Appendx for Solvng Asset Prcng Models when the Prce-Dvdend Functon s Analytc Ovdu L. Caln Yu Chen Thomas F. Cosmano and Alex A. Hmonas January 3, 5 Ths appendx provdes proofs of some results stated n our paper. Frst, we prove the exstence and unqueness of the prce-dvdend functon Proposton to the ntegral equaton 6 n the vector space S Defnton. Second, we calculate the system of lnear equatons for the coeffcents of the power seres for the prce-dvdend functon. Fnally, we use Cauchy s ntegral formula to bound all the dervatves of the prce-expected dvdend functon 3 so that we can calculate the error terms 4, 5 and 6. Journal of Economc Lterature Classfcaton Numbers: G, G3, C63, D5 Key Words: Analytcty, Asset prcng. Department of Mathematcs, Eastern Mchgan Unversty. Department of Mathematcs, Idaho State Unversty, Pocatello, ID 839 Department of Fnance, Mendoza College of Busness, Unversty of Notre Dame, E-mal: thomas.f.cosmano.@nd.edu, Phone: 574 63-578, Fax: 574 63-555. Department of Mathematcs, Unversty of Notre Dame, Notre Dame, IN 46556.
APPENDIX We use the followng notaton n the proofs: Cn = n = φ = φ n n =,,,..., Φx = π x e s K 3 = K e x K + σ γ φ α+φ α, K 4 = K 3 [ + Φ φk σ for ds for all x R, θ = x + σ + φ α γ, ], K 5 = K 4 e φ K σ + φk θ and K 6 = π K e σ γ + σ [ γ+k ]. Q s defned n 8 whch satsfes the ntegral equaton 9. A seres of lemmas are now proved whch are used n the proof of Proposton. s x a e Lemma : Let a > and ϕx = x+a Φ a σ. σ ds for all x R. Then ϕx Proof: Ths follows from the fact that ϕ has a unque global maxmum at x =. Lemma 3: For any real numbers A and k wth k, we have e σ s A +k s ds e k σ +k A [ + Φkσ]. Proof: Note that σ s A + ks = σ s A kσ + k σ + ka. e σ s A +k s ds = = e k σ e ka e k σ +k A e σ s A ks ds + e = e k σ +k A A+kσ e σ s A +ks ds σ s A+kσ ds + e ka e σ s A+kσ ds + = e k σ +k A + A+kσ e s σ ds + e s σ ds A kσ A kσ e s σ ds By Lemma, we acheve e σ s A +k s ds e k σ +k A [ + Φkσ]. e σ s A kσ ds e σ s A kσ ds
Lemma 4: For any f S, the functon T f defned by T fx = K 3e φk x fse σ [s φx+θ] ds les n the vector space S. Thus, T defnes a lnear transformaton from the vector space S to tself. Proof: By the defnton of S, we can fnd postve constants M and k such that fx Me k x. Usng Lemma 3, we deduce T fx K 3e φk x MK 3e φk x fs e σ [s φx+θ] ds e σ [s φx+θ] +k s ds MK 3 [ + Φkσ]e φk x+ k σ +k φ x + θ MK 3 e k σ +k θ [ + Φkσ]e φ k+ K x. The contnuty of T f can be easly checked by applyng the standard argument n real analyss; for example, see Theorem 56 n Kaplan 956. Now construct a sequence of functons { Q n n =,,,... } by settng Q = and Q n+ x = K + K 3e φk x and set Q = n= [Q n+ Q n ]. Recall that Cn = n = φ = φ n Cn < φ, and c lm n Cn = φ. Lemma 5: Q n+ x Q n x K K4 n e σ K n =,,,.... Q n se σ [s φx+θ] ds satsfes a Cn + = φ Cn +, b n = C + K θ n = C e Cn K x for
Proof: These nequaltes can be easly proven by nducton on n. Lemma 6: The seres Q = n= [Q n+ Q n ] s unformly convergent on any bounded closed nterval. Proof: Fx a bounded closed nterval [a, b]. For any x [a, b], by Lemma 5 we see [Q n+ x Q n x] K n= n= K4 n e σ K n = C + K θ n = C e Cn K a + b. The rato test together wth Weerstrass M-test see Theorem 3 Kaplan 956 mples the unform convergence of the seres Q = n= [Q n+ Q n ] on [a, b], provded lm K 4e σ K Cn + K θ Cn e [Cn+ Cn] K a + b = K 4 e σ n Lemma 7: The functon Q les n the vector space S. K φ + φk θ = K 5 <. Proof: The contnuty of Q follows from Theorem 3 n Kaplan 956 and Lemma 6. By Lemma 5, we also have Qx K n= K n= K4 n e σ K n = C + K θ n = C e Cn K x K4 n e σ K n = C + K θ n = C e φk x. 6. Lemma 8: For any x R, we have lm n [Qs Q ns]e σ [s φx+θ] ds = Proof: Ths s an mmedate consequence of Theorem 3 n Kaplan 956 and Lemma 3
Proof of Proposton : Equvalently, we are gong to show that the equaton 9 has a unque soluton Q n the vector space S. Exstence Applyng the lmt as n approaches to the equaton shows that the functon Q = n= [Q n+ Q n ] = lm n Q n s a soluton to the equaton 9 n the vector space S. Unqueness Suppose that Q S s another soluton to the equaton 9. Then Q and Q satsfy the functonal equaton Q Q = T Q Q. By the defnton of S, we can also fnd postve constants M and k such that Qx Qx Me k x. Usng the results n the proof of Lemma 4, we see Qx Qx T Q Q x MK 3 e k σ +k θ [ + Φkσ]e φ k+c K x for all x R. Applyng ths computaton n tmes gves rse to Qx Qx { n } MK3 n e φ k+c K σ + φ k+c K θ [ + Φσ φ k + σc K ] e φ n k+cn K x. Note that lm K 3e φ n k+cn K σ + φ n k+cn K θ [ + Φσ φ n k + σcn K ] n = K 3 e φ K σ + φk θ [ + Φ ] φk σ = K 5 <. We can fnd a postve nteger N and a postve real number δ < so that for any n N, K 3 e φ n k+cn K σ + φ n k+cn K θ [ + Φσ φ n k + σcn K ] δ. Then for all n N, we have Qx Qx { N } MK3 N e φ k+c K σ + φ k+c K θ [ + Φσ φ k + σc K ] δ n N e φ n k+cn K x. 4
Snce lm n δ n N = and lm n e φ n k+cn K x = e φ K x, we obtan Qx = Qx for all x R. Ths completes the proof of Proposton Solvng for the coeffcents n the analytc Prce-Dvdend Functon The ntegral equaton 6 may be wrtten as P x = K e K x + K e K x σ γ Ix. where Ix = s e γs σ P x + φx + sds. To solve the ntegral equaton we post that P x s analytc n whch case t has the functonal form. Consequently we may wrte P x + φx + s as P x + φx + s = e K x +φx+s x + φx + s x k. Usng the Bnomal theorem 6 we may rewrte ths equaton as P x + φx + s = e K x +φx+s Substtutng ths result nto Ix we obtan Ix = e K x +φx x + φx x k s. x + φx x k γ where γ = e γ+k s s σ s ds. The evaluaton of the ntegral γ entals the use of a change of varable followed by the evaluaton of all the moments of the normal dstrbuton. γ = K 3 a where K 7 = π e σ [ γ+k ] and a j= σ [ γ + K ] j σ j + j /Γ[ j+ ]. We can now wrte Ix as K 7 e K x +φx n k the ntegral equaton becomes e K x P x = K + K 6 e K x +φx 5 x + φx x k a = j x + φx x k a so that x x k.
Next we fnd the undetermned coeffcents b, =,... n. In order to do ths, we collect all the functons of x on the left hand sde of ths equaton and use Taylor s theorem to take an n th order Taylor expanson around x. Ths produces a system of n + lnear equatons n the varables b, =,... n. Defne the followng functons qx = K x + φx, r,k x = x x + φx k, and w,k x = expqxr,k x n l= l! wl,k x x x l. Substtutng for w,k x nto the ntegral equaton gves us K + K 6 l= l! wl,k x x x l a = x x k. Fnally, equate the coeffcents on the left and rght hand sde of ths equaton to yeld b = K + K 6 w,k x a and b l = K 6 l! wl,k x a, where l =,..., n. These Equatons are an n + system of lnear equatons n the b s. Analytc error of the Taylor polynomal approxmaton seres The complex functon Qz = e K z P z s analytc and t s expressble as a Taylor Qz = Q k z z z k for any complex number z, k! where z = x. We may use Cauchy s ntegral formula to estmate the error of the Taylor polynomal approxmaton Qz n Q k z z z k! k near z = z. Wrte C r for the crcle of radus r > centered at z = z n the complex plane. Cauchy s ntegral formula see Corollary 5.9 n Conway 973 gves Q k z = k! π C r Qz dz for k =,,,.... z z k+ Each pont z = x + y on C r satsfes x r x x + r and r y r. Set A = K n= K4 n e σ K n = C + K θ n = C. 6
Usng the result n the proof of Lemma 7, we obtan Qs Ae φk s for all s R, and Qz K + K 3 e φkx+y = K + K 3e φk φ x+ y σ K + AK 3e φk φ x + y σ Qs e [s φx θ φy] σ ds Qse σ [s φx θ] ds e σ [s φx θ] + φk s ds K + AK 3 e φk x + φ K x + φ y σ + φ K σ + φk θ [ + Φ = K + AK 3 e φk x + θ + φ y σ + φ K σ [ + Φ φk σ K + AK 3 e φk r+ x + θ + φ r σ + φ K σ Defne B r = K + AK 3 e φk r+ x + θ + φ r σ + φ K σ ] [ + Φ φk σ [ + Φ φk σ ] φk σ ]. ]. Cauchy s ntegral formula for Q k z together wth the above estmate to Qz gves rse to Q k z k! Qz dz π z z k+ = k! π Qz + re θ π re θ dθ re θ k+ k! π C r π Qz + re θ r k dθ B rk! r k. Ths bound allows us to estmate the analytc error 4. FOOTNOTES 6 The bnomal theorem states that we may wrte x +φx+s x k = k φx x k s. x + 7