RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS 1-6 CHAPTER 1

Transcription

1 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 HENRIK HULT, FILIP LINDSKOG ABSTRACT. The current collection of solutions to the exercises in the first part of the book Risk and Portfolio Analysis: principles and methods is not yet fully complete. Please inform us if you spot any errors. Exercise.. Arbitrage in bond prices CHAPTER a The cash flow of Bond D can be generated by the portfolio consisting of 06/200 units of Bond C, 6/02 units of Bond B and 6 2/02/00 units of Bond A. The price of the portfolio is =.3747, 02 whereas the price of Bond D is.55. Thus, short selling Bond D and buying the above bond portfolio create a profit of $0.753 b The cash flow of Bond A implies d = The cash flow of bond D and linear interpolation of discount factors provide the equations.55 = 6d + 6d d 3, d 2 = d + d 3 d t 3 t t 2 t that can be solved for d 2 and d 3. The result is d 2 = and d 3 = Finally, the cash flow of Bond E provides the equation = 4d + 4d 2 + 4d d 4 which implies d 4 = By Theorem. ii we have proved the absence of arbitrage. Exercise.2. Put-call parity a The payoff function f can be written as f x = x+k x + x K 2 +. Replacing x by S T and takeing expectation with respect to the forward probability Q yields the collar forward price E Q [ f S T ] = E Q [S T + K S T + S T K 2 + ] = E Q [S T ] + E Q [K S T + ] E Q [S T K 2 + ], where the three expectations above, in the order they appear, correspond to the forward price of the underlying asset, of the put option payoff and on the call option payoff. b The payoff function of the risk reversal is gx = x A + B x +. If S 0 denotes the current spot price, then the statement that both options are out of the money means that Date: 28 November 204.

2 2 HENRIK HULT, FILIP LINDSKOG S 0 < A and S 0 > B, i.e. B < A. Set K = B and K 2 = A and notice that draw the figure! f x = x gx. In particular, the forward prices are related as Exercise.3. Sports betting E Q [gs T ] = E Q [S T ] E Q [ f S T ]. The best available odds are 4.70 on Everton, 3.70 on draw, and.95 on Manchester City. Using these odds and betting 23 on Everton, 27 on draw and 53 on Manchester City is an arbitrage opportunity. Indeed, placing these bets costs = 997 and pays = 00. if the outcome is Everton = if the outcome is draw, and = if the outcome is Manchester City. Exercise.4. Lognormal model a Straightforward computations give E[e az IZ > b}] = e az e z2 /2 dz = e a2 /2 e z a2 /2 dz = e a2 /2 e w2 /2 dw b 2π b 2π b a 2π = e a2 /2 Φa b, where in the last step we used the relation Φx = Φ x. b From the result in a we obtain [ E[R c + ] = E e µ+z ci Z > logc µ }] [ = e µ E e Z I Z > logc µ }] µ logc cφ = e µ+ 2 /2 Φ + µ logc µ logc cφ. Similarly, [ E[R c 2 +] = E e 2µ+2Z 2ce µ+z + c 2 I = e 2µ+2 2 Φ 2 + µ logc 2ce µ+ 2 /2 Φ + µ logc Z > logc µ + c 2 Φ Combining the expressions for E[R c + ] and E[R c 2 +] gives }] µ logc VarR c + = E[R c 2 +] E[R c + ] 2 = e 2µ+2 2 Φ 2 + µ logc 2ce µ+ 2 /2 Φ + µ logc µ logc + c 2 Φ e µ+ 2 /2 Φ + µ logc µ logc 2. cφ.

3 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 3 Similarly, [ E[RR c + ] = E e 2µ+2Z ce µ+z I Z > logc µ }] = e 2µ+2 2 Φ 2 + µ logc ce µ+ 2 /2 Φ + µ logc which leads to CovR,R c + = E[RR c + ] E[R]E[R c + ] = e 2µ+2 2 Φ 2 + µ logc ce µ+ 2 /2 Φ + µ logc e µ+ 2 /2 e µ+ 2 /2 Φ + µ logc µ logc cφ. Finally, [ E[R c + R d + ] = E R 2 c + dr + cdi Z > logd µ }] This leads to = e 2µ+2 2 Φ 2 + µ logd c + de µ+ 2 /2 Φ µ logd + cdφ. + µ logd CovR c +,R d + = E[R c + R d + ] E[R c + ]E[R d + ] = e 2µ+2 2 Φ 2 + µ logd c + de µ+ 2 /2 Φ + µ logd µ logd + cdφ e µ+ 2 /2 Φ + µ logc µ logc cφ e µ+ 2 /2 Φ + µ logd µ logd dφ Exercise.5. Risky bonds a We obtain the risk-free rates from the cash flows of bonds A and B by solving 98 = 00e r, 04 = 5e r + 05e 2r 2. The discount factors are d = e r = 0.98 and d 2 = e 2r and the zero rates are r and r The credit spreads are obtained by solving 93 = 00e r +s, 98 = 0e r +s + 0e 2r 2+s 2 for s and s 2. We find that s and s Let I and I 2 denote the default indicators for the risky bank over a one- and two-year period, respectively. Denote the corresponding unknown default probabilities by q and q 2 and solve 93 = 00e r q, 98 = 0e r q + 0e 2r 2 q 2 for q and q 2. We find that q = 5/ and q

4 4 HENRIK HULT, FILIP LINDSKOG b For the investor it only makes sense to invest in Bond C and Bond D since the investor believes that they cannot default and, hence, are underpriced. Let w [0,0 4 ] be the amount invested in Bond C. Notice that at time the investor invests any cash flow in Bond D which at that time is a -year bond with the random price 0e r+s, where r and s are independent and normally distributed. Given the assumption of no defaults the cash flow at time is w w 0 98 and the random cash flow at time 2 is w er+s w 98 er+s + 0. Since r +s is normally distributed with mean 0.6 and variance the expected value of e r+s is e The expected cash flow at time 2 can be written c 0 + c w with 0 < c In particular, the expected cash flow is maximized by choosing w = 0 4 : all money goes into Bond C. c Here it is assumed that the investor has a correct view of how the market will price Bond D at time but is wrong in assuming that the bonds are non-defaultable. The default probabilities provided by the market prices at time 0 are correct. Investing everything in Bond C at time 0 and, at time, re-investing everything in Bond D gives that cash flow 0 4 I 00/93 at time and 0 4 I 2 e r+s 00/93 at time 2, where r and s are independent and independent of the default indicators. Plot the corresponding distribution function and simulate from the distribution of I 2,r,s to produce a histogram illustration the distribution of the terminal value of the investors strategy under the above assumptions. CHAPTER 3 Exercise 3.. Annuity a Let τ be the random year of death of the policy holder, where τ = means death of the policy holder within one year from today. The annuity contract gives the policy holder the cash flow C k } k, where C k = 0 k < y, c k y,τ > k, 0 k y,τ k. The value today V 0 of the annuity cash flow is [ ] V 0 = E ce kr k Iτ > k} = c k=y k=y e kr k Pτ > k. From the Gompertz-Makeham formula for the mortality rate, µ 0 x + u = A + Re αx+u, at age x + u of an age-x policy holder we find that k } Pτ > k = exp A + Re αx+u du = exp 0 Ak Reαx e αk }. α

5 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 5 b Here, c = 5000, y =, r k = 0.04 for all k, x = 65, A = 0.002, R = e 2, and α = 0.2. Inserting the numerical values into V 0 n = c n k=y e kr k exp Ak Reαx e αk }. α gives V $5,067 and V 0 30 $5,05. The corresponding values of the truncated, at n = 200 and n = 30, annuity payments to a hypothetical immortal policy holder are approximately $22, 476 and $85, 65, respectively. Notice the effect of the mortality rate and that possible annuity payments beyond the age of 95 for the mortal 65-year-old policy holder do not affect the current price of the annuity. Exercise 3.2. Hedging with index futures a If the day-to-day interest is deterministic and r t,t = r 0, for all t, then 00/B 0 = e r 0,+ +r t,t = e 365r 0,, from which r 0, = /365logB 0 /00 follows. With Y 0 = 00/B 0, the leverage of the futures strategy that corresponds to the quadratic hedge is h = CovS T K +,S T Y 0 VarS T Y 0 = CovS T K +,S T. Y 0 VarS T Since S T = explog W}, the solution to Exercise.4 gives formulas for CovS T K +,S T and VarS T. Inserting numerical values gives h = The variance of the hedging error is VarS T K + hy 0 S T = VarS T K + CorS T K +,S T 2. We can compute the correlation CorS T K +,S T = CovS T K +,S T VarST K + VarS T by inserting the expressions obtained in the solution to Exercise.4. The computations result in the value for the standard deviation of the hedging error. The quadratic hedge implies a bond position that pays h 0 = E[S T K + ] hy 0 E[S T ] at time T in one year. Equivalently, w 0 = h 0 /Y 0 is the investment in the risk-free bond that corresponds to the quadratic hedge. Again using the expressions from the solution of Exercise.4 to compute h 0 and w 0 gives h 0 and w 0 = b With h and h 0 from a, the hedging error is h S T + h 0 S T 0 +, S T = 00e W, where W is standard normally distributed. Therefore, a sample W,...,W n } from N0, is easily transformed into a sample from the distribution of the hedging error. c Now the time-t value of the long futures strategy with unit leverage is Y S T, where Y = e Z and S T = 00e W with Z and W being independent and standard

6 6 HENRIK HULT, FILIP LINDSKOG normally distributed. Moreover, the money market account is included as a hedging instrument. From the independence of Y and S T we find that CovS T K +,Y S T = E[Y ]CovS T K +,S T, CovY S T,S T = E[Y ]VarS T, VarY S T = VarY VarS T + VarY E[S T ] 2 + VarS T E[Y ] 2. So the covariance matrix of the hedging instruments Y S T,S T is VarY ST CovY S Σ Z = T,Y CovY S T,Y VarY Similarly, the covariances between liability and hedging instruments are CovST K Σ LZ = +,Y S T CovS T K +,Y 0 Using Proposition 3.2 we get the positions in the stochastic hedging instruments: h,h 2 = 0.352, , where the first hedging instrument is the futures strategy with unit leverage and the second instrument is on dollar invested in the money market account. The standard deviation of the hedging error is only slightly higher than before. The position in the zero coupon bond is 0.02 number of bonds with face value 00. Exercise 3.3. Leverage and margin calls a An arbitrage portfolio is obtained by adopting the following strategy: At time 0 take a short position in h forward contracts and a long position in h futures contracts. The net payment is 0. 2 At t = you receive hf F 0 from the futures contract. Put this into the money market account with zero interest rate. If it is negative you borrow the same amount. The net cash flow is then again zero. 3 At t = 2 the forward contracts gives hg 0 S 2, the futures contract hs 2 F and the money market account hf F 0. Thus the total payoff is hg 0 F 0 > 0. b If hf F 0 < K you have to borrow at the high interest rate R. Then you need to pay back the interest at t = 2, which is [hf 0 F K]e R. Thus, the total payoff at t = 2 is V 2 = hg 0 F 0 [hf 0 F K]e R IhF F 0 < K} = hg 0 F 0 [hf 0 F K] + e R c The expected value of V 2 is given by E[V 2 ] = hg 0 F 0 he R E[F 0 K/h F + ]. The last expectation is identified as the forward price of a put option on F with strike F 0 K/h which can be written as P 0 F 0 K/h, where with P 0 x = xφ d 2 F 0 Φ d, d = logf 0/x + 2 /2, d 2 = d.

7 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 7 The maximum expected value is reached for h = 455 number of forward contracts. Exercise 3.4. Immunization The discount factors corresponding to the cash flow times 0.5,,.5 and 2 years from today are determined as the solution to the equation system equation.3 on page = 00d = 2d d.55 = 6d d + 06d = 4d d + 4d d 2. The solution is d 0.5,d,d.5,d ,0.9680,0.948, The relation r t = logd t /t transforms the discount factors are transformed into the zero-rates r 0.5,r,r.5,r , , , Other zero-rates are assumed to be given by linear interpolation from those above. In particular, the 20-month rate is r 5/3 = r.5 + r 2 r.5 5/ It follows from Remark 3. on page 72 that it is sufficient to use to bonds to make the aggregate position immune against parallel shift in the zero-rate curve, as long as one of the bonds has a duration shorter than that of the liability, D = 5/3, and the other has a duration longer than that of the liability. Here, D C = P C d d d D D = P C d d d d A solution to the immunization problem is therefore hc = h D P C P D D D D C where P = 0 5 e r 5/3 5/ Exercise 3.5. Delta hedging with futures Exercise 4.2. Sports betting PD PD D D PP C D D C CHAPTER The initial capital is V 0 = 00 British pounds and the prices of the contracts paying pound if the gamblers selected result were to happen are q = /2.50, q x = /3.25 and q 2 = /2.70, respectively. The gambler believes p = p x = p 2 = /3. Buying one contract of each type costs q + q x + q 2 =.078 > and pays at maturity whatever happens: a synthetic risk-free bond. Here, this amounts to a guaranteed loss. Anyway, we have a risk-free asset with return R 0 = /q + q x + q 2 <. The three risky assets have returns R = q I Chelsea }, R x = q x I draw }, R 2 = q 2 I Liverpool }. However, the risky returns are linearly dependent: one of them can be expressed as a a constant minus a linear combination of the other two. In particular, the covariance matrix,

8 8 HENRIK HULT, FILIP LINDSKOG of the return vector is not invertible. Therefore we select two of the three risky assets as our risky assets and consider the trade-off problem with the above risk-free asset the investment problem 4.7 on page 92. Set where µ = µ µ 2, Σ = Σ, Σ,2 Σ 2, Σ 2,2 µ = E[R x ] = p x /q x, µ 2 = E[R 2 ] = p 2 /q 2, Σ, = VarR x = p x p x /q 2 x, Σ 2,2 = VarR 2 = p 2 p 2 /q 2 2, Σ,2 = CovR x,r 2 = p x p 2 /q x q 2, Σ 2, = Σ,2. The solution w and w 0, where w = w,w 2 T, to the trade-off problem 4.7 with trade-off parameter c is see 4.8 on page 92 w = V 0 c Σ µ R 0, w 0 = V 0 w T. Since we have chosen R x and R 2 as risky returns, the solution in terms of the capital invested in the outcomes Chelsea, draw and Liverpool is The numerical values are w w 2 = c q w = w 0, q + q x + q 2 q x w x = w 0 + w q + q x + q, 2 q 2 w 2 = w 0 + w q + q x + q ,, w 0 = 00 c In particular, for c =, w,w x,w ,33.32, As c highly risk-averse, w,w x,w ,28.54, For c = / smallest trade-off parameter that corresponds to long positions, w,w x,w 2 0,72.3, The expected value and the variance of the optimal portfolio are E[V ] = w 0R 0 + w T µ, VarV = w T w. Plotting the pairs VarV,E[V ] for c > / illustrates the efficient portfolio frontier. Exercise 4.3. Uncorrelated returns The investment problem and its solution are given by 4.9 and 4., respectively, with R 0 =, V 0 = 0,000 and 0 V 0 = 30. The matrix Σ is a diagonal matrix with diagonal entries Σ k,k = k 2. The numerical solution is approximately w 0,w,w 2,w 3,w 4,w ,67,335,224,68,34. Exercise 4.4. Hedging a zero-coupon bond a The portfolio value in six months is V 6 = wr+w 0 L, where L = $0,000, w+w 0 V 0 = $9,700 and R = 0,000 9,50 e µ+z/4,

9 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 9 where µ = 0.06, = 0.05 and Z is standard normally distributed. Since E[V 6 ] can be increased by increasing w 0 without increasing VarV 6, w+w 0 = V 0 for the optimal investment. Therefore, the hedging problem amounts to Since minimize w 2 VarR, subject to we[r] +V 0 w L 0. E[R] = 0,000 9,50 e µ/4+/42 /2 = the constraint is equivalent to w L V 0 /E[R] = and w is chosen as w = L V 0 /E[R] = in order to minimize the portfolio variance. b For a lognormal random variable e a+bz, E[e a+bz ] = e a+b2 /2 and E[e 2a+bZ ] = e 2a+2b2 and therefore Vare a+bz = e 2a+b e b. Here, this gives VarR = and VarR = The efficient frontier is illustrated by plotting the pairs w VarR,wE[R] +V 0 L for varying values of w. Exercise 4.5. Hedging stocks with options Exercise 4.6. Credit rating migration a Let w and w 2 = V 0 w, w [0,V 0 ], be the amounts invested in the two bonds, where V 0 = $0,000. The value of the bond portfolio in one year is V = w e 0.06 s +0.02Z + 0,000 w e 0.06 s Z = 00e Z w e s + 0,000 w where s,s 2 is independent of the standard normal Z and has a distribution specified by Table 4.. The expected value E[V ] and variance E[V 2 ] E[V ] 2 of the portfolio are computed from and E[V ] = 00e /2 E[V 2 ] = 00 2 e /2 4 i, j= 4 i, j= w e r i + 0,000 w w e r i + 0,000 w where r i,r j denotes the i, j entry in Table 4.. Exercise 4.7. Insurer s asset allocation e s 2, e r j Ps,s 2 = r i,r j e r j 2 Ps,s 2 = r i,r j, We want the solution to a convex optimization problem of the type 2.: here f w = 2 wt Σw + 2 L 2w T Σ L,R, g w g,0 = w T µ +.3E[L], g 2 w g 2,0 = w T.2E[L].

10 0 HENRIK HULT, FILIP LINDSKOG The sufficient conditions for an optimal solution in Proposition 2. translate into w = Σ Σ L,R + λ µ λ 2, T w = T Σ Σ L,R + λ T Σ µ λ 2 T Σ =.2E[L], µ T w = µ T Σ Σ L,R + λ µ T Σ µ λ 2 µ T Σ =.3E[L] so, with a = µ T Σ µ, b = µ T Σ and c = T Σ, b c λ.2e[l] = T Σ Σ L,R a b λ 2.3E[L] µ T Σ Σ L,R which gives λ,λ , Since the λ k s are positive we have obtained an optimal solution to the optimisation problem without any constraints on long/short positions. However, the solution is w w w which shows that the solution luckily corresponds to an optimal solution to the optimisation problem with a requirement of only long positions. If the first asset is uncorrelated with the liability, then Σ L,R = 0, and again we obtain positive λ k s, λ,λ , , and the solution w w 2 w In particular, the first asset becomes less attractive when it cannot be used to hedge the liability. CHAPTER 5. Exercise 5.. Credit Default Swap a First observe that buying the defaultable bond and the CDS results in a risk-free payoff of $00 in 6 months which costs of $98. Therefore, a rational investor would not invest in the risk-free bond which costs $99. Let w and w 2 be the amounts invested in the defaultable bond and in the CDS, respectively. The problem to solve is maximize E [ uw R + w 2 R 2 ] subject to w + w 2 00, w,w 2 0, where R = 00 I/96 and R 2 = 00I/2 with if the bond issuer defaults, I = 0 otherwise. Since ux = x and PI = = 0.02, E [ uw R + w 2 R 2 ] = w w

11 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 Since E [ uw R + w 2 R 2 ] is increasing in both w and w 2, w + w 2 = 00 for the optimal solution. Therefore the problem to solve simplifies into maximize w w subject to w [0,00] which gives w = $ / $98.04 and w 2 $.96. b Exercise 5. b does not make sense. It may read as follows instead: Another investor is an expected-utility maximizer with utility function ux = x β for β 0,, and invests $00 in long positions in the defaultable bond and the risk-free bond. Also this investor believes that the default probability is 0.02 and decides to invest less than $50 dollars in the defaultable bond. What can be said about β? Here, let w and w 2 denote the amounts invested in the defaultable bond and in the risk-free bond, respectively. Notice that E[uw R + w 2 R 2 ] = w w 00 β 2 00 β 2 + w Since E[uw R + w 2 R 2 ] is increasing in both w and w 2, w + w 2 = 00 for the optimal solution. Therefore the problem to solve simplifies into maximize w w subject to w [0,00] β w Denoting the above objective function by gw, setting g w = 0 and solving for w yield w β = 00 49/32/β + 49/32 /β /32. Observe that w β is increasing in β 0, with lim β w β = 00 and lim β 0 w β = 7/49. Setting w β = 50 and solving for β gives β = + log49/32 log32/65 = We conclude that the parameter β of the investor s utility function satisfies β < β. Exercise 5.2. Bets on credit rating Introduce indicator variables X,X 2,X 3,X 4 with X = if the rating is Excellent in 6 months and zero otherwise, and similarly for X 2, X 3 and X 4 if the rating is Good, Poor or Default, respectively. We have four contracts with current and future values given by S0 =,50 S 6 = 0,000X, S0 2 = 800 S2 6 = 0,000X 2, S0 3 = 700 S3 6 = 0,000X 3, S0 4 = 50 S4 6 = 0,000X 4. Let q k = S0 k /0,000 denote the reciprocal odds of outcome k. The optimization problem to solve can be formulated as follows. maximize E[uw q X + w 2 q 2 X 2 + w 3 q 3 X 3 + w 4 q 4 X 4] subject to w + w 2 + w 3 + w 4 0,000, w,w 2,w 3,w 4 0, β

12 2 HENRIK HULT, FILIP LINDSKOG where ux = γx /γ /γ with γ = 2.5. We identify the investment problem as the Horse race problem 5.2 on page 39 and therefore the solution is given by 5.3 on page 40, i.e. q k p k /q k 2.5 w k = 0,000 4 j= q jp j /q j 2.5 with p = 0., p 2 = 0.80, p 3 = 0.08, p 4 = 0.0. Inserting the numerical values of the p k s and q k s gives in dollars w 00., w , w , w Exercise 5.3. Hedging with electricity futures Exercise 5.4. Optimal payoff function The optimal payoff is given by, see 5.6, hx = u λ qx, px where λ is such that V 0 = B 0 u λ qx qxdx. px Here u y = y γ τ/γ. Since qx/px = expλθ θx}, λ must satisfy the equation V 0 = γ λ γ e γλθ θx τ qxdx B 0 = ] [λ γ e γλθ e γθx qxdx τ γ = [ ] λ γ e γλθ+λθγ τ. γ Therefore, and the resulting optimal payoff is hx = γ = γ [ λ γ V0 = [ V0 γ + τ B 0 [ V0 γ + τ B 0 B 0 γ + τ Exercise 6.. Convexity and subadditivity ] e γλθ Λθγ ] e γλθ Λθγ γλθ+γθx τ ] e γθx Λθγ τ. CHAPTER 6 Subadditivity and positive homogeneity imply that, for any λ [0,], ρλx + λx 2 ρλx + ρ λx 2 = λρx + λρx 2. Positive homogeneity and convexity imply that ρx + X 2 = ρ 2 2 X + 2 X 2 = 2ρ 2 X + 2 X 2 ρx + ρx 2. Exercise 6.2. Stop-loss reinsurance

13 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 3 Since L = mins,fs p we find that F L FS p = and F L FS p ε < 0.95 for ε > 0. In particular, We conclude that p = F S Exercise 6.3. Quantile bound FL 0.99 = minm : F Lm 0.99} = FS p. Exercise 6.4. Tail conditional median Exercise 6.5. Production planning Exercise 6.6. Risky bonds 0.99 FS 0.95 gives FL 0.99 = F S a Let I and I 2 be the default indicators for the two issuers. They are assumed to be independent and identically distributed, I = with probability p and I = 0 with probability p. The returns of the two bonds are then given by The expected value of R k is and the variance is R k = 05 P k I k = R 0 q I k, k =,2. µ = E[R k ] = R 0 q E[I k] = R 0 p =.05, k =,2, q 2 = V R k = V R0 q I k = R2 0 p p = q 2 Since the default indicators are independent, so are the returns R and R 2, which implies CovR,R 2 = 0. Let µ T = µ, µ be the mean vector of R,R 2 T and Σ be the covariance matrix given by 2 0 Σ = 0 2 Let w 0 be the amount invested in the risk-free bond and w T = w,w 2 the amounts invested in the two defaultable bonds, respectively. The objective is to solve. maximize w 0 R 0 + w T µ, subject to w T Σw V , w 0 + w T V 0, w 0 0,w 0,w 2 0. Here V 0 = 0 6 is the initial capital and V 0 0 = If we, for now, ignore the short-selling constraints, then the sufficient conditions for optimality are R 0 λ 2 = 0 and µ + λ Σw + λ 2 = 0, 2 w T Σw V and w 0 + w T V 0 3 λ 0 and λ 2 0, 4 λ w T Σw V = 0 and λ 2w 0 + w T V 0 = 0.

14 4 HENRIK HULT, FILIP LINDSKOG Assuming λ > 0 and λ 2 > 0 leads to λ 2 = R 0, w = λ Σ µ R 0, by and using the first condition in 4 gives Then we solve for λ which gives and We can compute V = λ 2 µ R 0 T Σ µ R 0. λ = V 0 0 µ R 0 T Σ µ R 0 /2 V 0 0 w = /2 Σ µ R 0, µ R 0 T Σ µ R 0 w 0 = V 0 w w 2. and putting in the numerical values gives Σ / 2 0 = 0 / 2, w = w 2 = , w 0 = Since the solution to the optimization problem without short-selling constraints actually satisfies the short-selling constraints we conclude that this is the optimal solution to the problem. b The mean and standard deviation of the optimal portfolio are given by w 0 R 0 + w + w 2 µ = 05023, w T Σw = c Let X = V V 0 R 0 be the net worth. Then the discounted loss is The distribution of I + I 2 is given by L = X/R 0 = R 0 w 0 R 0 + w R + R R 0 = 0 6 w 0 w R 0 R + R 2 = 0 6 w 0 w R 0 R 0 q I + I 2 = 0 6 w 0 2w + w q q I + I 2. PI + I 2 = 0 = p 2, PI + I 2 = = 2p p, PI + I 2 = 2 = p 2,

15 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 5 and the quantile function is therefore given by 0, if u p 2, FI +I 2 u =, if p 2 < u p 2, 2, if p 2 < u. The Value-at-Risk is then given by VaR u X = F u = L 0 6 w 0 2w q + w = 0 6 w 0 2w + w q q q F I +I 2 u 0, if p 2 u,, if p 2 u < p 2, 2, if u < p 2. With u = 0.05 and p = we have p 2 = < 0.05 and p 2 = and therefore VaR 0.05 X = 0 6 w 0 2w = 5500 q The Expected Shortfall can be computed as ES 0.05 X = 0.05 = VaR u Xdu 0 6 w 0 2w q = w 2p p 2 p q d Let I = with probability p = 0.9 and I 2 unchanged. Then and F PI + I 2 = 0 = p p = , PI + I 2 = = p p + p p = , PI + I 2 = 2 = p p = , I +I 2 u = The Value-at-Risk is then given by VaR u X = F u = L 0 6 w 0 2w q 0, if u p p,, if p p < u p p, 2, if p p < u. + w = 0 6 w 0 2w + w q q In particular, with u = 0.05, VaR 0.05 X = q F I +I 2 u 0, if p p u,, if p p u < p p, 2, if u < p p. 0 6 w 0 2w + w q q =

16 6 HENRIK HULT, FILIP LINDSKOG The Expected Shortfall is given by ES 0.05 X = VaR u Xdu = 0 6 w0 2 w/ q + = Exercise 6.7. Leverage and margin calls a Recall from the solution to Exercise 3.3 that w 2p p p p 0.05 q V 2 = hg 0 F 0 [hf 0 F K]e R IhF F 0 < K} = hg 0 F 0 [hf 0 F K] + e R, where F = F 0 exp 2 } 2 + Z with Z standard normal, F 0 = 99.95, G 0 = 00, = 0.6, = /2, R = 0.24, K = 0 4 and h = 455. We find that PhF 0 F K > For p PhF 0 F K > 0, VaR p V 2 = F V 2 p = e R F hf 0 F K p hg 0 F 0 = e R hf F 0 F p K hg 0 F 0 = e R hf 0 + F F p K hg 0 F 0 = e R hf 0 + FF p K hg 0 F 0, where FF p = F 0 exp 2 } 2 + Φ p. Summing up, we have found that, for p 0.05, where VaR p V 2 = a 0 + a FF p, a 0 = e R hf 0 K hg 0 F 0, a = e R h. b Since ES p V 2 = p p 0 VaR uv 2 du and F F p = expm+sφ p} with m = logf 0 2 /2 and s =, Example 6.5 yields, for p 0.05, with coefficients a 0 and a as in part a. Exercise 6.8. Risk and diversification See Figure. ES p V 2 = a 0 + a ΦΦ p se m+s2 /2 = a 0 + a F 0 ΦΦ p

17 RISK AND PORTFOLIO ANALYSIS: SOLUTIONS TO EXERCISES IN CHAPTERS -6 7 VaR Index ES Index VaR and ES Index FIGURE. Plots of n VaR 0.05 V n V 0 and n ES 0.05 V n V 0.