DSpace at VNU: Regularization for the Inverse Problem of Finding the Purely Time-Dependent Volatility

Vietnam J Math DOI 10.1007/s10013-015-0164-9 Regularization for the Inverse Problem of Finding the Purely Time-Dependent Volatility Dang Duc Trong1 · Dinh Ngoc Thanh1 · Nguyen Nhu Lan2 Received: 29 January 2014 / Accepted: 10 April 2015 © Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2015 Abstract We consider the inverse problem of finding the volatility σ ∈ Lρ (0, T ) such that t UBS (X, K, r, t, σ (τ )dτ ) = u(t), ≤ t ≤ T , where UBS is the Black–Scholes formula and u(t) is the observable fair price of an European call option The problem is ill-posed Using the residual method, we shall regularize the problem An explicit error estimate is given Keywords Calibration · Volatility · Ill-posed · Regularization Mathematics Subject Classification (2010) 35R30 · 65J20 · 91B24 Introduction In the classical stochastic finance theory, the stock price X is assumed to satisfy an Ito process (see [19, p 13]) d X = μX dt + σ X dW, where μ is the stock drift, σ = σ (t, X ) is the stock volatility, W is a Wiener process Denote by C(X, K, r, t, X ) the fair price of an European call option, where X is the current price of the stock, K is the strike price, r is the continuously compounded interest rate, t is time As Nguyen Nhu Lan nnlancantho@gmail.com Department of Mathematics and Computer Science, Ho Chi Minh City University of Science, Vietnam National University, 227 Nguyen Van Cu, District 5, Ho Chi Minh City, Vietnam Tay Do University, Can Tho, Vietnam D.D Trong et al shown by the Black–Scholes theory [21, p 49], if the stock price X satisfies an Ito process then the fair price C satisfies the (generalized) Black–Scholes equation ∂ 2C ∂C ∂C + X σ (t, X ) = + rX ∂t ∂X ∂X In the latter equation, the quantity σ (t, X ) plays a crucial role in option pricing The volatility is used to measure the uncertainty of the future return of the stocks [20, p 377] If the function is known, we can calculate the option price C(X, K, r, t, X ) The problem is called the direct problem of option pricing However, the volatility σ (t, X ) is a market parameter and not directly observable Hence, it is also called the implied (or implicit) volatility In the primitive Black–Scholes model, the volatilities are assumed to be constants However, in practice, the observation shows that they are not constant over time Therefore, the market is concerned with the problem of identifying of non-constant volatility [20, p 741] We have the problem of calibrating the implied volatility σ (t, X ) from the fair price of the option It is called the inverse problem of option pricing The problem is ill-posed and studied in many papers [2–15, 18, 19, 24] Among these papers, the method of integral equations (Green function) is used in [2–4, 19] The mollification method is used in [12] In other papers [7, 15, 18], the authors use the Tikhonov method (or minimization method) to regularize the problem In the present paper, we deal with the case σ = σ (t) We put a(t) = σ (t) and call it the volatility function (see [14]) In this case, the dynamic of stock prices satisfies a generalized geometric Brownian motion d X = μX dt + σ (t)X dW We also note that similar purely t-dependent models can be seen in many papers For example, the value of an annually callable bond satisfies the Hull–White model dr = (α(t) − β(t))r(t) + σ (t)dW If the stock prices are modeled by the generalized geometric Brownian motion then the solution of the partial differential equation of C can be represented through the Black– Scholes function [17, pp 71–72] XΦ(d1 ) − Ke−rt Φ(d2 ), s > 0, s = 0, (X − Ke−rt )+ , UBS (X, K, r, t, s) = where ν = ln X K and ν + rt + 2s , √ s √ Φ(z) = √ 2π Now, we can give a precise form of the inverse problem d1 = d2 = d1 − s, z −∞ e−x /2 dx Calibration Problem Let u(t) be the known fair price of the market We consider the problem of finding the unknown volatility function a(t) from u(t) satisfying t UBS X, K, r, t, a(τ )dτ = u(t), ≤ t ≤ T If we put for short t N0 (a)(t) = UBS X, K, r, t, a(τ )dτ then N0 is an operator N0 : a → N0 (a) (1) Regularization for the Inverse Problem of Finding the Purely The case of purely t-dependent volatility σ = σ (t) is studied only in a few papers as [10, 14, 15, 23], especially, in [14, 23] Since the problem is ill-posed, these papers studied the regularization of the problem in many common spaces C[0, T ], L2 (0, T ), Lp (0, T ) (p ≥ 1) We will discuss the details later From now on, we shall denote by uex the fair price data, which is a function in the range of the operator N0 Since N0 (a)(t) is increasing with respect to t, the operator N0 is not surjective on the set of positive functions W = C + [0, T ] ⊂ C[0, T ] or W = Lλ,+ (0, T ) = {v ∈ Lλ (0, T ) : v(x) ≥ for x ∈ (0, T )} Let V be a normed space such that N0 (W ) ⊂ V The problem of finding conditions of a datum u ∈ V such that u ∈ N0 (W ) is called the problem of solvability Moreover, in practical situations, the exact datum uex is the result of a measurement Hence, we often not have the exact value of the datum uex We only get a noise uε such that uε − uex W ≤ ε The noise uε is often not in N0 (W ), i.e., the problem N0 (a) = uε possibly has no solution Therefore, the unsolvability often happens for real data Even if uε ∈ N0 (W ), corresponding solutions might have large error (see the next section) Hence, the problem of stability seems to be more important than the one of solvability In the present paper, we only consider the regularization for the problem We recall the definition of a regularization scheme in the present context Since our problem is nonlinear, we choose the classical definition in [22, Ch 2, p 43] In fact, let (U , · U ) and (Y , · Y ) be two normed spaces and A is an operator from U into Y with A−1 : A(U ) → U not continuous Let wex ∈ A(U ) be the exact datum, vex ∈ U be the corresponding exact solution satisfying A(vex ) = wex and let wε ∈ Y be an inexact (but real) datum of wex From wε ∈ Y , we have to find a “stable solution” vε ∈ U (called a regularization solution) for the problem To this end, we consider some problems Let δ1 > and O be a neighborhood of wex (in the topology of Y ) The first problem is the one of convergence analysis (or consistency problem) which is of finding an operator R : O × (0, δ1 ] → U , called a regularization scheme such that there exists a positive function δ = δ(ε), limε→0+ δ(ε) = 0, satisfying R(wε , δ(ε)) − vex U ≤ε for all wε ∈ O, wex − wε Y ≤ δ(ε) As shown in [22, pp 44–45], the latter property is fulfilled if two following properties hold (i) (ii) (continuous property) for each δ ∈ (0, δ1 ], R(·, δ) is continuous on O (inverse property) R(Av, δ) → v as δ → 0+ for every Av ∈ O The second problem is the one of convergence rate which is of establishing error estimate between the regularization solution and the exact solution As known, the problem is not easy Readers can see [16, 23] for some nontrivial estimates We turn to discuss our problem In [14], the authors studied three versions: the C-space case with U = Y = C[0, T ], the Lp -case with U = Lp (0, T ), Y = Lq (0, T ), and the L2 space case with U = Y = L2 (0, T ) As shown in the paper, in all cases, the problems are ill-posed in the Hadamard sense [1, p 17] A regularization for the case U = Y = C[0, T ] was not established in [14] In [23], using an a priori information, the authors constructed a regularization scheme for the continuous case Moreover, they also considered the problem D.D Trong et al ε of convergence rate for the C-space case and obtained the rate O ln optimal in the sense that there exists a C0 > satisfying ˜ R(w, δ(ε)) − vex sup {w: wex −w Y which has the property limε→0+ δ(ε) = In the Lp -case, the authors of [14] considered only convergence analysis of regularization problem Moreover, the regularization is descriptive Hence, to get the consistency result, a priori information in the solution was used The authors studied the outer operator N (S)(t) = UBS (X, K, r, t, S(t)) with S in the set ˜ ) ≤ S(t ˜ ) ≤ κ, t1 , t2 ∈ [0, T ], t1 < t2 Dκ+ := S˜ ∈ L∞ (0, T ) : ≤ S(t For a datum uδ ∈ Lq (0, T ), uδ (t) ≥ (0 ≤ t ≤ T ), the authors found a minimizer S δ ∈ Dκ+ of the extremal problem ˜ − uδ N (S) Lq (0,T ) subject to S˜ ∈ Dκ+ → min, If N (S) = u and uδn → u in Lq (0, T ) then the authors proved that S δn → S in Lp (0, T ) as n → ∞ The problem of convergence rate was not studied in this case Finally, in the L2 -case, N0 is an operator in the Hilbert space L2 (0, T ) In [14], using the Tikhonov method, the authors constructed the regularization solution aαδ as a minimizer of the extremal problem N0 (a) ˜ − uδ L2 (0,T ) + α a˜ − a ∗ L2 (0,T ) → min, subject to a˜ ∈ D(N0 ), where a ∗ ∈ D(N0 ) is a fixed initial guess datum and D(N0 ) := a˜ ∈ L2 (0, T ) : ess inf a˜ ≥ c0 > 0≤t≤T The latter condition is crucial in the L2 -results of [14] It was mentioned there “constraints of the form a(t) ≥ or ≤ a(t) ≤ c < ∞ a.e on [0, T ] are not able to stabilize the solution process sufficiently” Let L > and G : L2 (0, T ) → L2 (0, T ) be the Fr´echet derivative of N0 at a ∈ D(N0 ) such that L a˜ − a 2L2 (0,T ) From the√general theory of the Tikhonov regularization, to obtain the rate convergence of order O( ε), we need [14, Proposition 5.3] the source condition N0 (a) ˜ − N0 (a) − G(a˜ − a) L2 (0,T ) ≤ there exists a function w ∈ L2 (0, T ) satisfying a − a ∗ = G∗ w and the closeness condition w condition implies the condition a − a ∗ ∈ H (0, T ), where m(t) = ∂UBS (X,K,r,t,S(t)) , ∂s L2 (0,T ) < L In the context of our problem, the source a(T ) − a ∗ (T ) = 0, (a − a ∗ ) ∈ L2 (0, T ), m and the closeness condition has the form (a − a ∗ ) m < L2 (0,T ) L Regularization for the Inverse Problem of Finding the Purely Since the function a is unknown, it is not easy to find an initial guess datum a ∗ such that a(T ) − a ∗ (T ) = and that the other conditions hold In the present paper, we shall consider the problem of consistency and the problem of convergence rate for the case U = Lλ (0, T ), Y = Lp (0, T ) (1 ≤ λ ≤ p < ∞) We note that, in this case, Lp (0, T ) (p = 2) is not a Hilbert space, hence the standard Tikhonov method (as in [14, 15]) could not be applied easily to solve the problem Moreover, we also −γ give explicitly error estimates of order O ln 1ε (γ > 0, see Corollary 1) in which we not use the strict conditions as σ (t) ≥ c0 > 0, the source condition and the closeness condition In the present paper, we not obtain an optimal result for rate convergence But, from the optimal result in [23], the logarithmic rate in the present work is reasonable We will study the optimal rate convergence in a future work The remaining part of the paper is divided into five sections In Section 2, we decompose the problem into two problems called the outer and the inner ones, give some properties of Black–Scholes formula and prove the instability of the inverse problems Section gives the existence and the regularization of the outer problem The regularization of whole problem and rate of convergence are presented in Section In Section 5, we give some numerical experiments Finally, we give conclusion of our paper in Section Preliminary Results 2.1 Some Properties of the Black–Scholes Formula The following properties are presented in [23], but for convenience, we recall it here without proofs Since X, K, r are market observable constants, we can denote (for short) k(t, s) = UBS (X, K, r, t, s) The function UBS is defined in the domain X > 0, K > 0, r ≥ 0, t ≥ 0, s > By direct calculation, one has (see its proof, e.g., in [14]) (ν + rt)2 X ν + rt s ∂UBS exp − = √ − − ∂s 2s 2π s >0 and lim UBS (X, K, r, t, s) = X, s→+∞ lim UBS (X, K, r, t, s) = (X − Ke−rt )+ s→0+ From the latter limit, we shall put k(t, 0) = UBS (X, K, r, t, 0) := (X −Ke−rt )+ We easily get (see [23] for proofs) Lemma For t, s1 , s2 , s ≥ 0, k(t, s1 ) < k(t, s2 ) if and only if ≤ s1 < s2 Hence, k(t, s1 ) = k(t, s2 ) if and only if s1 = s2 b) (X − Ke−rt )+ ≤ k(t, s) < X c) If an α satisfies (X − Ke−rt )+ ≤ α < X then there exists uniquely a ξ(t) ≥ such that a) k(t, ξ(t)) = α D.D Trong et al Now, we consider the problem of finding a function S (Lebesgue) measurable on [0, T ] such that k(t, S ) = ψ The solution will be denoted by S = S(ψ) From Lemma 1, we have Proposition Put D = {v ∈ L∞ (0, T ) : (X − Ke−rt )+ ≤ v(t) < X a.e.} Let ψ : [0, T ] → R be a measurable function The problem k(t, S ) = ψ has a unique measurable solution S = S(ψ)(t) if and only if ψ ∈ D 2.2 The Ill-posedness of the Calibration Problem We shall consider the ill-posedness of N0 as an operator from Lλ (0, T ) to Lp (0, T ) Let t a0 ∈ Lλ (0, T ), u0 ∈ Lp (0, T ) satisfy N0 (a0 ) = u0 Then k t, a0 (τ )dτ = u0 (t) We put un (t) = k t, t an (τ )dτ where an (t) = a0 (t) + hn (t) and hn (t) = 0, ≤ t ≤ T − n1 , n3 , T − n1 ≤ t ≤ T For each t ∈ [0, T ), one has limn→∞ un (t) = u0 (t) From Lemma 1, one has ≤ un (t) < X Using the Lebesgue dominated convergence theorem, one has lim n→∞ un − u0 p = 0, and lim n→∞ an − a0 λ = hn λ = n2/λ = ∞ This follows that the calibration problem is unstable at u0 Hence, it is ill-posed 2.3 The Outer and the Inner Problems As mentioned in Introduction, the functions a(t), u(t), which satisfy the equation N0 (a) = u, are called the exact solution and the exact data, respectively Hence, we also denote a = aex , u = uex As shown in [14, 23], the problem can be divided into two problems The outer problem is From a function u ∈ Y , we find a function S(u) : [0, T ] → [0, ∞) such that k(t, S(u)(t)) = u(t) and the inner problem is From the solution of the outer problem S(u)(t), we find the (positive) volatility function t a(t) such that a(τ )dτ = S(u)(t) The Outer Problem In the section, we shall construct a regularization scheme for the outer problem The section is divided into four subsections In Section 3.1, we shall give a definition of the approximation R(w, δ) and compare with other regularization methods in [14, 23] In the two Sections 3.2, 3.3, we shall prove the consistency of our approximation In fact, in Section 3.2, we shall prove that R(w, δ) satisfies the property (i), i.e., continuous with respect to w, in the definition of regularization scheme in Introduction The inverse property (ii) is investigated in Section 3.3 Finally, in Section 3.4, we shall consider the problem of convergence rate Regularization for the Inverse Problem of Finding the Purely 3.1 Definition of Regularization Scheme for the Outer Problem Let v ∈ Lλ,+ (0, T ) = {v ∈ Lλ (0, T ) : v ≥ 0} We put J (v)(t) = k(t, v(t)) Let D be as in Proposition As shown in Lemma 1, J : Lλ,+ (0, T ) → D ⊂ Lp (0, T ) is injective As noted in Section 1, for δ > 0, we shall construct R(·, δ) : D → Lλ,+ (0, T ) such that limδ→0+ R(J (v), δ) = v In the section, we shall use the method of residuals [1, p 31] to regularize the outer problem Hence, our regularization for (CP) can be called a modified method of residuals Precisely, let δ > and let w ∈ D Put Bw,δ = {v˜ ∈ Lλ,+ (0, T ) ∩ L∞ (0, T ) : w − J (v) ˜ ∞ ≤ δ} From w, we construct a (minimizer) function R(w, δ) := ϕ such that ϕ ∈ Bw,δ and ϕ(t) ≤ v(t) ˜ for every t ∈ [0, T ], v˜ ∈ Bw,δ (2) The definition defines uniquely the function ϕ We shall prove the existence of the minimizer Before going to details of the construction, we have some comments In [14], the authors gave a result of convergence analysis for the outer problem in Lp -spaces In fact, with the datum w, they used the approximation R(w) as minimizer of the problem J (v) − w Lq (0,T ) −→ subject to κ = {v ∈ L∞ (0, T ) : ≤ v(t1 ) ≤ v(t2 ) ≤ κ for ≤ t1 ≤ t2 ≤ T } v ∈ D+ The algorithm depends on the constant κ which is an available a priori information As mentioned, the regularization is descriptive Similarly, in [23], we used the formula R(w, δ) = min{κ, max{S(w), δ}} From the mathematical point of view, it might be better if we have not to use the a priori constant κ This is the reason of our algorithm Now, we give the closed-form of R(w, δ) Put wδ+ (t) = max{(X − Ke−rt )+ , w(t) − δ} The definition of wδ+ implies wδ+ ∈ D From Proposition 1, we can get the function S(wδ+ ) such that k(t, S(wδ+ )) = wδ+ We shall prove that R(w, δ) = S(wδ+ ) in the following result Lemma Let δ > and let w ∈ D Then i) S(wδ+ ) ∈ Bw,δ , S(wδ+ )(t) ≤ v˜ for every t ∈ [0, T ] and v˜ ∈ Bw,δ Hence the function S(wδ+ ) satisfies (2); ii) Therefore, S(wδ+1 )(t) ≥ S(wδ+ )(t) for all t ∈ (0, T ) and < δ1 < δ Proof We first prove S wδ+ ∈ Bw,δ For every t ∈ [0, T ], we have ≤ w(t) − wδ+ (t) ≤ δ and J (S(wδ+ ))(t) = k(t, S(wδ+ )(t)) = wδ+ (t) + So, w − J (S(wδ )) ∞ ≤ δ It follows that S(wδ+ ) ∈ Bw,δ Now, we prove S(wδ+ )(t) ≤ v˜ for every v˜ ∈ Bw,δ We have k(t, v(t)) ˜ ≥ w(t) − δ for all t ∈ (0, T ) Lemma implies k(t, v(t)) ˜ ≥ (X − Ke−rt )+ Using both inequalities gives k(t, v(t)) ˜ ≥ wδ+ (t) = k(t, S(wδ+ )(t)) ˜ for all for every t ∈ (0, T ) By the monotonicity of k(t, ·), we get ≤ S(wδ+ )(t) ≤ v(t) ≤ t ≤ T This implies that S(wδ+ ) is a solution of (2) Finally, we verify Part ii) In fact, we have w − J (S(wδ+1 )) ∞ ≤ δ1 D.D Trong et al But < δ1 ≤ δ, hence S(wδ+1 ) ∈ Bw,δ From Part (i) of the lemma, we have S(wδ+ )(t) ≤ S(wδ+1 )(t) which gives the last desired inequality This completes the proof of Lemma 3.2 The Continuity of R(w, δ) For a fixed δ, to prove the continuity of R(w, δ) with respect to w ∈ D, we state and prove some lemmas Lemma Let √ 2(X + K) s ≥ m := max 2|ν|, 2|ν + rT |, + max ,1 √ 2π Then there exists a positive number < δ0 < such that if the inequality k(t, s) ≤ X − δ holds for a δ ∈ (0, δ0 ), then ln δ m ≤ s ≤ Kδ ≡ + Proof We can rewrite the inequality k(t, s) ≤ X − δ as X √ 2π ∞ e−t /2 d1 ∞ K dt + √ 2π e−t −d2 /2 dt ≥ δ Put β0 = max{|ν|, |ν + rT |} We have for s ≥ m2 ≥ 4β02 √ √ β0 s s −√ ≥ − 2 s This implies √ √ √ 1 s s − , −d2 = −d1 + s ≥ − d1 ≥ 2 2 So, we have X+K ∞ e−t /2 dt ≥ δ √ √ s 2π −2 Putting z = √ s − 12 , we have ∞ /2 dt = z It follows that or For ∞ e−z /2 − z e−t z X+K X + K e−z /2 ≥ √ √ z 2π 2π ∞ √ √ e−t /2 e−z /2 dt ≤ z t2 2(X+K) √ ,1 2π e−t /8 ln dt ≥ δ ≥ δ √ s−1)2 /8 , we shall get e−( s ≤ 1+ /2 z 2(X + K)e−( s−1) √ √ 2π ( s − 1) s ≥ m ≥ + max δ ≥ δ This implies By choosing δ0 > small enough, we shall get the desired inequality for all < δ ≤ δ0 This completes the proof of Lemma Regularization for the Inverse Problem of Finding the Purely Lemma Let α, L satisfy < α < L For every s1 , s2 ∈ [0, L], one has |s1 − s2 | ≤ α + M(α, L, t)|k(t, s1 ) − k(t, s2 )|, where √ (ν + rt) s 2π s (ν + rt)2 exp + + α≤s≤L X 2s M(α, L, t) = max and ν = ln X K Proof We consider three cases Case ≤ s1 , s2 ≤ α |s1 − s2 | ≤ α ≤ α + M(α, L, t)|k(t, s1 ) − k(t, s2 )| Case α ≤ s1 , s2 ≤ L We have k(t, s1 ) − k(t, s2 ) = ∂k ∂s (t, c)(s1 − s2 ), where α ≤ min{s1 , s2 } ≤ c ≤ max{s1 , s2 } ≤ L Hence, ≤ −1 ∂k ∂s (c) ≤ M(α, L, t) This gives |s1 − s2 | ≤ α + M(α, L, t)|k(t, s1 ) − k(t, s2 )| Case ≤ s1 ≤ α ≤ s2 ≤ L One has |s1 − s2 | = α − s1 + s2 − α ≤ α + M(α, L, t)|k(t, s2 ) − k(t, α)| But, from Lemma 1, one has k(t, s1 ) ≤ k(t, α) ≤ k(t, s2 ) This implies that |k(t, α) − k(t, s2 )| ≤ |k(t, s1 ) − k(t, s2 )| Hence, |s1 − s2 | ≤ α + M(α, L, t)|k(t, s1 ) − k(t, s2 )| This completes the proof of Lemma Proposition Let α be as in Lemma and let δ > For two functions u, w ∈ D one has |S(u+ δ )(t)| ≤ Kδ = + ln 1δ , + |S(u+ δ )(t) − S(wδ )(t)| ≤ α + M(α, Kδ , t)|u(t) − w(t)| and + S(u+ δ ) − S(wδ ) λ ≤ αT 1/λ + max M(α, Kδ , t)T 0≤t≤T p−λ p u−w p As a consequence, R(·, δ) : D → Lλ (0, T ) is continuous Here, we recall that D ⊂ Lp (0, T ) is defined in Proposition Proof For u, v ∈ Lp (0, T ) and t ∈ (0, T ) fixed, we give the estimate of + |R(u, δ)(t) − R(v, δ)(t)| = |S(u+ δ )(t) − S(vδ )(t)| D.D Trong et al + + + One has k(t, S(u+ δ )(t)) − k(t, S(vδ )(t)) = uδ (t) − vδ (t) For δ small enough, we obtain + + ≤ uδ (t), vδ (t) ≤ X − δ From Lemma 3, we shall get ≤ S(u+ δ )(t), S(vδ+ )(t) ≤ Kδ = + ln δ Using Lemma 3, one has + + + |S(u+ δ )(t) − S(vδ )(t)| ≤ α + M(α, Kδ , t)|uδ (t) − vδ (t)| + On the other hand, |u+ δ (t) − vδ (t)| ≤ |u(t) − v(t)| Hence, + |S(u+ δ )(t) − S(vδ )(t)| ≤ α + M(α, Kδ , t)|u(t) − v(t)| Now, we prove the continuity of R(·, δ) One has + |S(u+ δ ) − S(wδ )(t)| ≤ α + M(α, Kδ , t)|u(t) − w(t)| This implies R(u, δ) − R(w, δ) λ ≤ αT λ + max M(α, Kδ , t)T p−λ p u−w 0≤t≤T p So, for wn ∈ D ⊂ Lp (0, T ) such that limn→∞ wn −w p = 0, we get after letting n → ∞, α → that lim supn→∞ R(wn , δ) − R(w, δ) λ = It follows that R(·, δ) is continuous at w ∈ D This completes the proof of Proposition 3.3 The Inverse Property of R(w, δ) We shall prove that R(w, δ) is a regularization scheme In fact, we prove Proposition Let vex ∈ Lλ,+ (0, T ) Then R(J (vex ), δ)(t) ↑ vex (t) as δ ↓ and lim R(J (vex ), δ) − vex δ→0 λ = 0, where we recall that J (v)(t) = k(t, v(t)) Proof Putting wex = J (vex ), we get wex ∈ D and vex = S(wex ) From the definition + + of R, we have R(wex , δ) = S(wex,δ ) For < δ1 < δ, Lemma implies S(wex,δ )(t) ≥ + + S(wex,δ )(t) for all t ∈ (0, T ) Hence, limδ↓0 S(wex,δ ) exists for every t ∈ (0, T ) Putting + + ), we claim that ψ = vex In fact, wex − J (S(wex,δ )) ∞ ≤ δ, by ψ(t) = limδ↓0 S(wex,δ Lemma Let δ → 0, we get wex − J (ψ) ∞ = This implies k(t, vex (t)) = k(t, ψ(t)) Using Lemma 1, we get vex = ψ Now, the Lebesgue monotone convergence theorem implies lim R(J (vex ), δ) − vex δ→0 λ This completes the proof of Proposition + = lim S(wex,δ ) − vex δ→0 λ = Regularization for the Inverse Problem of Finding the Purely 3.4 Error Estimates We have Proposition Let L, ε, δ, α > and let vex ∈ Lλ,+ (0, T ), wex = J (vex ) Let wε ∈ D be such that wε − wex p ≤ ε Then 1/λ R(wε , δ) − vex λ ≤ 2αT 1/λ + {vex (t)>L} +δ max M(α, L, t)T |vex (t)|λ dt p−λ p 0≤t≤T + ε max M(α, Kδ , t)T p−λ p 0≤t≤T So, if we choose δ = δ(ε), α = α(δ), L = L(δ) such that limε→0 ε max0≤t≤T M(α, Kδ , t) = 0, limδ→0 α(δ) = 0, limδ→0 L(δ) = ∞, and limδ→0 δM(α(δ), L(δ), t) = for t ∈ (0, T ) then lim R(wε , δ(ε)) − vex λ ε→0 = Proof We have 1/λ R(J (vex ), δ) − vex λ ≤ {vex (t)>L} + |vex (t)|λ dt {0≤vex (t)≤L} + |S(wex )(t) − S(wex,δ )(t)|λ dt 1/λ + )(t) ≤ vex (t) ≤ L Hence, Proposition gives If ≤ vex (t) ≤ L then ≤ S(wex,δ + )| ≤ α + δM(α, L, t) |S(wex )(t) − S(wex,δ So, {0≤vex (t)≤L} + |S(wex )(t) − S(wex,δ )(t)|λ dt 1/λ ≤ αT 1/λ + δ max M λ (α, L, t)T 0≤t≤T This implies 1/λ R(J (vex ), δ) − vex λ ≤ {vex (t)>L} |vex (t)|λ dt +αT 1/λ + δ max M(α, L, t)T p−λ p 0≤t≤T On the other hand, R(w, δ) − R(J (vex ), δ) λ + = S(wδ+ ) − S(wex,δ ) ≤ αT 1/λ λ + + max M(α, Kδ , t) wδ+ − wex,δ 0≤t≤T ≤ αT 1/λ + max M(α, Kδ , t)εT 0≤t≤T p−λ p λ p−λ p D.D Trong et al The above estimates give 1/λ R(w, δ) − vex λ ≤ 2αT 1/λ + |vex (t)|λ dt {vex (t)>L} +δ max M(α, L, t)T p−λ p 0≤t≤T + ε max M(α, Kδ , t)T p−λ p 0≤t≤T This completes the proof of Proposition We need an estimate for M(α, L, t) In fact, we have Lemma Put ν0 = max{|ν|, |ν + rt|}, K1 = √ 2π X exp √ ≤ M(α, L, t) ≤ K1 L exp ν0 Then ν0 L + 2α √ + (ν+rt) + Proof We have M(α, L, t) = maxα≤s≤L X2πs exp (ν+rt) 2s t ≤ T , one has |ν + rt| ≤ max{|ν|, |ν + rT |} = ν0 Hence, √ ν02 2π L ν0 L ≤ M(α, L, t) ≤ exp + + X 2α s But, for ≤ This completes the proof of Lemma Theorem Let the assumptions of Proposition hold Let < ρ < 1/2, δ = ε ρ Put V (L) = {vex > L} = {t ∈ (0, T ) : vex (t) > L} Then there exist a constant C3 > and L = L(ε) > such that limε→0 L(ε) = ∞ and R(wε , δ) − vex where β(ε) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ Proof We choose α = ν2 C3 + ln 1ε C3 ln 1ε V (L(ε)) |vex (t)| ≤ β(ε), λ dt 1/λ if V (L(ε)) = ∅, if V (L(ε)) = ∅ C1 ln λ ε , L = C2 ln ε , where the constants C1 , C2 > satisfy ν2 < 2C01 + 2r < 1, 2C01 + C82 < r Using Proposition and Lemma 5, we can compute directly to get the desired estimate This completes the proof of Theorem Corollary Let the assumptions in Theorem hold If there exist μ, K2 > such that < λμ < and ≤ vex (t) ≤ (T K −t)μ , then there exists a constant C4 > such that R(wε , δ) − vex λ C4 ≤ ln Proof The assumption on vex implies that L ≤ |vex (t)|λ dt ≤ vex (t)≥L T T− K2 L 1/μ 1−λμ λ K2 (T −t)μ ε when vex (t) ≥ L So, we have K2λ K2λ dt = λμ (T − t) −λμ + K2 L −λμ+1 Regularization for the Inverse Problem of Finding the Purely Direct computation gives the desired conclusion The Inner Problem As mentioned, the inner problem is the one of recovering a derivative from its integral Precisely, we consider the problem of recovering the function aex (t) from the approximate value Sε (t) = R(uε , δ(ε)) of Sex := S(uex )(t) The problem is ill-posed and many textbooks presented well-known regularization schemes Here, we give only a simple version t of recovering the function aex (t) We recall that Sex (t) = aex (τ )dτ We first have Lemma Let < h < 2h < T and let θ ∈ [0, h] For f ∈ Lλ (0, T ), put f (t + θ) f (t − θ) fθ (t) = and ωf (h) = sup0≤θ≤h fθ − f λ for ≤ t ≤ T − h, for T − h ≤ t ≤ T , Then limh→0+ ωf (h) = Proof Let ε > Since C [0, T ] is dense in Lλ (0, T ), there exists a continuous function g ∈ C[0, T ] such that f − g λ ≤ 6ε We note that gθ − fθ λ λ T −h = T |g(t + θ) − f (t + θ)|λ dt + T −h T −h+θ = |g(t) − f (t)|λ dt + θ ≤ g−f T −θ |g(t − θ) − f (t − θ)|λ dt |g(t) − f (t)|λ dt T −h−θ λ λ Since λ ≥ 1, we have gθ − f θ λ ≤2 g−f λ + g − gθ λ Using the latter inequality, we have f − fθ λ ≤ f −g λ + gθ − f θ λ ≤ f − g λ + g − gθ λ ≤ ε + ωg (h) Since limh→0 ωg (h) = 0, we can find an h0 > such that f − fθ ≤ θ ≤ h ≤ h0 This completes the proof of Lemma λ ≤ ε for every Theorem Let α0 ∈ (0, 1], ≤ λ ≤ p and let the assumptions of Theorem hold In addition, we assume that aex ∈ Lλ (0, T ) For h ∈ (0, T ) such that < h < 2h < T , put aεh (t) = Sε (t+h)−Sε (t) , h Sε (t)−Sε (t−h) , h ≤ t ≤ T − h, T − h < t ≤ T Then aεh(ε) − aex λ ≤ 31−1/λ 4T p−λ p 1/λ (β(ε))(1−α0 )λ + 2ωaλex (β(ε))α0 , where β(ε) is as in Theorem and h() = (())0 Moreover, if aex is a Hăolder function satisfying |aex (t1 ) − aex (t2 )| ≤ K2 |t1 − t2 |γ for t1 , t2 ∈ [0, T ], D.D Trong et al where < γ ≤ 1, then we can choose α0 = aεh(ε) − aex λ γ +1 ≤ 31−1/λ 4T to get p−λ p 1/λ + 2K2λ T γ (β(ε)) γ +1 If, in addition, we have the assumptions of Corollary then ⎞ γ (1−λμ) ⎛ λ(1+γ ) h(ε) ⎠ ⎝ aε − aex λ ≤ O ln 1ε Proof We have for ≤ t ≤ T − h |aεh (t) − aex (t)| ≤ (|Sε (t + h) − Sex (t + h)| + |Sε (t) − Sex (t)|) h + |aex (t + hθ ) − aex (t)|dθ a+b+c λ Using the inequality T −h ≤ a λ +bλ +cλ for a, b, c ≥ 0, λ ≥ 1, we get |aεh (t) − aex (t)|λ dt ≤ I1 + I2 + I3 , where 3λ−1 hλ 3λ−1 I2 = hλ T −h I1 = |Sε (t + h) − Sex (t + h)|λ dt ≤ T −h |Sε (t) − Sex (t)|λ dt ≤ T −h I3 = 3λ−1 p−λ p p−λ p β λ (ε), β λ (ε), λ |aex (t + hθ ) − aex (t)|dθ dt ≤ 3λ−1 ωλ (h) The estimates of I1 , I2 , I3 imply that T −h 3λ−1 T hλ 3λ−1 T hλ ⎛ |aεh (t) − aex (t)|λ dt ≤ λ−1 ⎝ 2T p−λ p ⎞ β λ (ε) hλ + ωaλex (h)⎠ Similarly, for T − h < t ≤ T , we have |aεh (t) − aex (t)| ≤ (|Sε (t) − Sex (t)| + |Sε (t − h) − Sex (t − h)|) h + |Sex (t) − Sex (t − h) − haex (t)| h It follows that ⎛ T T −h |aεh (t) − aex (t)|λ dt ≤3 We get in view of the above inequalities T |aεh (t) − aex (t)|λ dt ≤ λ−1 ⎝ 2T ⎛ λ−1 ⎝ 4T p−λ p ⎞ β λ (ε) hλ p−λ p + ωaλex (h)⎠ ⎞ β λ (ε) hλ + 2ωaλex (h)⎠ Regularization for the Inverse Problem of Finding the Purely By choosing h = h(ε) = (β(ε))α0 , we get T |aεh(ε) (t) − aex (t)|λ dt ≤ 3λ−1 4T p−λ p (β(ε))(1−α0 )λ + 2ωaλex (β(ε))α0 ) Since aex is in Lλ (0, T ), Lemma gives lim ωaex (h) = h↓0 Noting that β(ε) → as ε → 0, we get the desired result Moreover, if aex is a Hăolder function as in the statement of the theorem then |aex (t1 ) − aex (t2 )| ≤ K2 |t1 − t2 |γ ≤ K2 hγ , In this case, we have |(aex )θ (t) − aex (t)| ≤ K2 T |θ |γ t1 , t2 ∈ [0, T ], |t1 − t2 | ≤ h It follows that |(aex )θ (t) − aex (t)|λ dt ≤ K2λ T |θ |λγ Therefore, ωaλex (h) ≤ sup T 0≤θ≤h So, if α0 = 1+γ , h = (β(ε)) γ +1 then T |aθ (t) − a(t)|λ dt ≤ K2λ T hλγ |aεh(ε) (t) − aex (t)|λ dt ≤ 3λ−1 4T p−λ p + 2K2λ T λγ (β(ε)) γ +1 This implies aεh(ε) − aex λ ≤ 31−1/λ 4T p−λ p 1/λ + 2K2λ T γ (β(ε)) γ +1 Now, if we have the assumptions of Corollary 1, then we can choose β(ε) = ln C4 1−λμ λ ε to get the final result This completes the proof of Theorem Table Error between exact solution and approximate solution ε = 10−2 ti ε = 10−3 Error ti ε = 10−4 Error ti Error 0.02703 0.00752 0.00778 0.25000 0.03922 0.16667 0.04076 0.12500 0.02793 0.50000 0.06152 0.33333 0.03171 0.25000 0.02713 0.75000 0.03585 0.50000 0.03505 0.37500 0.02725 0.66667 0.04647 0.50000 0.02914 0.83333 0.03926 0.62500 0.02785 0.75000 0.02820 0.87500 0.03222 D.D Trong et al Numerical Results From the results of the previous sections, we have many ways to establish a computation program Here, we give a rough algorithm The given data are K, X, r, T , the noise level ε, and the noise fair option price u(t) Since u ∈ D (see Proposition 1), we can replace u by min{u, X} The algorithm is divided into three steps Step We can choose h = h(ε) = β(ε) ∼ O ln ε as in Theorems and We divide the interval [0, T ] into n subintervals by ti = (i − 1)T /n, i = 1, , n + 1, where T n := β(ε) Step We construct the function R(u, δ)(t) for t = ti , ≤ i ≤ n + Let δ = δ(ε) = ε ρ be as in Theorem We can find Si := R(u, δ)(ti ) by solving the equation k(ti , x) = max{(X − Kerti )+ , u(ti ) − δ} (i = 1, , n + 1) Step We denote = Si+1 − Si n = (Si+1 − Si ) ti+1 − ti T for i = 1, , n The output is an approximation of the implied volatility a(t) at ti = (i − 1)T /n, i = 1, , n + 1, i.e., (a(t1 ), , a(tn+1 )) ≈ (a1 , , an+1 ) For example, we choose T = 1, X = 100, K = 101, r = 0.05 and a volatility function a(t) = 0.2 + 0.02 (1 − t)0.1 We use blsprice (Price, Strike, Rate, Time, Volatility) in MATLAB R2013a to calculate u(ti ) Testing with ε = 10−2 , 10−3 , 10−4 and error at points ti is as Table In Figs and 2, we plot the exact solution and the approximate solution In left panels, the graph of exact outer solution is blue and the graph of approximate outer solution is green Right panels are graphs of exact (red) and approximate (blue) volatility functions (a) Outer solution Fig ε = 10−3 (b) Volatility function Regularization for the Inverse Problem of Finding the Purely (a) Outer solution (b) Volatility function Fig ε = 10−4 Conclusion The regularization problem of finding a purely time-dependent volatility is considered in a lot of papers [14–16, 23] Our paper is devoted to the Lp -case of the problem In [14], the authors considered the case and established the convergence analysis for a descriptive regularization with p = But the regularization needed an available a priori constant, and the rate of convergence was not considered Now, in our paper, we construct a regularization scheme in which we not use the a priori constant We also give an analysis for convergence rate of our regularization and some numerical experiments Acknowledgments The authors are grateful to three anonymous referees for their precious suggestions leading to the improvement version of our paper References Baumeister, J.: Stable solution of inverse problems Friedr Vieweg & Son (1987) 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A computational scheme for uncertain volatility model in option pricing Appl Numer Math 59, 1754–1767 (2009) ... properties of Black–Scholes formula and prove the instability of the inverse problems Section gives the existence and the regularization of the outer problem The regularization of whole problem and rate... precise form of the inverse problem d1 = d2 = d1 − s, z −∞ e−x /2 dx Calibration Problem Let u(t) be the known fair price of the market We consider the problem of finding the unknown volatility. .. over time Therefore, the market is concerned with the problem of identifying of non-constant volatility [20, p 741] We have the problem of calibrating the implied volatility σ (t, X ) from the fair