t=RVt−1/2[Rt−μt−1−(RVt−Σt−1)δ],

and

ft−1(RVt)=2−pn/2|σVσ′|−p/2Γn(p2)−1|Wt|p−n−12exp(−Tr((σVσ′)−1Wt)2),

where

Wt=σ−1(RVt−Σt−1)(σ−1)′+pV.

E.2 Model characteristics

It follows from (14) that the conditional variance can be expressed as

Σt=ω+βΣt−1β′+αRVtα′=ω+βΣt−1β′+α{Σt−1+σ[Wt−pV]σ′}α′=ω+βΣt−1β′+αΣt−1α′+ασWtσ′α′−pασVσ′α′=ω−pασVσ′α′+βΣt−1β′+αΣt−1α′+ασWtσ′α′.

Given that

Σt=ω−pασVσ′α′+βΣt−1β′+αΣt−1α′+ασWtσ′α′,

the model is well defined (in the sense that the support of distribution of Σt is the symmetric positive definite real matrix) whenever the following condition is fulfilled:

ω−pασVσ′α′≥0,

and we can express ω as

ω=pασVσ′α′+γγ′.

Further, given that

Et−1[Σt]=ω+βΣt−1β′+αΣt−1α′,

the variance matrix is covariance-stationary if all the eigenvalues of ββ′+αα′ are <1.

E.3 Proof of Proposition 4, the conditional expectation of log returns

Et−1[exp(e′iRt)]=exp(rf+λie′i Σt−1ei)

Et−1[exp(e′iRt)]=Et−1[exp(e′iμt−1+e′i(RVt−Σt−1)δ+e′iRVt1/2z1,t)]=Et−1[exp(e′iμt−1+e′iσ[Wt−pV]σ′δ+e′iRVt1/2z1,t)]=Et−1[exp(e′iμt−1+e′iσ[Wt−pV]σ′δ+12e′iRVtei)]=Et−1[exp(e′iμt−1+e′iσ[Wt−pV]σ′δ+12e′i(Σt−1+σ[Wt−pV]σ′)ei)]=Et−1[exp(e′iμt−1+e′iσ[Wt−pV]σ′δ+12(e′iΣt−1ei+e′iσ[Wt−pV]σ′ei))]

=Et−1[exp(e′iμt−1+e′iσWtσ′δ−pe′iσVσ′δ+12e′i Σt−1ei+12e′iσWtσ′ei−p2e′iσVσ′ei)]=Et−1[exp(e′iμt−1+12e′i Σt−1ei−pe′iσVσ′δ−p2e′iσVσ′ei+e′iσWtσ′δ+12e′iσWtσ′ei)]=Et−1[exp(e′iμt−1+12e′i Σt−1ei−pe′iσVσ′δ−p2e′iσVσ′ei+Tr((σ′δe′iσ+12σ′eie′iσ)Wt))]=Et−1[exp(e′iμt−1+12e′i Σt−1ei−pe′iσVσ′δ−p2e′iσVσ′ei−p2ln[det(In−2(σ′δe′iσ+12σ′eie′iσ)V)])].

Hence

e′iμt−1=rf+(λi−12)e′i Σt−1ei+pe′iσVσ′δ+p2e′iσVσ′ei+p2ln[det(In−2(σ′δe′iσ+12σ′eie′iσ)V)].

E.4 Proof of the moment-generating functions

E.4.1 Proof of Proposition 5

Et−1[exp(u′Rt+Tr(θΣt))]=exp(A(u, θ)+Tr(B(u, θ)Σt−1))

with

A(u,θ)=∑i=1nui(rf+pe′iσVσ′δ+p2e′iσVσ′ei+p2ln[det(In−2(σ′δe′iσ+12σ′eie′iσ)V)])−p2ln[det(In−2(12σ′uu′σ+σ′δu′σ+σ′α′θασ)V)]−pu′σVσ′δ−p2u′σVσ′u+Tr(θ(ω−pασVσ′α′))

B(u, θ)=β′θβ+α′θα+12uu′+∑i=1neiui(λi−12)e′i.

In the following, we provide more details on the proof of that result:

Et−1[exp(u′Rt+Tr(θΣt))]   =Et−1[exp(u′(μt−1+(RVt−Σt−1)δ+RVt1/2z1,t)+Tr[θ(ω−pασVσ′α′+βΣt−1β′+αΣt−1α′+ασWtσ′α′)])]   =Et−1[exp(u′(μt−1+σWtσ′δ−pσVσ′δ)+12u′RVtu+Tr[θ(ω−pασVσ′α′+βΣt−1β′+αΣt−1α′+ασWtσ′α′)])]   =Et−1[exp(u′(μt−1+σWtσ′δ−pσVσ′δ)+12u′Σt−1u+12u′σWtσ′u−p2u′σVσ′u+Tr[θ(ω−pασVσ′α′+βΣt−1β′+αΣt−1α′+ασWtσ′α′)])]   =Et−1[exp(u′μt−1−pu′σVσ′δ−p2u′σVσ′u+Tr(θ(ω−pασVσ′α′))+Tr((β′θβ+α′θα+12uu′)Σt−1)+Tr[(12σ′uu′σ+σ′δu′σ+σ′α′θασ)Wt])]

=exp(u′μt−1−pu′σVσ′δ−p2u′σVσ′u+Tr(θ(ω−pασVσ′α′))+Tr((β′θβ+α′θα+12uu′)Σt−1)−p2ln[det(In−2(12σ′uu′σ+σ′δu′σ+σ′α′θασ)V)])=exp(∑i=1nuie′iμt−1−pu′σVσ′δ−p2u′σVσ′u+Tr(θ(ω−pασVσ′α′))+Tr((β′θβ+α′θα+12uu′)Σt−1)−p2ln[det(In−2(12σ′uu′σ+σ′δu′σ+σ′α′θασ)V)])=exp(∑i=1nui(rf+(λi−12)e′i Σt−1ei+pe′iσVσ′δ+p2e′iσVσ′ei+p2ln[det(In−2(σ′δe′iσ+12σ′eie′iσ)V)])−pu′σVσ′δ−p2u′σVσ′u+Tr(θ(ω−pασVσ′α′))+Tr((β′θβ+α′θα+12uu′)Σt−1)−p2ln[det(In−2(12σ′uu′σ+σ′δu′σ+σ′α′θασ)V)] )=exp(∑i=1nui(rf+pe′iσVσ′δ+p2e′iσVσ′ei+p2ln[det(In−2(σ′δe′iσ+12σ′eie′iσ)V)])−p2ln[det(In−2(12σ′uu′σ+σ′δu′σ+σ′α′θασ)V)]−pu′σVσ′δ−p2u′σVσ′u+Tr(θ(ω−pασVσ′α′))+Tr[(β′θβ+α′θα+12uu′+∑i=1neiui(λi−12)e′i)Σt−1]).

E.4.2 Proof of Proposition 6

Et[exp(u′∑i=1τ+1Rt+i)]=Et[exp(u′Rt+1)Et+1[exp(u′∑i=2τ+1Rt+i)]]=Et[exp(u′Rt+1)Et+1[exp(u′∑j=1τRt+1+i−1)]]=Et[exp(u′Rt+1+C(u; τ)+Tr(D(u; τ)Σt+1))]=exp(C(u; τ))Et[exp(u′Rt+1+Tr(D(u; τ)Σt+1))]=exp(C(u; τ)+A(u,D(u; τ))+Tr(B(u,D(u; τ))Σt)).

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The online version of this article (DOI: 10.1515/snde-2014-0034) offers supplementary material, available to authorized users.

Published Online: 2015-3-7
Published in Print: 2015-12-1

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