MATH11158 May 2018 MAT-MSc-OMF Optimization Methods in Finance 1. Risk Neutral Probability Measure and Pricing. Let S0, S1, . . . , Sn be securities and Ω = {ω1, ω2, . . . , ωm} be a state space describing the possi- ble “states of the world” at time t = 1 (each ωj has a positive probability of occurring). By Si1(ωj) we denote the price/value of security S i at time t = 1 in state ωj . By S i 0 we denote the price of security S i at time t = 0 (present). Security S0 is the risk-free asset paying interest rate r ≥ 0 between time 0 and 1. Finally, assume that S00 = 1 and S 0 1(ωj) = 1 + r = R for all j. (a) A risk-neutral probability measure on Ω is a vector p = (p1, . . . , pm) ⊤ satisfying certain properties. List the properties. [5 marks] (b) Define Type A and Type B arbitrage using the notation established above. [5 marks] (c) State the First Fundamental Theorem of Asset Pricing. [4 marks] (d) Assume n = 2 (three securities, including the risk-free asset), m = 2 (two states), S10 = 9, S 1 1(ω1) = 9, S 1 1(ω2) = 12, S 2 1(ω1) = 22, S 2 1(ω2) = 11 and r = 0.1 (i.e. R = 1 + r = 1.1). Either by using a no-arbitrage/replication argument or by calculating the risk neutral probability measure directly, find S20 (that is, the price of the 2nd security at time t = 0). What is the risk neutral probabiliy measure? Is there arbitrage? [11 marks] Comment: Material covered in lecture and a numerical problem similar to one solved in class. 1 Solution: Solution: (a) (i) pj > 0 for all j = 1, 2, . . . ,m (ii) ∑ j pj = 1 (iii) Si0 = 1 R ∑m j=1 pjS i 1(ωj), for all i = 0, 1, . . . ,m (b) Type A arbitrage: A trading strategy (a portfolio y0, . . . , yn) having posi- tive initial cashflow ( ∑ i S i 0yi < 0) and no risk of loss later on ( ∑n i=1 S i 1(ωj)yi ≥ 0 for all j = 1, 2, . . . ,m). Type B arbitrage: A trading strategy (a portfolio y0, . . . , yn) having non- negative initial cashflow ( ∑ i S i 0yi ≤ 0) and positive probability of profit (= positive cashflow) in the future (there exists j such that ∑n i=1 S i 1(ωj)yi > 0). (c) There is no arbitrage (i.e., no type A and no type B) if and only if there exists a Risk Neutral Probability Measure. (d) We form a portfolio P of ∆ shares of S1 and B amount of S0 (cash). At time 1, the value of the portfolio is given by{ P1(ω1) = S 1 1(ω1)∆ + S 0 1(ω1)B = 9∆+ 1.1B, P1(ω2) = S 1 1(ω2)∆ + S 0 1(ω2)B = 12∆+ 1.1B. Note that we know the value of the portfolio at time 0: P0 = S 1 0∆+B = 9∆+B. Let us now choose ∆ and B so as to replicate the price/value of asset S2 at time 1. That is, we require that the following holds: P1(ω1) = S 2 1(ω1), P1(ω2) = S 2 1(ω2). This leads to the following system of 2 equations in 2 unknowns (∆, B): 9∆ + 1.1B = 22, 12∆+ 1.1B = 11. The solution is: ∆∗ = − 113 = −3 23 and B∗ = 50. Now, the portfolio is indistinguishable from S2 at time 1; hence, the price of the portfolio at time 0 (something we know) should be the same as the price of S2 at time 0 (something we do not know). So, we conclude that S20 = P0 = 9∆ ∗ +B∗ = −9× 113 + 50 = 17 The risk neutral probabilities p1, p2 need to satisfy conditions (a)(iii) above, so for cash S0 and asset S1 we have 1 = 11.11.1(p1 + p2) 9 = 11.1 (9p1 + 12p2) the first equation gives p1 + p2 = 1 ⇒ p2 = 1 − p1, substituting into the second equation gives 9.9 = 9p1 + 12(1 − p1) = 12 − 3p1 and thus p1 = 0.7, p2 = 0.3. As the price for the second asset S 2 0 we get S20 = 1 1.1 (22p1 + 11p2) = 1 1.1 (15.4 + 3.3) = 17 as before. Since p1, p2 > 0 these form a risk-neutral probability measure and hence there is no arbitrage. 2 2. Mean Absolute Deviation. (a) Assume that X is a space of random variables (which we take as repre- senting loss of an investment). What is the definition of a coherent risk measure ρ(X) on X ? [4 marks] (b) Show that ρ(X) = IE[X ] is a coherent risk measure. [2 marks] (c) Show that ρ(X) = IE[X ]+MAD[X ] satisfies all criteria for a coherent risk measure apart from monotonicity (Note: MAD[X ] := IE[|X − IE[X ]|] is mean absolute deviation). [6 marks] (d) Given possible investment options A,B with returns given by random variables rA, rB (where we use the convention that £1 invested in asset i will yield £ri at the end of the investment period). Let xi denote the fraction of the total investment to be invested in option i, and X = rAxA+ rBxB be the random variable giving the return on one unit of investment. Assume that the joint return (rA, rB) for the assets will take values (0.98, 0.94) and (1.03, 1.09) with probabilities 0.4 and 0.6 respectively. Consider the following variant of the Markowitz model that uses MAD[X ] as a risk measure: min x MAD[X ], s.t. IE[X ] ≥ 1.02, xA + xB = 1, xA ≥ 0, xB ≥ 0. Express MAD[X ] in terms of the decision variables xA, xB and show how this objective can be modelled as minimizing a linear objective subject to linear constraints. [13 marks] Solution: (a) A coherent risk measure ρ(X) has to satisfy • Monotonicity: If X,Y are random variables representing loss with X(ξ) ≤ Y (ξ) then ρ(X) ≤ ρ(Y ) • Sub-additivity: If X,Y are random variables, then ρ(X + Y ) ≤ ρ(X) + ρ(Y ) • Translation Equivariance: If X is a random variable representing loss, c ∈ IR then ρ(X + c) = ρ(X) + c • Positive homogenity: If X is a random variable, α ≥ 0 then ρ(αX) = αρ(X) [4 marks] 3 (b) Monotonicity, Translation Equivariance and Positive Homogenity follow from elemental properties of the Expectation. [1 mark] For subadditivity we even have IE(X + Y ) = IE(X) + IE(Y ) so ρ(X + Y ) ≤ ρ(X) + ρ(Y ) is satisfied. [1 mark] (c) Sub-additivity: ρ(X + Y ) = IE[X + Y ] + IE[|X + Y − IE[X + Y ]|] ≤ IE[X ] + IE[Y ] + IE[|X − IE[X ]|+ |Y − IE[Y ]], using |a+ b| ≤ |a|+ |b| = ρ(X) + ρ(Y ) [2 marks] Translation Equivariance: ρ(X + c) = IE[X + c] + IE[|X + c− IE[X + c]|] = IE[X ] + c+ IE[|X − IE[X ]|] = ρ(X) + c [2 marks] Positive Homogenity: ρ(αX) = IE[αX ] + IE[|αX − IE[αX ]|] = αIE[X ] + IE[α|X − IE[X ]|] = αρ(X), for α ≥ 0. [2 marks] (d) Introduce the additional variables R1, R2 and ER to denote the return in each possible future scenario and the expected return. These can be defined by the additional linear constraints R1 = 0.98xA + 0.94xB (1) R2 = 1.03xA + 1.09xB (2) ER = 0.4R1 + 0.6R2 (3) and introduce variables AD1, AD2 that denote the absolute deviation of the return in each scenario from the expected return: ADi = |Ri − ER|. [7 marks] These can be modelled by the linear constraints ADA ≥ RA − ER, ADA ≥ ER−RA (4) ADB ≥ RB − ER, ADB ≥ ER−RA (5) Thus the objective minMAD[X ] can be modelled as min 0.4ADA + 0.6ADB, s.t. (1), (2), (3), (4), (5). [4 marks] Note that since ADA, ADB are minimized, at least one side of each of the constraints (4),(5) must be active at the solution, ensuring that (4),(5) is indeed equivalent to ADi = |ri − ER|. [2 marks] 4 3. Scenario Generation (a) State three different ways of generating a set of discrete scenarios that approximate a (known) continuous distribution. Give at least two advan- tages and disadvantages for each. [6 marks] (b) Assume you want to find 4 scenarios that match a bivariate normal dis- tribution with the following mean and variance( h1 h2 ) ∼ N (µ,Σ), µ = ( 0 1 ) , Σ = ( 2 1 1 4 ) , by Moment Matching. Assume you want to match the five quantities IE[h1], IE[h2], IE[h 2 1], IE[h 2 2], IE[h1h2]. Write down the moment matching optimization problem. State clearly what are the variables to be optimised over. [8 marks] (c) How would you express the following constraints (i) Each scenario should have at least a 10% probability (ii) There must be two scenarios with an h1 value below or equal to 0 and two with an h1 value greater than or equal to 0. [3 marks] (d) Assume you want to approximate the bivariate uniform distribution on [−2, 2]×[−2, 2] by the 5 discrete scenarios (0, 0), (1, 1), (1,−1), (−1, 1), (−1,−1). If you are looking for the best match using Wasserstein/Kantorovich distances (i) How would you work out the probabilities for each
of these scenarios? A diagram might be helpful. [4 marks] (ii) Give the optimal values for the probabilities. It is sufficient to give geometric arguments to work out the exact values. [4 marks] Solution: (a) Possibilities include: Moment Matching, Historical Data, Wasserstein/Kantorovich/Clustering, Construction by hand. (b) The moment matching problem is min pi,x ∑ N i=1(fi(x, pi) −Mi)2 s.t. eTpi = 1, pi ≥ 0 where Mi are the values of the statistical properties (moments) to be matched and fi(x, pi) are formulas to calculate the i-th statistical property from the scenario data. For the given case IE[h1] M1 = 0 f1(x, pi) = ∑4 k=1 pikxk,1 IE[h2] M2 = 1 f2(x, pi) = ∑4 k=1 pikxk,2 IE[h21] M3 = 2 f3(x, pi) = ∑4 k=1 pikx 2 k,1 IE[h21] M4 = 5 f4(x, pi) = ∑4 k=1 pikx 2 k,2 IE[h1h2] M5 = 1 f5(x, pi) = ∑4 k=1 pikxk,1xk,2 5 Note that Var[h2] = IE[(h2 − IE[h2])2] = IE[(h2 − 1)2] = IE[h22 − 2h2 + 1] = IE[h22]− 2IE[h2] + 1. ⇒ IE[h22] = Var[h2] + 2IE[h2]− 1 = 4 + 1 = 5 also Cov[h1, h2] = IE[(h1 − IE[h1])(h2 − IE[h2])] = IE[(h1 − 0)(h2 − 1)] = IE[h1h2 − h1] = IE[h1h2]− IE[h1] = IE[h1h2] The moment matching optimization problem is then minpi,x ( ∑4 k=1 pikxk,1) 2 + ( ∑ 4 k=1pikxk,2 − 1)2 +( ∑ 4 k=1pikx 2 k,1 − 2)2 + ( ∑ 4 k=1pikx 2 k,2 − 5)2 +( ∑ 4 k=1pikxk,1xk,2 − 1)2 s.t. pi1 + pi2 + pi3 + pi4 + pi5 = 1, pik ≥ 0 (c) (i) Add the constraints pii ≥ 0.1. (d) (i) The probabilities would equal the total probability originating from each scenarios Voronoi region. You would need to work out the Voronoi region and determine its area as a fraction of the support of U([−2, 2]× [−2, 2]) (ii) Below is a plot of the support of the uniform distribution, the sce- narios and their Voronoi regions x1 x2 −1 0 1 −1 0 1 The central region is a square with side length √ 2, hence its area is 2. The area of the total support region [−2, 2]× [−2, 2] shown in green is 16. Each of the four corner Voronoi regions must have the same area which can therefore be determined as 16−24 = 3.5 The probabilities are thus pi(0,0) = 2 16 = 1 8 , pi(1,1) = pi(−1,1) = pi(1,−1) = pi(−1,−1) = 3.5 16 = 7 32 6
