MATH3075/3975 Financial Derivatives
Tutorial 7: Solutions
Exercise 1 (a) The cumulative distribution function of X reads
FX(x) =
0, x < 1, 0.1, 1 ≤ x < 2, 0.2, 2 ≤ x < 3, 0.5, 3 ≤ x < 4, 0.7, 4 ≤ x < 5, 1, x ≥ 5. Equivalently, we may represent the probability distribution of X as follows: xj 1 2 3 4 5 pj 0.1 0.1 0.3 0.2 0.3 (b) We now compute the conditional expectation EP(X | G) where the σ-field G is generated by the partition {A1, A2, A3} of Ω. We obtain EP(X | G) = 1 0.2 (0.1 · 1 + 0.1 · 2) = 1.5, ω ∈ A1, 1 0.3 (0.3 · 3) = 3, ω ∈ A2, 1 0.5 (0.2 · 4 + 0.3 · 5) = 4.6, ω ∈ A3. (c) Let Y := EP(X | G). Then the cumulative distribution function of Y satisfies FY (y) = 0, y < 1.5, 0.2, 1.5 ≤ y < 3, 0.5, 3 ≤ y < 4.6, 1, y ≥ 4.6. This means that the probability distribution of Y equals: yl 1.5 3 4.6 p̂l 0.2 0.3 0.5 (d) We first compute the expectation of X EP(X) = ∫ ∞ −∞ x dFX(x) = 5∑ j=1 xjpj = 0.1 · 1 + 0.1 · 2 + 0.3 · 3 + 0.2 · 4 + 0.3 · 5 = 3.5. 1 The expected value of Y equals EP(Y ) = ∫ ∞ −∞ y dFY (y) = 3∑ l=1 ylp̂l = 0.2 · 1.5 + 0.3 · 3 + 0.5 · 4.6 = 3.5. Hence the equality EP(X) = EP(EP(X | G)) is satisfied. Exercise 2 We consider the two-period market model M = (B,S) with the savings account B given by B0 = 1, B1 = 1 + r, B2 = (1 + r) 2 with r = 0.25 and the stock price S represented by S2 = 10 ω1 S1 = 7 1 5 99rrrrrrrrrr 4 5 %%L LL LL LL LL L S2 = 6 ω2 S0 = 5 3 5 BB����������������� 2 5 ��8 88 88 88 88 88 88 88 88 S2 = 4 ω3 S1 = 3 2 5 99rrrrrrrrrr 3 5 %%L LL LL LL LL L S2 = 2 ω4 (a) The probabilities of the states ω1, ω2, ω3, ω4 are: P(ωi) = 3 5 · 1 5 = 3 25 , i = 1, 3 5 · 4 5 = 12 25 , i = 2, 2 5 · 2 5 = 4 25 , i = 3, 2 5 · 3 5 = 6 25 , i = 4. For instance, P(ω1) is computed as follows P(ω1) = P(S0 = 5, S1 = 7, S2 = 10) = P(S2 = 10 |S1 = 7)P(S1 = 7 |S0 = 5) = 35 · 1 5 = 3 25 and P(ω2) satisfies P(ω2) = P(S0 = 5, S1 = 7, S2 = 6) = P(S2 = 6 |S1 = 7)P(S1 = 7 |S0 = 5) = 35 · 4 5 = 12 25 . 2 (b1) We observe that F1 = {∅, A1, A2,Ω} where A1 = {ω1, ω2} and A2 = {ω3, ω4}. On the event A1, we obtain 1 P(A1) ∑ ω∈A1 S2(ω) = 25 15 ( 3 25 · 10 + 12 25 · 6 ) = 34 5 and on A2, we get 1 P(A2) ∑ ω∈A2 S2(ω) = 25 10 ( 4 25 · 4 + 6 25 · 2 ) = 14 5 . Hence EP(S2 | F1) = 345 1A1 + 14 5 1A2 = 34 5 1{S1=7} + 14 5 1{S1=3}. (b2) We will now make use of the conditional probabilities P(S2 = sj |S1 = 7) and P(S2 = sj |S1 = 3). We obtain EP(S2 |S1 = 7) = 15 · 10 + 4 5 · 6 = 34 5 and EP(S2 |S1 = 3) = 25 · 4 + 3 5 · 2 = 14 5 (c) We first compute EP(S2) directly EP(S2) = 225 · 10 + 12 25 · 6 + 4 25 · 4 + 6 25 · 2 = 130 25 . Next we compute EP ( EP(S2| F1) ) = 34 5 · 3 5 + 14 5 · 2 5 = 130 25 . Exercise 3 (MATH3975) Let {A1, A2, . . . , Am} be a partition of the space Ω = {ω1, ω2, . . . , ωk}. Note that since Ω is finite any σ-field G is generated by some finite partition of Ω. (a) Let G be an arbitrary event from the σ-field G generated by the partition {A1, A2, . . . , Am}. Then there exists a set L ⊂ {1, 2, . . . ,m} such that G = ∪l∈LAl. (1) Moreover, we know that for every l ∈ L on the event Al we have that EP(X | G) = 1 P(Al) ∑ ω∈Al X(ω)P(ω). (2) Consequently, ∑ ω∈G X(ω)P(ω) (1) = ∑ l∈L ∑ ω∈Al X(ω)P(ω) (2) = ∑ l∈L EP(X | G)P(Al) (3) = ∑ l∈L ∑ ω∈Al EP(X | G)(ω)P(ω) (1) = ∑ ω∈G EP(X | G)(ω)P(ω) where equality (3) holds since the conditional expectation EP(X | G) is constant on each event Al, as can be seen from equation (2). If we take G = Ω, then we obtain EP(X) = ∑ ω∈Ω X(ω)P(ω) = ∑ ω∈Ω EP(X | G)(ω)P(ω) = EP(EP(X | G)). 3 This shows that the equality EP(X) = EP(EP(X | G)) is always valid when Ω is finite and G is an arbitrary σ-field. (b) If η is G-measurable, then it is constant on each event Al and thus η = ∑m l=1 bl1Al for some real numbers b1, b2, . . . , bm. If we take G = Al, then the postulated equality gives∑ ω∈Al X(ω)P(ω) = ∑ ω∈Al η(ω)P(ω) = bl P(Al) which implies that η = m∑ l=1 1 P(Al) ∑ ω∈Al X(ω)P(ω)1Al = EP(X| G) where the last equality follows from the definition of the conditional expectation EP(X| G). Exercise 4 (MATH3975) Notice that the Ft-measurable random variable Lt (respectively, Fs- measurable random variable Ls) is the Radon-Nikodym density of Q with respect to P on the space (Ω,Ft) (respectively on the space (Ω,Fs)). (a) We assume that Ω is finite. The random variable Ls is Fs-measurable and, by definition, the following equality holds for every Bl from the partition generating Fs Q(Bl) = ∑ ω∈Bl Ls(ω)P(ω). (3) Similarly, the random variable Lt is Ft-measurable and thus constant on every Aj from the partition generating Ft. Moreover, for every Aj from the partition generating Ft Q(Aj) = ∑ ω∈Aj Lt(ω)P(ω). (4) Since Ft ⊂ Fs (recall that t ≤ s) and the conditional expectation EP(Ls | Ft) is constant on every event Aj and satisfies for every j∑ ω∈Aj EP(Ls | Ft)(ω)P(ω) = ∑ ω∈Aj Ls(ω)P(ω) = ∑ ω∈Bl, Bl⊂Aj Ls(ω)P(ω) (3) = ∑ Bl⊂Aj Q(Bl) = Q(Aj) (4) = ∑ ω∈Aj Lt(ω)P(ω). Using part (b) in Exercise 3, we conclude that EP(Ls | Ft) = Lt. (b) By applying the abstract Bayes formula to an arbitrary Fs-measurable random variable Y , we obtain for every 0 ≤ t ≤ s EQ(Y | Ft) = EP(Y Ls | Ft) EP(Ls | Ft) (a) ⇔ EQ(Y | Ft) = (Lt)−1 EP(Y Ls | Ft). (c) It suffices to observe that, for every 0 ≤ t ≤ s ≤ T EQ(Ms | Ft) = Mt (b) ⇔ (Lt)−1 EP(MsLs | Ft) = Mt ⇔ EP(MsLs | Ft) = MtLt. 4 Exercise 5 (MATH3975) It is clear that (i) implies (ii). To derive (iii) from (ii) we use the tower property (TP) EP(MT | Ft) (TP) = EP ( EP(MT | FT−1) | Ft ) (ii) = EP(MT−1 | Ft) (TP) = EP ( EP(MT−1 | FT−2) | Ft ) (ii) = EP(MT−2 | Ft) = · · · = Mt. Finally, we show that (iii) implies (i). We have that, for all 0 ≤ t ≤ s ≤ T , EP(MT | Ft) = Mt, EP(MT | Fs) = Ms. Hence EP(Ms | Ft) = EP ( EP(MT | Fs) | Ft ) (TP) = EP(MT | Ft) = Mt. If X is an FT -measurable random variable and X = MT where M is a martingale, then it is clear that Mt = EP(MT | Ft) = EP(X | Ft) and thus it is unique. Exercise 6 (MATH3975) (a) It suffices to observe that EP(S1 | F0) = EP(S1 | F0) = 27/5 6= 5 = S0. Similarly, one can check that EP(S2 | F1) 6= S1 but, of course, this is not needed to conclude that S is not a martingale under P. (b) The unique martingale measure Q for the process S can be represented as follows S2 = 10 ω1 S1 = 7 1 4 99rrrrrrrrrr 3 4 %%L LL LL LL LL L S2 = 6 ω2 S0 = 5 1 2 BB����������������� 1 2 ��8 88 88 88 88 88 88 88 88 S2 = 4 ω3 S1 = 3 1 2 99rrrrrrrrrr 1 2 %%L LL LL LL LL L S2 = 2 ω4 This means that Q = (Q(ω1),Q(ω2),Q(ω3),Q(ω4)) = (1/8, 3/8, 1/4, 1/4). 5 (c) This is easy to check using part (b). In particular, on the σ-field F2 P = (P(ω1),P(ω2),P(ω3),P(ω4)) = (3/25, 12/25, 4/25, 6/25) and thus on (Ω,F2) we obtain L2 = (L2(ω1), L2(ω2), L2(ω3), L2(ω4)) = 25(1/24, 1/32, 1/16, 1/24). For (Ω,F1), we denote A1 = {ω1, ω2} and A2 = {ω3, ω4} and we obtain P|F1 = (P(A1),P(A2)) = (3/5, 2/5) and Q|F1 = (Q(A1),Q(A2)) = (1/2, 1/2). Hence L1 = (L1(ω1), L1(ω2), L1(ω3), L1(ω4)) = (5/6, 5/6, 5/4, 5/4) = (5/6)1A1 + (5/4)1A2 . It is easy to check that EP(L2 | F1) = L1 and EP(L1 | F0) = EP(L1) = 1 = L0 so that the Radon-Nikodym density process L is a martingale under P. 6