Large losses - probability minimizing approach Micha l Barski Faculty of Mathematics and Computer Science, University of Leipzig, Germany Faculty of Mathematics, Cardinal Stefan Wyszyn´ski University in Warsaw, Poland Michal.Barski@math.uni-leipzig.de 6 January 14, 2016 1 0 2 Abstract n a Theprobabilityminimizingproblemoflargelossesofportfolioindiscreteandcontinuous J time models is studied. This gives a generalizationof quantile hedging presented in [3]. 3 1 Key words: quantile hedging, shortfall risk, transaction costs, risk measures. ] F AMS Subject Classification: 60G42, 91B28, 91B24, 91B30. M . GEL Classification Numbers: G11, G13 n i f - q 1 Introduction [ 1 Let (St) be a d-dimensional semimartingale on a filtered probability space (Ω, ,( t),P) which F F v represents the stock prices. We denote by the set of all martingale measures, it means that 8 Q Q if Q P and (S ) is a martingale with respect to Q. Let H be a measurable random 8 t ∈ Q ∼ F 3 variablecalledcontingent claim. Itisknownthatonsuchmarketwehavetwoprices: thebuyer’s 3 price u = inf EQ[H] and the seller’s price u = sup EQ[H], which usually are different. 0 A natubral queQs∈tiQon arises : what price from thesso calleQd∈Qarbitrage-free interval [u ,u ] should . b s 1 be chosen? This problem was a motivation for introducing risk measures on financial markets. 0 6 Various approaches were presented to answer this question, see for instance [1],[2],[3],[4],[6]. 1 In [3] Föllmer, and Leukert study the quantile hedging problem. They define a random : v variable ϕ connected with the strategy (x,π) by: x,π i X x,π X ar ϕx,π = 1{XTx,π≥H}+ HT 1{XTx,π α then u(H X) = α − 2. on Ac we have X = 0. NoticethatonthesetAwehaveX = 0ifH u 1(α)andX = H u 1(α)ifH > u 1(α). Thus − − − ≤ − on A we have X = (H u 1(α))+. Since X = 0 on Ac we obtain that X = 1 (H u 1(α))+. − A − − − In other words X if it is of the form X = 1 (H u 1(α))+ for some A such that A − ∈ S − ∈ F A u(H) α . ⊇ { ≤ } 3 Lemma 3.3 For each X L0 such that EQ[X] x there exists a random variable Z ∈ + ≤ 0 ∈ S such that EQ[Z] EQ[X] and ≤ P[u((H X)+) α] = P[u((H Z)+) α]. − ≤ − ≤ Proof : Let us define A := ω : u(H X)+ α . Then Z := 1 (H u 1(α))+ and we A − { − ≤ } − ∈ S have P[u((H Z)+) α] = P[u((H 1 (H u 1(α))+)+) α] A − − ≤ − − ≤ = P[ω A:u((H (H u 1(α))+)+) α]+P[ω Ac :u(H) α] − ∈ − − ≤ ∈ ≤ = P[ω A u(H) α : u(H) α]+P[ω A u(H) > α :u(u 1(α))) α] − ∈ ∩{ ≤ } ≤ ∈ ∩{ } ≤ = P(A). On the set Ac holds Z = 0 X. On A if u(H) α then Z = 0 X and if u(H) > α then ≤ ≤ ≤ Z = H u 1(α) X. Thus we have Z X and EQ[Z] EQ[X]. (cid:3) − − ≤ ≤ ≤ Remark 3.4 The above calculations show that for any X = 1 (H u 1(α))+ holds B − − ∈ S P(u(H X)+ α) = P(B). − ≤ Using lemma 3.3 and remark 3.4 we can reformulate theorem 3.1 in the following form. Theorem 3.5 If there exists set A˜ u(H) α which is a solution of the problem : ⊇ { ≤ } P(A) max (3.5.1) −→ EQ[1 (H u 1(α))+] x (3.5.2) A − 0 − ≤ then the replicating strategy for 1 (H u 1(α))+ is optimal. A˜ − − Proof : Indeed, by lemma 3.3 the problem P[u(H X)+ α] max, EQ[X] x , X L0 − ≤ −→ ≤ 0 ∈ + can be replaced by P[u(H X)+ α] max, EQ[X] x , X . 0 − ≤ −→ ≤ ∈ S However, by remark 3.4 we know that for X = 1 (H u 1(α))+ we have A − − ∈ S P[u(H X)+ α] = P(A) and the required formulation is obtained. (cid:3) − ≤ Remark 3.6 Let us consider the optimizing problem from theorem 3.5 given by 3.5.1 and 3.5.2 but without the requirement that A˜ u(H) α . Notice that if P(u(H) α) > 0 then the ⊇ { ≤ } ≤ solution A˜ must contain u(H) α . Suppose the contrary and define A˜˜ := A˜ u(H) { ≤ } ∪ { ≤ α . Then EQ[1 (H u 1(α))+] = EQ[1 (H u 1(α))+] x and P(A˜˜) > P(A˜) what is a } A˜˜ − − A˜ − − ≤ 0 contradiction. This shows that that the requirement A˜ u(H) α in the theorem 3.5 can be ⊇ { ≤ } dropped. 4 In some particular cases the existence and construction of the set A˜ can be solved by using Neyman-Pearsonlemma. TothisendletusintroduceameasureQ¯ whichisabsolutelycontinuous with respect to Q by: dQ¯ (H u 1(α))+ − = − . dQ EQ[(H u 1(α))+] − − Then set A˜ solves the following problem P(A) max −→ x Q¯(A) 0 . ≤ EQ[(H u 1(α))+] − − To make the paper self-contained we present a part of the Neyman-Pearson lemma. Let P and 1 P be two probability measures such that there exists density dP1. 2 dP2 Lemma 3.7 If there exists constant β such that P dP1 β = γ then P dP1 β P (B) 2{dP2 ≥ } 1{dP2 ≥ } ≥ 1 for any set B satisfying P (B) γ. 2 ≤ Proof : Let B be a set satisfying P (B) γ and denote B˜ := dP1 β . Then we have 2 ≤ {dP2 ≥ } P (B˜) P(B)= (1 1 )dP = (1 1 )dP + (1 1 )dP 1 − ZΩ B˜ − B 1 ZddPP12≥β B˜ − B 1 ZddPP21<β B˜ − B 1 (1 1 )βdP 1 βdP ≥ ZddPP21≥β B˜ − B 2 −ZddPP12<β B 2 = β dP dP = β(γ P (B)) 0. 2 2 2 (cid:18)ZB˜ −ZB (cid:19) − ≥ (cid:3) This lemma is useful for the Black-Scholes model since there the condition Q¯ dP β = {dQ¯ ≥ } x0 is satisfied. However, in case of discrete Ω this condition no longer holds. This EQ[(H u−1(α))+] will be−shown in the example of the CRR model. 4 Black-Scholes model Here we follow an example presented in [3]. The stock price S is given by t dS = S (µdt+σdW ), S = s, t t t 0 where µ and σ > 0 are constants and W is a standard Brownian motion. For this model t St = se(µ−12σ2)t+σWt. and the unique martingale measure Q is given by dQ = e−µσWT−12(σµ)2T. dP Moreover, the process W = W + µt is a Brownian motion with respect to Q. Notice that the t∗ t σ density of the martingale measure can be expressed in term of S , namely T dQ µ = cS−σ2, where c is some constant. dP T 5 We study a risk minimizing problem for a European call option with strike K. Recall that the problem is reduced to constructing set A˜ being a solution of P(A) max −→ x Q¯(A) 0 , ≤ EQ[(S K u 1(α))+] T − − − where measure Q¯ is as in the previous section : dQ¯ (S K u 1(α))+ T − = − − . dQ EQ[(S K u 1(α))+] T − − − Notice, that the superscript ”+ ” above can be dropped since for any a,b,c 0 holds ((a ≥ − b)+ c)+ = (a b c)+. According to Neyman-Pearson lemma we are searching for the set A˜ − − − of the form: dP dP µ c = c (S K u 1(α))+ = Sσ2 c c (S K u 1(α))+ , dQ¯ ≥ 1 dQ ≥ 2 T − − − T ≥ · 2 T − − − (cid:26) (cid:27) (cid:26) (cid:27) (cid:26) (cid:27) where c ,c are nonnegative constants such that 1 2 EQ[1 (S K u 1(α))+] = x . (4.0.3) A˜ T − − − 0 Let us consider two cases. 1) µ σ2 ≤ µ Then the function x xσ2 is concave and has 0 in 0 and thus the solution is given by A˜ = {ST ≤ c3} =−→{WT∗ ≤ c4}, where c3 and c4 s.t. c3 = seσc4−12σ2T are constant numbers satisfying 3.0.3. The optimal strategy is a strategy which replicates the following contingent claim: 1 (S K u 1(α))+ = 1 (S K u 1(α))+ A˜ T − − − {ST≤c3} T − − − = (S K u 1(α))+ (S c )+ (c K u 1(α))1 T − − − − T − 3 − 3 − − − {ST>c3} and the corresponding probability is equal c µT P(A˜)= P(W c )= Φ 4− σ . T∗ ≤ 4 √T (cid:18) (cid:19) For calculating constants c and c from 4.0.3 we use formula for pricing European call option. 3 4 EQ (S K u 1(α))+ (S c )+ (c K u 1(α))1 = T − − − − T − 3 − 3− − − {ST>c3} h i c +σT c sΦ(d¯+) (K +u−1(α))Φ(d¯ ) sΦ − 4 +c3Φ 4 − − − √T − √T − (cid:18) (cid:19) (cid:18) (cid:19) (c K u 1(α))Q W > c = 3 − T∗ 4 − − { } c +σT c sΦ(d¯+) (K +u−1(α))Φ(d¯ ) sΦ − 4 +(K +u−1(α))Φ 4 = x0, − − − √T − √T (cid:18) (cid:19) (cid:18) (cid:19) 6 where d¯ = 1 ln K+u−1(α) +1σ√T and Φ stands for the distribution function of the + −σ√T s 2 N(0,1) distribution. (cid:16) (cid:17) 2) µ > σ2 µ In this case the function x xσ2 is convex and therefore our solution is of the form −→ A˜= S c = W < c W > c T 5 T 6 T∗ 7 T∗ 8 { }∪{ } { }∪{ } µ wherec5 < c6 are two solutions of the equation xσ2 = c¯(x K u−1(α))+, wherec¯is a constant number s.t. 4.0.3 holds. Constants c7,c8 are given by c5−= s−eσc7−21σ2T,c6 = seσc8−12σ2T. The optimal strategy is a strategy which replicates the following contingent claim: 1 (S K u 1(α))+ = A˜ T − − − (S K u 1(α))+ (S c )+ (c K u 1(α))1 +(S c )++ T − − − − T − 5 − 5− − − {ST>c5} T − 6 (c K u 1(α))1 6 − − − {ST>c6} and the corresponding probability is equal c µT c µT P(A˜) = P(W < c )+P(W > c )= Φ 7− σ +Φ 8− σ . T∗ 7 T∗ 8 √T − √T (cid:18) (cid:19) (cid:18) (cid:19) Now we need to determine all necessary constants. Using the same methods as in the previous case we obtain EQ[1 (S K u 1(α))+]= sΦ(d¯ ) (K +u 1(α))Φ(d¯ ) A˜ T − − − + − − − − c c c c sΦ 7 +σ√T +sΦ 8 +σ√T + K +u 1(α) Φ 7 Φ 8 = x − 0 − √T − √T − √T − − √T (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19) (cid:16) (cid:17) (4.0.4) Summarizing, constants are determined by 4.0.4 and by the fact that c ,c are solutions of the 5 6 µ equation xσ2 = c¯(x K u−1(α))+, where c¯is a positive constant. − − 5 CRR model Let (S ) be a stock price given by n n=0,1,2,...,N S = S (1+ρ ), S = S, n+1 n n 0 where(ρ )isasequenceofindependentrandomvariablessuchthatp := P(ρ = u) = 1 P(ρ = n n n − d), where u > d,u > 0,d < 0. This means that at any time the price process S can increase to n the value S (1+u) or decrease to S (1+d). We assume that p (0,1). It is known, that the n n ∈ unique martingale measure for this model is given by p := d . ∗ u−d Let us study the risk minimizing problem for the call optio−n with strike K. Let us denote (S K¯)+ := (S K u 1(α))+ and consider two measures: the objective one P N N − − − − P(ω ) =pk(1 p)N k k − − 7 and the measure Q¯ (which is not necessarily a probability measure) given by Q¯(ω ) := (S(1+u)k(1+d)N k K¯)+p k(1 p )N k. k − ∗ ∗ − − − Here ω means an elementary event for which the number of jumps upwards is equal to k. Our k aim is to find set A˜ which solves: P(A) max −→ Q¯(A) x . 0 ≤ For the CRR model existence of the required set A˜ is clear since Ω is finite. However we want to find it explicitly. Unfortunately, the Neyman-Pearson lemma for the measures P and Q¯ can not be applied here since Ω is discrete and the condition dP x Q¯ a(H u 1(α))+ = 0 for some a > 0 − dQ ≥ − E[(H u 1(α))+] (cid:26) (cid:27) − − is very rarely satisfied. The first way of constructing A˜, which seems to be natural, is to find a constant a¯ such that dP x a¯ = inf a :Q¯ a(H u 1(α))+ 0 . dQ ≥ − − ≤ EQ[(H u 1(α))+] (cid:26) n o − − (cid:27) and then expect that dP A¯= a¯(H u 1(α))+ − dQ ≥ − (cid:26) (cid:27) is a solution. Unfortunately, this is not a right construction as shown in the example below. Example Let Ω = ω ,ω ,ω and P and Q are two measures given by p = 7 ,p = 4 ,p = 4 and { 1 2 3} 1 15 2 15 3 15 q = 4 ,q = 3 ,q = 3 . We want to maximize P(A) subject to the condition Q(A) x = 6 . 1 10 2 10 3 10 ≤ 0 10 We have p1 = 63,p2 = 48,p3 = 48 and the above construction gives A˜ = ω . However q1 54 q2 54 q3 54 { 1} Q( ω ,ω ) = 6 and P( ω ,ω ) = 8 > 7 = P(ω ). { 2 3} 10 { 2 3} 15 15 1 Below we present a lemma which provides construction o A˜ when measures satisfy some particular condition. It turns out that this condition is satisfied by a significant number of cases in the hedging problem of call option. Lemma 5.1 Let Ω = ω ,ω ,...,ω and measures P and Q (not necessary probabilistic) satisfy 1 2 n { } the following conditions: p p p ... p > 0 < q q q ... q 1 2 3 n 1 2 3 n ≥ ≥ ≥ ≥ ≤ ≤ ≤ ≤ andγ beafixedconstant. LetA˜= ω ,ω ,...,ω , wherethenumberk issuchthatQ(ω ,ω ,...,ω ) 1 2 k 1 2 k { } ≤ γ and Q(ω ,ω ,...,ω ,ω ) > γ. Then P(A˜) P(A) for any set A satisfying Q(A) γ. 1 2 k k+1 ≥ ≤ Proof : Let B Ω s.t. Q(B) γ. ⊆ ≤ 1) First assume that A˜ B = . Then B k and we have P(ω˜) P(ω) for each ω˜ A˜ and ∩ ∅ | |≤ ≥ ∈ ω B. As a consequence ∈ P(A˜) = P(ω) P(ω) = P(B). ≥ ωX∈A˜ ωX∈B 8 2) If A˜ B = then by (1) A˜ A˜ B is a solution of ∩ 6 ∅ \{ ∩ } P(A) max −→ Q(A) γ Q( A˜ B ) ≤ − { ∩ } and so P(A˜ A˜ B ) P(B A˜ B ). As a consequence P(A˜) P(B). (cid:3) \{ ∩ } ≥ \{ ∩ } ≥ Since P(ω ) increases with k if p > 1 and decreases if p < 1, the only point to apply the k 2 2 lemma is to state the monotonicity of the measure Q¯. In fact we are interested in monotonicity of Q¯ only on the set where it is strictly positive. Let us denote a := Q¯(ω )= (S(1+u)k(1+d)N k K¯)+p k(1 p )N k, k k − ∗ ∗ − − − (S(1+u)k+1(1+d)N k 1 K¯)+ − − b := − , k (S(1+u)k(1+d)N k K¯)+ − − 1+u q := , 1+d where the sequence b is well defined under convention that a = for a 0. Then Q¯(ω ) is increasing if aka+k1 ≥ 1kfor each k = 0,1,...,N −1. This conditio0n is∞equivalen≥t to that bk ≥ k1−p∗p∗ for each k = 0,1,...,N 1. But now note that the sequence b is decreasing. To see that one k − can calculate that b k+1 1 (q 1)2 0. b ≤ ⇐⇒ − ≥ k The last condition is always satisfied. Thus Q¯(ω ) is increasing if k (S(1+u)N K¯)+ 1 p ∗ b = − − . N−1 (S(1+u)N−1(1+d) K¯)+ ≥ p∗ − Note that this case includes the situations when p 1. ∗ ≥ 2 By analogous arguments one can obtain condition under which Q¯(ω ) is decreasing. This is the k case when the bk¯ ≤ 1−p∗p∗, where k¯ is the minimal k for which bk 6= ∞. Indeed, then we have bk ≤ 1−p∗p∗ for all k ≥ k¯ what implies that ak+1 < ak for k ≥ k¯. Before summarizing the above consideration let us introduce the following notation A := ω Ω s.t. the number of jumps upwards is equal to k . k { ∈ } for the set containing all elements ω . The following lemma is a consequence of lemma 5.1. k Lemma 5.2 1)(P increasing, Q¯ decreasing) wLehterk¯e=themninu{mkb:erbkk6=is∞s.t}..Q¯If(Ap ≥ 21Aand bk¯..≤. A1−p∗p∗ t)henxA˜=anAdNQ¯(∪AAN−A1∪...∪..A. NA−k∪BN)−>k−x1, N N 1 N k 0 N N 1 N k 1 0 ∪ − ∪ ∪ − ≤ ∪ − ∪ ∪ − − and the set B contains maximal number of any elements from the set A such that N k 1 N k 1 Q¯(B ) x− − Q¯(A A ... A ). − − N k 1 0 N N 1 N k 2)(P d−ec−reas≤ing, Q¯−increas∪ing) − ∪ ∪ − If p ≤ 12 and (S(1(+Su(1)N+−u1)N(1−+Kd¯))+K¯)+ ≥ 1−p∗p∗ (for example when p∗ ≥ 12) then A˜ = A0 ∪A1 ∪...∪ A B , where the number−k is s.t. Q¯(A A ... A ) x and Q¯(A A ... A )> k k+1 0 1 k 0 0 1 k+1 ∪ ∪ ∪ ∪ ≤ ∪ ∪ ∪ x and the set B contains maximal number of any elements from the set A such that 0 k+1 k+1 Q¯(B ) x Q¯(A A ... A ). k+1 0 0 1 k ≤ − ∪ ∪ ∪ 9 Example As an application of lemma 5.2 we study a risk minimizing problem for a call option with strike K = 600 in a 3-period model with parameters : S = 1000, u = 0,1, d = 0,2, p = 1. Price 0 − 4 at time 0 of the option is u = EQ[(S 600)+] = 398 7 . Assume that we have only x = 150 0 3 − 27 0 and α = 5 is a level of acceptable losses measured by u(x) = √x. We denote by ωabc, where a,b,c u,d elementary events with interpretation of a,b,c as a history of the price process. ∈ { } For example ωudu means the event where the price process moved up in the first and the third period and moved down in the second one. Since we can not hedge the original contingent claim H = (S 600)+: 3 − H(ωuuu)= 731, H(ωuud) = H(ωudu) = H(ωduu) = 368, H(ωudd) = H(ωdud)= H(ωddu) = 104, H(ωddd)= 0, we have to hedge H˜ = 1 (S 625)+. Since p = 1 and p = 2, we can apply lemma 5.2(2) for A˜ 3− 4 ∗ 3 construction of A˜. Below we present three possible right candidates for H˜. H˜(ωuuu) =0, H˜(ωddd)= 0, H˜(ωddu)= H˜(ωdud) = H˜(ωudd) = 79 and H˜(ωuud)= H˜(ωudu)= 343, H˜(ωddu)= 0 n o or H˜(ωuud) = 0, H˜(ωudu) =H˜(ωddu) = 343 n o or H˜(ωuud) = 343, H˜(ωudu)= 0, H˜(ωddu) = 343 n o Moreover, P(A˜) = 3 3+ 1 3 2 3+ 1 2 3 2 = 15. 4 4 · 4 · 4 · 4 · 16 (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) 6 Incomplete markets Now let us consider the case when the equivalent martingale measure is not unique. This means that the market is incomplete and not every contingent claim can be replicated. We preserve x,π all assumptions from previous section. Recall that the wealth process X is a supermartingale t with respect to each martingale measure Q . In this case theorem which describes optimal ∈ Q strategy is of the form: Theorem 6.1 Assume that there exists set A˜ which is a solution of the problem: P(A) max −→ sup EQ[1 (H u 1(α))+] x . A − 0 − ≤ Q ∈Q Then the strategy which hedges the contingent claim 1 (H u 1(α))+ is optimal. A˜ − − Proof : Let us consider an arbitrary admissible strategy (x,π), where x x . We will show that 0 P(u(H Xx,π)+ α) P(A˜). ≤ Notice,−thatTfor an≤y a,b≤,c 0wehave (a b)+ c b (a c)+ andthus u((H Xx,π)+) α Xx,π (H u 1(α≥))+. As a cons−equen≤ce fo⇐r⇒any≥Q − we obtain − T ≤ ⇐⇒ T ≥ − − ∈Q EQ[1{u((H−XTx,π)+)≤α}(H −XTx,π)+] ≤ EQ[1{u((H−XTx,π)+)≤α}XTx,π] EQ[Xx,π] x x , ≤ T ≤ ≤ 0 10

Large losses - probability minimizing approach

Similar Large losses - probability minimizing approach