15.450 Exam Preparation, Practice Problem Solutions

M.I.T. Sloan School of Management 15.450 Professor Leonid Kogan Solutions for Practice Problems 1. Consider a 3-period model with t = 0, 1, 2, 3. There are a stock and a risk-free asset. The initial stock price is $4 and the stock price doubles with probability 2/3 and drops to one-half with probability 1/3 each period. The risk-free rate is 1/4. (a) Compute the risk-neutral probability at each node. Solution: Let q denote the risk-neutral probability of up-node and 1 − q denote risk-neutral probability of the down-node. Then by the definition of the risk-neutral probabilities, � � 1 Q St+1 St = E 1+r � � �� 1 1 St = q (2St ) + (1 − q) 1+r 2 with r = 14 . Solving this equation gives q = 12 . Notice that this calculation holds true at every non-terminal node. We conclude that the risk-neutral probability at each node is given by 12 probability of up-node and 12 probability of down-node. (b) Compute the Radon-Nikodym derivative (dQ/dP) of the risk-neutral measure with respect to the physical measure at each node. Solution: The original (physical) measure assigns probabilities pu = 23 and pd = 13 to the up- and down-node, respectively. The risk-neutral measure assigns probabilities qu = 12 and qd = 12 . The Radon-Nikodym derivative at each node is a random variable that takes on the value � � 1/2 3 dQ qu (u) = = = dP pu 2/3 4 in the up-node and the value � dQ dP � (d) = qd 1/2 3 = = pd 1/3 2 in the down-node. Again, notice that this calculation is valid at each and every node. (c) Compute the state-price density at each node. Solution: Fix the current node at time t and let the state-price density at this node be denoted by πt . Denoting the values of state-price densities at the childen nodes by 1 πt+1 (u) and πt+1 (d), we have the following pricing equations: 2 πt+1 (u) 1 πt+1 (d) 1 · 2St + · St 3 πt 3 πt 2 � � 4 πt+1 (u) 1 πt+1 (d) = + St 3 πt 6 πt 2 πt+1 (u) 5 1 πt+1 (d) 5 1 = · + · 3 πt 4 3 πt 4 � � 5 πt+1 (u) 5 πt+1 (d) = + 6 πt 12 πt St = Solving this system of equations in the unknowns πt+1 (u) πt and πt+1 (d) πt , we get the solution πt+1 (u) 3 = πt 5 πt+1 (d) 6 = πt 5 Now, starting at the initial node at time t = 0 and setting π0 = 1 allows us to solve for the state-price density at every node recursively. This calculation leads us to conclude that at a node ω whose history consists of i up-movements and j down-movements, the state-price density is given by � �i � �j 3 6 πt (ω) = 5 5 (d) Price a lookback option with payoff at t = 3 equal to (max0≤t≤3 St ) − S3 using risk-neutral probability. Solution: The following binomial tree describes the evolution of stock price and the bold-face numbers next to the final stock price are the payoffs from the lookback option: 2 $32, $0 $16 $8, $8 $8 $8, $0 $4 $2, $6 $4 $8, $0 $4 $2, $2 $2 $2, $2 $1 $0.5, $3.5 Recalling from part (a) that the risk-neutral probabilities of an up-movement and a � �3 down-movement are both 12 , all terminal nodes have the same Q-probability of 12 = 1 . Therefore the price of the lookback option is given by 8 �3 1 E Q [D3 ] 1+r � �3 � � 4 1 7 · 8+6+2+2+ 5 8 2 64 1 43 · · 125 8 2 172 125 � C = = = = (e) Price the lookback option using state-price density and compare your answer to (d). Solution: Using the state-price density we computed in part (c), we can calculate the price of the lookback option as C = E [π3 D3 ] � = P (ω) π3 (ω) D3 (ω) ω where the sum is across all the possible time-3 nodes ω. But note that for any ω, denoting the number of up-movements by i and the number of down-movements by j, 3 we have P (ω) π3 (ω) = = = = �� � � � � �� � � � � i j i j 1 6 2 3 3 3 5 5 � �i � �j 2 2 5 5 � �i+j 2 5 � �3 2 5 Therefore, C = � P (ω) π3 (ω) D3 (ω) ω � �3 � 2 = D3 (ω) 5 ω � �3 � � 2 7 = 8+6+2+2+ 5 2 172 = 125 Of course, we get the same answer as we did in part (d) using the risk-neutral proba­ bilities. 2. Show that, under the risk-neutral measure, the discounted gain process t Ĝt = Pt � D s + Bt s=1 Bs � � is a martingale (i.e. EtQ Ĝt+1 = Ĝt ) from the definition of risk-neutral measure in lecture notes � T � � B t Pt = EtQ Du Bu u=t+1 That is the reason why the risk-neutral measure is also called the ”equivalent martingale measure” (EMM). � � Solution: We want to show EtQ Ĝt+1 = Ĝt . 4 EtQ � � t+1 � � � Ds Q Pt+1 Ĝt+1 = Et + Bt+1 s=1 Bs Now recall that � Pt+1 = EtQ+1 � T � Bt+1 Du B u u=t+2 Substituting this into the first expression, we have � � t+1 � � � D P t+1 s EtQ Ĝt+1 = EtQ + Bt+1 s=1 Bs � � T � t+1 � � Bt+1 � Ds 1 Q Q = Et E Du + Bt+1 t+1 u=t+2 Bu Bs s=1 � � T � t+1 � � Du � Ds Q = EtQ Et+1 + Bs Bu u=t+2 s=1 � T � � Ds = EtQ Bs s=1 � T � t � � D Ds u = EtQ + Bu Bs u=t+1 s=1 � T � t � 1 Q � Bt Ds = Et Du + Bt Bu Bs u=t+1 s=1 t Pt � D s = + Bt s=1 Bs = Ĝt This shows that Ĝt is a martingale under the risk-neutral measure. 3. Consider the following model of interest rates. Under the physical probability measure P, the short-term interest rate is exp(rt ), where rt follows drt = −θ(rt − r) dt + σr dZt , where Zt is a Brownian motion. Assume that the SPD is given by � � t � πt = exp − ru − 0 0 5 t 1 2 η du − 2 u � t � ηu dZu 0 where ηt is stochastic, and follows dηt = −κ(ηt − η) dt + ση dZtη where Ztη is a Brownian motion independent of Zt . (a) Derive the dynamics of the interest rate under the risk-neutral probability Q. Solution: From the form of SPD πt , the price of risk at time t is ηt . Therefore under the risk-neutral measure Q, dZtP = dZtQ − ηt dt and the shock that drives the price of risk process, dZtη remains unchanged under Q. Thus, the dynamics of the risk-free rate rt can be written as drt = −θ (rt − r̄) dt + σr dZt � � Q = −θ (rt − r̄) dt + σr dZt − ηt dt = (−θ (rt − r̄) − σr ηt ) dt + σr dZtQ (b) Compute the spot interest rates for all maturities. (Hint: look for bond prices in the form P (t, T ) = exp(a(T − t) + b(T − t)rt + c(T − t)ηt )). Solution: Suppose the current time is t and assume that the bond matures at time T . Denote its price P (t, T ). By the definition of risk-neutral measure Q, � � � T �� Q P (t, T ) = Et exp − rs ds t Now let us guess the bond price P (t, T ) in the functional form P (t, T ) = exp (a (T − t) + b (T − t) rt + c (T − t) ηt ) On the one hand, by the definition of the risk-neutral measure, we know that � � Q dP (t, T ) Et = rt dt P (t, T ) On the other hand, we can calculate the same expression using Ito’s Lemma: dP (t, T ) = − (a� (T − t) + b� (T − t) rt + c� (T − t) ηt ) dt P (t, T ) � � +b (T − t) (−θ (rt − r̄) − σr ηt ) dt + σr dZtQ � � +c (T − t) −κ (ηt − η̄) dt + ση dZtη,Q 1 1 + (b (T − t))2 · σr2 dt + (c (T − t))2 · ση2 dt 2 2 6 and therefore, � � Q dP (t, T ) Et = − (a� (T − t) + b� (T − t) rt + c� (T − t) ηt ) dt P (t, T ) +b (T − t) (−θ (rt − r̄) − σr ηt ) dt + c (T − t) (−κ (ηt − η̄)) dt � 1� + (b (T − t))2 σr2 + (c (T − t))2 ση2 dt 2 Equating this expression to rt dt, we get a system of three ODEs (match the coefficients on constant, rt , and ηt terms): 1 1 0 = −a� (τ ) + θ¯ rb (τ ) + σr2 (b (τ ))2 + ση2 (c (τ ))2 2 2 � 0 = −b (τ ) − 1 − θb (τ ) 0 = −c� (τ ) − σr b (τ ) − κc (τ ) Also, we have a terminal condition that P (T, T ) = 1, so that a (0) = 0, b (0) = 0, and c (0) = 0. Note that the second equation is an autonomous equation in b (τ ) and it is straight­ forward to solve. Given the solution for b (τ ) , we can then solve the third equation for c (τ ). Then, finally, we can solve for a (τ ) from the first equation. The solutions are � 1� b (τ ) = − 1 − e−θτ θ� � � � σr 1 � 1 � −θτ −κτ −κτ c (τ ) = 1−e − e −e θ κ κ−θ and a (τ ) is not reported here for simplicity (it’s not difficult to compute, but the resulting expression is long and messy). The solutions for a (τ ), b (τ ), and c (τ ) complete the characterization of bond price P (t, T ) = exp (a (T − t) + b (T − t) rt + c (T − t) ηt ) (c) Compute the instantaneous expected rate of return on a zero-coupon bond with time to maturity τ . Solution: Suppose T − t = τ , i.e., time to maturity is equal to τ . Then, � � � � � � Q dP (t, T ) P dP (t, T ) P dP (t, T ) Et − rt dt = Et − Et P (t, T ) P (t, T ) P (t, T ) = b (τ ) σr ηt dt To understand this calculation, recall that the only difference between the two instantaneous drifts is dZt = dZtQ − ηt dt and the coefficient in front of dZt in 7 dP (t,T ) P (t,T ) is b (T − t) σr . (d) Show that the slope of the term structure of interest rates predicts the excess returns on long-term bonds. Discuss the intuition. Show that more volatility in the price of risk, η, means more predictability in bond returns. Solution: First, let us discuss what we mean by predictability. In the case that returns are serially i.i.d and independent from any other random variables, there is no predictability because there is no other piece of information that allows us to “predict” the returns. In this question, we have shown in part (b) that the bond prices are of the form P (t, T ) = exp (a (T − t) + b (T − t) rt + c (T − t) ηt ) and in part (c) that the expected excess return on a bond with time to maturity τ = T − t is given by � � dP (t, T ) Et − rt dt = b (T − t) σr ηt dt P (t, T ) When ηt is high, we can see from the first equation that the term structure of interest rates is steep and from the second equation that excess returns on longterm bonds are high. Therefore, when the term structure is steeper, and excess returns on long-term bonds are higher. The intuition is as follows: times of high price of risk ηt are typically thought of as recessions when people are more riskaverse. High price of risk drives down the price of bonds, resulting in high excess returns (the terminal value of the bond is constant at $1 at maturity). Another observation we can make is that more dispersion in the distribution of ηt means more predictability in excess bond returns. Note that in the extreme case when ηt is a constant, our problem reduces to a setting where returns are i.i.d over time and we do not have any predictability (because expected excess bond returns would equal b (T − t) σr η̄, and excess bond returns would be random noise plus this mean). As ηt becomes more variable, the fraction of variation in the excess bond returns due to variation in the mean of the excess returns, b (T − t) σr ηt , becomes larger and we have more predictability. We can show that the variance σ2 of the stationary distribution of ηt is 2κη , so more variability in ηt could be due to higher volatility of shocks (higher ση ) or lower rate of mean-reversion (lower κ). 4. Suppose that uncertainty in the model is described by two independent Brownian motions, Z1,t and Z2,t . Assume that there exists one risky asset, paying no dividends, following the process dSt = µ(Xt ) dt + σ dZ1,t St where dXt = −θXt dt + dZ2,t The risk-free interest rate is constant at r. 8 (a) What is the price of risk of the Brownian motion Z1,t ? Solution: Let the price of risk for Z1,t and Z2,t be ηt = must have � � µ (Xt ) St − σSt 0 η = rSt � η1,t η2,t �T . Then we This gives µ (Xt ) St − ση1,t St = rSt whereas there is no constraint for η2,t . Hence, the price of risk of Z1,t is η1,t = µ (Xt ) St − rSt µ (Xt ) − r = σSt σ (b) Give an example of a valid SPD in this model. Solution: dπt � = −rdt − ηt dZt , π0 = 1 πt Any η2,t such that ηt satisfies the Novikov’s condition will be allowed. The simplest example would be to let η2,t = 0. (c) Suppose that the price of risk of the second Brownian motion, Z2,t , is zero. Char­ acterize the SPD in this model. Solution: dπt = −rdt − η1,t dZ1,t πt Let yt = ln πt , by Ito’s lemma, dyt = 1 1 dπt − (dπt )2 πt 2πt 1 2 = −rdt − η1,t dZ1,t − η1,t dt 2 So yt 1 = −rt − 2 1 πt = e−rt− 2 �t 0 � t 2 η1,t dt � − 0 2 dt− η1.t t η1,t dZ1,t 0 �t 0 η1,t dZ1,t (d) Derive the price of a European Call option on the risky asset in this model, with maturity T and strike price K. Solution: Existence of SPD implies existence of risk-neutral measure. Under risk-neutral measure, all traded assets must have drift r and the volatility is unchanged. So under risk-neutral measure, dSt = rdt + σdZ1Q,t St 9 The physical drift does not matter at all. � � Ct = EtQ e−r(T −t) [ST − K]+ Standard Black-Scholes formula applies. So Ct = St N (d1 ) − Ke−r(T −t) N (d2 ) where d1,2 � � ln (S/K) + r ± 12 σ 2 T √ = σ T 5. Consider a European call option on a stock. The stock pays no dividends and the stock price follows an Ito process. Is it possible that, while the stock price declines between t1 and t2 > t1 , the price of the Call increases? Justify your answer. Solution: Yes, it is possible. For instance, if volatility becomes extremely large once price hits a low level, we could have the situation described in the question. The underlying reason is that call option is an increasing function of both volatility and stock price. So if price drop is accompanied by large increase in volatility, we may a higher call price. The easiest way to see this is through a 3-period binomial tree example: Suppose price at time-0 is 100. At time-1, it may go up to 120 or go down to 10. If price goes up to 120, it may go further up to 140 or down to 100 at time-2. If price goes down to 10, the volatility becomes huge. As a result, price may go up to 10000 or down to 1 at time-2. Let interest rate be 0 and consider an European call with strike at 150 and maturity of two periods. Let’s find the call option price at each node. If price goes up to 120 at time-1, then the call is worth 0 since, under no circumstances at time-2, the call will be in the money (both 140 and 100 are less than strike 150). If price goes down to 10, let’s find the call price by solving the contingent claim prices. Let the time-1 price of 1 unit of payoff at time-2 and state S2 = 10000 be q10000 and the time-1 price of 1 unit of payoff at time-2 and state S2 = 1 be q1 . Since interest rate is 0, we have q10000 + q1 = 1 Since the time-1 stock price is 10, we also have 10000q10000 + q1 = 10 Solving the simultaneous equation, we have q10000 = 1 1110 and q1 = 1111 1111 10 Since the call pays (10000 − 150) = 9850 dollars if price goes up to 10000 at time-2 and 0 dollars if price goes down to 1. We have the call price, at time-1 and state S1 = 10, C10 = 9850q10000 = 9850 ≈ 9. 1111 Let’s solve for time-0 contingent claim prices q120 and q10 of time-1 payoffs. Zero interest rate implies q120 + q10 = 1 Stock price movements implies 120q120 + 10q10 = 100 This gives q120 = 9 2 and q10 = 11 11 Hence, the call price at time-0 is C100 = C10 q10 ≈ declines at time-1 but call price increases. 18 11 << 9 = C10 . So stock price The same principle applies in continuous-time as long as we allow volatility to become large once stock price hits a low level. This is certainly possible since the stock price process follows a general Ito process. 6. Suppose that the stock price St follows a Geometric Brownian motion with parameters µ and σ. Compute � � E0 (ST )λ . Solution: Using Ito’s lemma, we have 1 dStλ = λStλ−1 dSt + λ (λ − 1) Stλ−2 (dSt )2 2 1 λ−1 = λSt (µSt dt + σSt dZt ) + λ (λ − 1) σ 2 Stλ dt � � � 2 � 1 λ 2 = St λ µ + (λ − 1) σ dt + λσdZt 2 Let Xt = Stλ . Then we have dXt = µX dt + σX dZt Xt where � µX σX 1 = λ µ + (λ − 1) σ 2 2 = λσ 11 � Hence, Xt also follows geometric Brownian motion. As a result, 1 2 XT = X0 e(µX − 2 σX )T +σX ZT Therefore, � � E STλ = E [XT ] = X0 eµX T 1 2 = S λ eλ(µ+ 2 (λ−1)σ )T 0 7. Suppose that, under P, the price of a stock paying no dividends follows dSt = µ(St ) dt + σ(St ) dZt St Assume that the SPD in this market satisfies dπt = −r dt − ηt dZt πt (a) How does ηt relate to r, µt , and σt ? Solution: By definition of SPD, we must have πt St as a martingale. By Ito’s lemma, d (πt St ) = πt dSt + St dπt + dSt dπt = µ (St ) πt St dt − rπt St dt − σ (St ) ηt πt St + [...] dZt = πt St (µ (St ) − r − σ (St ) ηt ) dt + [...] dZt To qualify for a martingale process, the drift of πt St must be 0. So µ (St ) − r − σ (St ) ηt = 0 µ (St ) − r ηt = σ (St ) (b) Suppose that there exists a derivative asset with price C(t, St ). Derive the instan­ taneous expected return on this derivative as a function of t and St . Solution: Recall that in general, if Xt is the price of a traded asset with the dynamics dXt = µX dt + σX dZt Xt then the expected excess return is given by � � � � P dXt Q dXt (µX − r) dt = E −E Xt Xt = σX · ηt 12 Now note that the price of our derivative asset C (t, St ) satisfies ∂C (t, St ) ∂C (t, St ) 1 ∂ 2 C (t, St ) dt + dSt + (dSt )2 2 ∂t ∂St 2 ∂St � � ∂C (t, St ) ∂C (t, St ) 1 ∂ 2 C (t, St ) 2 = + (µ (St ) St ) + (σ (St ) St ) dt ∂t ∂St 2 ∂St2 � � ∂C (t, St ) + (σ (St ) St ) dZt ∂St dC (t, St ) = and thus � ∂C (t, St ) ∂C (t, St ) 1 ∂ 2 C (t, St ) 2 + (µ (St ) St ) + (σ (St ) St ) dt ∂t ∂St 2 ∂St2 � � 1 ∂C (t, St ) + (σ (St ) St ) dZt C (t, St ) ∂St = µC (t, St ) dt + σC (t, St ) dZt dC (t, St ) 1 = C (t, St ) C (t, St ) � So we can express the expected excess return on this derivative as � � 1 ∂C (t, St ) ∂C (t, St ) 1 ∂ 2 C (t, St ) 2 µC (t, St )−r = + (µ (St ) St ) + (σ (St ) St ) −r C (t, St ) ∂t ∂St 2 ∂St2 or 1 σC (t, St ) ηt = C (t, St ) � � ∂C (t, St ) µ (St ) − r (σ (St ) St ) · ∂St σ (St ) (c) Derive the PDE on the price of the derivative C(t, S), assuming that its payoff is given by H(ST ) at time T . Solution: Under risk-neutral measure, all traded securities have instantaneous expected return r. So under risk-neutral measure, the stock has drift rSt and the derivative � � � � ∂C ∂C 1 ∂2C 2 2 Q dC Et = + rSt + σ (St ) St dt/C = rdt C ∂t ∂S 2 ∂S 2 We have a PDE ∂C ∂C 1 ∂2C + rSt + σ (St )2 St2 − rC = 0 ∂t ∂S 2 ∂S 2 (d) Suppose that there is another derivative trading, with a price D(t, St ) which does not satisfy the PDE you have derived above. Construct a trading strategy generating arbitrage profits using this derivative, the risk-free asset and the stock. Solution: Suppose for derivative D (t, St ), we have ∂D ∂D 1 ∂2D + rSt + σ (St )2 St2 − rD > 0 ∂t ∂S 2 ∂S 2 13 at some (t, St ). Then consider holding the derivative, shorting ∂D shares and ∂S financing the above position (D − ∂D S ) by borrowing at short rate. The instan­ ∂S t taneous gain is � � ∂D ∂D 1 ∂2D ∂D ∂D 2 dt + dSt + (dSt ) − dSt − D− St rdt ∂S �� 2 ∂S 2 ∂S ∂S �∂t �� �� �� �� � Gain from derivative Gain from stock Interest payment � � 2 ∂D 1∂ D ∂D 2 2 = dt + σ (St ) St dt − D − St rdt ∂t 2 ∂S 2 ∂S � � ∂D ∂D 1 ∂2D 2 2 = + rSt + σ (St ) St − rD dt > 0 ∂t ∂S 2 ∂S 2 This is a riskless arbitrage. We can form a similar riskless arbitrage if ∂D ∂D 1 ∂2D + rSt + σ (St )2 St2 − rD < 0 2 ∂t ∂S 2 ∂S 8. Consider a futures contract with price changing according to Ft+1 = Ft + λ + µt + σF εt , µt+1 = ρµt + σµ ut where εt and ut are independent IID N (0, 1) random variables. Assume that the interest rate is constant at r. Your objective is to construct an optimal strategy of trading futures between t = 0 and T to maximize the terminal objective � � E −e−αWT where WT is the terminal value of the portfolio. Assume the initial portfolio value of W0 . (a) Formulate the problem as a dynamic program. Describe the state vector, verify that it follows a controlled Markov process. Solution: The state vector for this problem is (t, Wt , µt ). Let θt be the control variable that represents the number of futures contracts at time t. Clearly, µt+1 follows a Markov process given the state variable µt . Furthermore, to verify that Wt is a controlled Markov process, note that Wt+1 = θt (Ft+1 − Ft ) + (1 + r) Wt = θt (λ + µt + �t+1 ) + (1 + r) Wt and as such, the conditional distribution of Wt+1 only depends on the state vari­ ables Wt and µt . 14 (b) Derive the value function at T and T − 1 and optimal trading strategy at T − 1 and T − 2. Solution: Let us start from t = T . The value function is simply given by J (T, WT , µT ) = − exp (−αWT ) Now, for t = T − 1, the Bellman equation says J (T − 1, WT −1 , µT −1 ) = max ET −1 [J (T, WT , µT )] θT −1 = max ET −1 [− exp (−αWT )] θT −1 = max ET −1 [− exp (−α (θT −1 (λ + µT −1 + �T ) + (1 + r) WT −1 ))] θT −1 Simple algebra leads to the first order condition θT∗ −1 = λ + µT −1 ασF2 Plugging this into the Bellman equation, we get the value function at time t = T − 1: � � 1 2 J (T − 1, WT −1 , µT −1 ) = − exp − 2 (λ + µT −1 ) − α (1 + r) WT −1 2σF For t = T − 2, the Bellman equation is given by J (T − 2, WT −2 , µT −2 ) = max ET −2 [J (T − 1, WT −1 , µT −1 )] θT −2 � � �� 1 = max ET −2 − exp − 2 (λ + µT −1 )2 − α (1 + r) WT −1 θT −2 2σF where FT −1 = FT −2 + λ + µT −2 + σF �T −1 µT −1 = ρµT −2 + σu uT −1 WT −1 = θT −2 (λ + µT −2 + �T −1 ) + (1 + r) WT −2 We can repeat a similar calculation as before and arrive at the optimal control θT∗ −2 = λ + µT −2 α (1 + r) σF2 9. Suppose you can trade two assets, a risk-free bond with interest rate r and a risky stock, paying no dividends, with price St . Assume St+1 = St × exp(µ + σεt ) where εt are IID N (0, 1) random variables. 15 Assume that whenever you buy the stock you must pay transaction costs, but you can sell stock without costs. Specifically, when you buy X dollars worth of stock, you must pay (1 + τ )X, so the fee is proportional, given by τ . Your objective is to figure out how to trade optimally to maximize the objective � � E −e−αWT where WT is the terminal value of the portfolio. (a) What should be the state vector for this problem? Formulate the problem as a dynamic program, verify the assumptions on the state vector and the payoff function. Solution: The state vector for this problem is (t, Xt , Bt ), where Xt and Bt rep­ resent the dollar value of stock holdings and bond holdings at the beginning of period t. Let λt be the control variable that represents the net increase in the dollar value of the stock holdings as a result of rebalancing at time t. So, for example, if λt > 0, then we invest more in the stock at time t. Note that we incur a transaction cost of τ λt in period t if λt > 0 and none otherwise. Under this definition, the dollar value of stock holdings at the end of the period t is Xt + λt . Then the dynamics of Xt and Bt are given by Xt+1 = exp (µ + σ�t+1 ) (Xt + λt ) Bt+1 = (1 + r) (Bt − λt − τ · max (λt , 0)) It is clear that Xt and Bt are controlled Markov processes because their conditional distributions of Xt+1 and Bt+1 only depend on the state variables Xt and Bt and the control variable λt . Furthermore, as such, these state variables capture all the relevant information regarding the dynamics of the variable of ultimate interest, Wt = Xt + Bt . (b) Write down the Bellman equation. Solution: The Bellman equation is given by J (t, Xt , Bt ) = max Et [J (t + 1, Xt+1 , Bt+1 )] λt where Xt+1 = exp (µ + σ�t+1 ) (Xt + λt ) Bt+1 = (1 + r) (Bt − λt − τ · max (λt , 0)) The Bellman equation in the terminal time period is simply J (T, XT , BT ) = − exp (−α (XT + BT )) 16 10. Suppose we observe returns on N independent trading strategies, rtn , n = 1, 2, t = 1, ..., T . Assume that returns are IID over time, and each strategy has normal distri­ bution: rtn ∼ N (µn , σ 2 ) Assume µ1 > µ2 . (a) Estimate the mean return on each strategy by maximum likelihood. Express µ �n as a function of observed returns on strategy n. Solution: It is a standard calculation (see, for example, lecture notes) to show that the maximum likelihood estimate of the mean of a normal distribution is the sample mean. Hence T 1� n µ̂n = r T t=1 t (b) Since returns are normally distributed, µ �n is also normally distributed. Describe its distribution. (In general, for arbitrary return distribution, µ �n is only approxi­ mately normal). Solution: Since rtn , t = 1, . . . , T , are drawn from IID normal distribution, the sample mean µ̂n is a sum of independent normal variables and is again �normally � 2 distributed. Since rtn ∼ N (µn , σ 2 ), the sample mean µ̂n is distributed N µn , σT . (c) What is the distribution of maxn (µ �n )? characterize it using the CDF function. Solution: Note that µ̂1 and µ̂2 are independent so that the cumulative distribu­ tion function (CDF) of max (µ̂1 , µ̂2 ), denoted Fmax (·), is given by Fmax (x) = P (max (µ̂1 , µ̂2 ) ≤ x) = P (µ̂1 ≤ x) · P (µ̂2 ≤ x) = F1 (x) · F2 (x) � �√ � �√ T (x − µ2 ) T (x − µ1 ) ·Φ = Φ σ σ � � σ2 where we use the fact that µ̂n ∼ N µn , T . (d) Suppose you are interested in identifying the strategy with the higher mean return. You pick the strategy with the higher estimated mean. What is the probability that you have made a mistake? Solution: We are interested in the probability that µ̂2 is greater than µ̂1 (so that we mistakenly infer that the second trading strategy � has2 �the higher mean). Recalling that µ̂1 and µ̂2 are independent and µ̂n ∼ N µn , σT , we know that � µ̂1 − µ̂2 ∼ N 17 2σ 2 µ 1 − µ2 , T � Furthermore, P (µ̂1 ≤ µ̂2 ) = P (µ̂1 − µ̂2 ≤ 0) � � T (µ1 − µ2 ) = 1−Φ 2σ 2 We see that this probability is decreasing in the distance between the true means µ1 − µ2 and decreasing in the number of observations T . On the other hand, this probability is increasing in σ 2 , reflecting the difficulty to estimate the mean when the distribution has large variance. 11. Suppose interest rate follows an AR(1) process rt − r = θ(rt−1 − r) + εt where εt are IID N (0, σ 2 ) random variables. You want to estimate the average rate, r, based on the sample rt , t = 0, 1, ..., T . Assume that we know the true value of θ. (a) Derive the estimate of r by maximum likelihood. Solution: It is easily seen that T � � � � � 2 L r, ¯ θ, σ |r1 , . . . , rT = f rt |r, ¯ θ, σ 2 ; r0 , . . . , rt−1 t=1 so that T � � � � � L r, ¯ θ, σ 2 |r0 , . . . , rT = ln f rt |r, ¯ θ, σ 2 ; r1 , . . . , rt−1 = t=1 T � t=1 � ln � 1 1 √ exp − 2 ((rt − r̄) − θ (rt−1 − r̄))2 2 2σ 2πσ �� In particular, the maximum likelihood estimate of r̄, call it r̂¯, satisfies the first order condition ∂L ∂r̄ T 1−θ � = ((rt − r̄) − θ (rt−1 − r̄)) σ 2 t=1 0 = It follows that T � 1 r̂¯ = (rt − θrt−1 ) T (1 − θ) t=1 18 (b) Show that this estimate is valid even if the shocks εt are not normally distributed, as long as the mean of εt is zero. Solution: Now we assume that �t are independent over time and E [�t ] = 0. Note that T r̄ˆ = � 1 (rt − θrt−1 ) T (1 − θ) t=1 T � 1 = ((1 − θ) r̄ + �t ) T (1 − θ) t=1 T � 1 = r̄ + �t T (1 − θ) t=1 By the Law of Large Numbers, T � 1 �t → 0 T (1 − θ) t=1 in probability, and hence we establish that r̂¯ converges in probability to the true value r̄ and therefore is consistent. (c) Treating εt as IID, derive the asymptotic variance of your estimator of r. Do not use Newey-West, derive the result from first principles. How does the answer depend on θ? Solution: For this part, we maintain that �t are IID over time, E [�t ] = 0, and V ar[�t ] = σ 2 . In the last part, we saw that T � 1 r̂¯ = r̄ + �t T (1 − θ) t=1 and hence √ � � T r̄ˆ − r̄ = T 1 1 � √ �t 1 − θ T t=1 By the Central Limit Theorem, T � � 1 � √ �t ⇒ N 0, σ 2 T t=1 and we conclude that √ � � T r̂¯ − r̄ ⇒ N 19 � 0, σ2 (1 − θ)2 � Note that the asymptotic variance of our estimator is increasing in θ. This makes sense, because in the limiting case where θ = 0, rt are actually IID over time and asymptotic variance of the sample mean is equal to the variance of the shocks. In the other limiting case where θ → 1, we are very close to the unit-root case (random walk) where we do not have mean-reversion and hence estimation of the long-run mean becomes increasingly difficult and imprecise. Moreover, calculating theoretical asymptotic variance of the estimator as in this question provides an alternative to constructing standard errors using the NeweyWest method. There are advantages and disadvantages to each method. The­ oretical asymptotic variance is a convenient way to see inner workings of the estimator (in our example, the dependence of asymptotic variance on the persis­ tence of AR(1) process) with excellent finite sample properties, assuming correct model specification. However, derivation of asymptotic variance of an estimator may not be so straightforward in more complex situations and standard errors obtained this way are more sensitive to model misspecification compared to the more model-independent Newey-West standard errors. 12. Suppose you observe two time series, Xt and Yt . You have a model for Yt : Yt+1 = ρYt + (a0 + a1 Xt ) εt+1 , t = 0, 1, ..., T where εt+1 ∼ N (0, 1), IID. Assume that the shocks εt are independent of the process Xt and the lagged values of Yt . There is no model for Xt . (a) Using the GMM framework, which moment condition can be used to estimate ρ? Solution: We first want to establish that E [(a0 + a1 Xt ) �t+1 |Yt ] = 0 To see why this is true, note that E [(a0 + a1 Xt ) �t+1 |Yt ] = = = = E [a0 + a1 Xt |Yt ] · E [�t+1 |Yt ] E [a0 + a1 Xt |Yt ] · E [�t+1 ] E [a0 + a1 Xt |Yt ] · 0 0 where the first equality follows from independence of the processes Xt and �t , the second equality follows from independence of �t+1 with the lagged values of Xt , and the third equality follows from the assumption that �t+1 ∼ N (0, 1). Now having derived this condition, our usual arguments now allow us to derive the moment conditions E [g (Yt ) (a0 + a1 Xt ) �t+1 ] = 0 for any function g. 20 (b) Argue why it is valid to estimate ρ using an OLS regression of Yt+1 on Yt . Solution: In particular, if we pick g (Yt ) = Yt , then the moment condition be­ comes E [Yt · (a0 + a1 Xt ) �t+1 ] = 0 and our sample analogue is ET [Yt · (a0 + a1 Xt ) �t+1 ] = 0 or ET [Yt · (Yt+1 − ρYt )] = 0 Note that the GMM estimate ρ̂ that solves this sample moment condition is also the OLS estimate from the regression Yt+1 = ρYt + ut where ut represents unspecified error terms. The reason why the GMM estimate of ρ coincides with the OLS estimate is that the sample moment condition for ρ̂ simply says that the residual term Yt+1 − ρY ˆ t is orthogonal to the regressor Yt , and this orthogonality between the fitted residuals and the regressors is the first order condition of the OLS estimate. Hence it is valid to estimate ρ̂ by simply running the regression Yt+1 = ρYt + ut to find the OLS estimate of ρ. (c) Suppose that the variance of the estimator ρ� is (1/T )σρ2 . Describe how you would test the hypothesis that ρ = 0. Solution: We are assuming that V ar (ρ̂) = 1 2 σ T ρ To test the hypothesis that ρ = 0, we make use of the χ2 test (refer to page 32 of Lecture Notes 8). We construct the test statistic as ξ= ρ̂2 ρ̂2 =T 2 V ar (ρ̂) σρ We reject the null hypothesis of ρ = 0 if the test statistic is sufficiently large, i.e., if ξ ≥ ξ¯ where the cutoff point ξ¯ is such that � � CDFχ2 (1) ξ¯ = 1 − α and α is the size of our hypothesis test. 21 (d) Write down the conditional log-likelihood function L(ρ, a0 , a1 ). Solution: Given the model Yt+1 = ρYt + (a0 + a1 Xt ) �t+1 where �t+1 are IID N (0, 1), the log-likelihood function is L (ρ, a0 , a1 ) = T � t=1 T � ln f (Yt |X0 , . . . , Xt−1 , Y0 , . . . , Yt−1 ; ρ, a0 , a1 ) � � 2 1 (Yt − ρYt−1 ) exp − 2 2σt2−1 2πσt−1 t=1 � � T 2 � � 1 � (Y − ρY ) t t−1 2 = − ln 2πσt−1 − 2 2 2σt−1 t=1 = ln �� � where the conditional standard deviation of Yt+1 , denoted σt2 , is given by σt2 = (a0 + a1 Xt )2 Therefore, T T � 1� T 1� � ln (a0 + a1 Xt−1 )2 − L (ρ, a0 , a1 ) = − ln (2π) − 2 t=1 2 t=1 2 � Yt − ρYt−1 a0 + a1 Xt−1 �2 (e) Suppose that the parameters a0 and a1 are known. Derive the maximum-likelihood estimate for ρ. Solution: We now suppose that a0 and a1 are known. Then the maximum likelihood estimate of ρ satisfies the first order condition ∂L ∂ρ �� � T � � Yt − ρYt−1 Yt−1 = a0 + a1 Xt−1 a0 + a1 Xt−1 t=1 � T � � Yt−1 (Yt − ρYt−1 ) = (a0 + a1 Xt−1 )2 t=1 0 = and hence �T ρ̂ = Yt−1 Yt t=1 (a0 +a1 Xt−1 )2 �T Yt2−1 t=1 (a0 +a1 Xt−1 )2 22 13. Suppose we observe a sequence of IID random variables Xt ≥ 0, t = 1, ..., T , with probability density pdf (X) = λe−λX , X ≥ 0 (a) Write down the log-likelihood function L(λ). Solution: The log-likelihood function is given by L (λ|X1 , . . . , XT ) = = T � t=1 T � ln p (Xt |λ) (ln λ − λXt ) t=1 = T ln λ − λ T � Xt t=1 �. (b) Compute the maximum likelihood estimate λ Solution: The maximum likelihood estimate λ̂ satisfies the first order condition ∂L ∂λ T T � = − Xt λ t=1 0 = and we get � λ̂ = T 1� Xt T t=1 �−1 �. (c) Derive the standard error for λ Solution: To compute the standard error of the maximum likelihood estimator λ̂, we resort to the general GMM standard errors (refer to page 28 of the Lecture Notes 8). We have � 2 � ∂ ln p (X | λ) t ˆ dˆ = Eˆ |λ = λ ∂λ2 T 1 � ∂ 2 ln p (Xt |λ) = |λ = λ̂ T t=1 ∂λ2 � T � 1� 1 = − T t=1 λ̂2 1 = − λ̂2 23 and �� � �2 ∂ ln p (Xt |λ) |λ = λ̂ Ŝ = Ê ∂λ �2 T � 1� 1 = − Xt ˆ T t=1 λ � � � � T T � � 1 2 1 1 = − Xt + Xt2 2 T t=1 λ̂ λ̂ T t=1 But we note that � λ̂ = T 1� Xt T t=1 �−1 and we can simplify Ŝ to Ŝ = − 1 λ̂2 � + T 1� 2 X T t=1 t � Finally, the variance of our estimator is given by � � V ar λ̂ = Ŝ T · dˆ2 where all the matrix multiplications simplify a great deal since we are working in the scalar case. In particular, putting together expressions for dˆ and Ŝ, we get � � �� T � � 1 4 1 1� 2 X V ar λ̂ = λ̂ − + T T t=1 t λ̂2 � � �� T � 1 1 = −λ̂2 + λ̂4 X2 T T t=1 t where � λ̂ = T 1� Xt T t=1 �−1 14. Suppose you observe a series of observations Xt , t = 1, ..., T . You need to fit a model Xt+1 = f (Xt , Xt−1 ; θ) + εt+1 where E[εt+1 |Xt , Xt−1 , ..., X1 ] = 0. Innovations εt+1 have zero mean conditionally on Xt , Xt−1 ,...,X1 . You also know that innovations εt+1 have constant conditional vari­ ance: E[ε2t+1 |Xt , Xt−1 , ..., X1 ] = σ 2 24 The parameter σ is not known. θ is the scalar parameter affecting the shape of the function f (Xt , Xt−1 ; θ). (a) Describe how to estimate the parameter θ using the quasi maximum likelihood approach. Derive the relevant equations. Solution: Assume that εt+1 |Xt , ...X1 follows Gaussian distribution N (0, σ 2 ). Then the logged likelihood is l (X3 , ..., XT ; θ, σ) = = T −1 � t=2 T −1 � t=2 ε2t+1 1 − ln 2π − ln σ − 2 2 2σ 1 (Xt+1 − f (Xt , Xt−1 ; θ))2 − ln 2π − ln σ − 2σ 2 2 To maximize the logged likelihood, we differentiate w.r.t θ and F.O.C gives T −1 1 � ∂f (Xt , Xt−1 ; θ) θ: 2 (Xt+1 − f (Xt , Xt−1 ; θ)) =0 σ t=2 ∂θ Upon simplifying, we have 0= � � � � ˆ T −1 � �� ∂f Xt , Xt−1 ; θˆ ∂f X , X ; θ � t t−1 Xt+1 − f Xt , Xt−1 ; θ̂ = ε̂t+1 ∂θ ∂θ t=2 T −1 � � t=2 The equivalent moment condition is � � ∂f (Xt , Xt−1 ; θ) E εt+1 =0 ∂θ which is valid. (b) Describe in detail how to use parametric bootstrap to estimate a 95% confidence interval for θ. Solution: Parametric bootstrap can be done through the following steps: (1)Estimate θ̂� using QMLE and obtain a sample of residuals ε̂t+1 = Xt+1 − � f Xt , Xt−1 ; θ̂ � � (i) (i) (2)Fix θ̂. Generate ith sample (for i = 1, 2, ..., N ) of X1 , ..., XT by drawing (i) (i) randomly from�the sample residuals with replacement to get ε̂t+1 and Xt+1 = � (i) (i) ˆ (i) f Xt , Xt−1 ; θ + ε̂t+1 . Note that we need to exclude the ”burn-in” sample and � � (i) (i) keep only last T observations. For ith sample, estimate θ̂(i) with X1 , ..., XT . (3) Get θ̂2.5% and θ̂97.5% (5th and 95th percentile) from the sample of estimates θ̂(i) , i = 1, 2, ..., N . � � � � �� (4) The bootstrapped confidence interval is given by θ̂ − θ̂97.5% − θ̂ , θ̂ − θ̂2.5% − θ̂ . 25 (c) Describe how to estimate the bias in your estimate of θ using parametric boot­ strap. Solution: The bias can be estimated using � � � � (i) E θ̂ − θ ≈ ÊN θ̂ − θ̂ where ÊN [·] represents the sample average of N bootstrapped estimates. (d) Derive the asymptotic standard error for θ� (large T ) using GMM standard error formulas. Solution: Recall that the moment condition for θ is � � ∂f (Xt , Xt−1 ; θ) E (Xt+1 − f (Xt , Xt−1 ; θ)) =0 ∂θ Asymptotic variance of θ̂ is 1 T � �−1 dˆ� Ŝ −1 dˆ . � dˆ = → = = �2 � � ∂ 2 f (Xt , Xt−1 ; θ) ∂f (Xt , Xt−1 ; θ) − ÊT (Xt+1 − f (Xt , Xt−1 ; θ)) ∂θ2 ∂θ � �2 � � ∂ 2 f (Xt , Xt−1 ; θ) ∂f (Xt , Xt−1 ; θ) E (Xt+1 − f (Xt , Xt−1 ; θ)) − ∂θ2 ∂θ � � �2 � � � ∂f (Xt , Xt−1 ; θ) ∂ 2 f (Xt , Xt−1 ; θ) −E E εt+1 ∂θ2 ∂θ �� �� �2 � �2 � ∂f (Xt , Xt−1 ; θ) ∂f (Xt , Xt−1 ; θ) = −E 0−E ∂θ ∂θ Since for j > 0, � � ∂f (Xt , Xt−1 ; θ) ∂f (Xt−j , Xt−j−1 ; θ) E εt+1 εt−j+1 ∂θ ∂θ � � ∂f (Xt , Xt−1 ; θ) ∂f (Xt−j , Xt−j−1 ; θ) = E E [εt+1 |Xt , Xt−1 , ...] εt−j+1 ∂θ ∂θ = 0 26 there is no autocorrelation. As a result, � Ŝ = → = = = �2 � ∂f (Xt , Xt−1 ; θ) ÊT (Xt+1 − f (Xt , Xt−1 ; θ)) ∂θ � �2 � � ∂f (Xt , Xt−1 ; θ) 2 E (Xt+1 − f (Xt , Xt−1 ; θ)) ∂θ � �2 � � ∂f (Xt , Xt−1 ; θ) E ε2t+1 ∂θ � �� � � ∂f (Xt , Xt−1 ; θ) 2 � 2 E E εt+1 |Xt , Xt−1 ∂θ �� �2 � ∂f (Xt , Xt−1 ; θ) σ2E ∂θ Thus, asymptotic variance is 2 1 2 σ /E T �� � ∂f (Xt ,Xt−1 ;θ) ∂θ �2 � . 15. Consider an estimator θ� for a scalar-valued parameter θ. Suppose you know, as a function of the true parameter value θ0 , the distribution function of the estimator, i.e., you know CDFθ�−θ0 (x) (In practice, you may be able to estimate the above CDF using bootstrap). Note that this CDF does not depend on model parameters. Based on the definition of the confidence interval, derive a formula for a confidence interval which covers the true parameter value with probability 95%. Solution: Since the CDF of θ̂ − θ0 is independent of the paramter θ0 , the 2.5 and 97.5 percentiles of the distribution, denoted as α2.5% and α97.5% , are fixed numbers independent of θ0 . As a result, � � Pr α2.5% < θ̂ − θ0 < α97.5% = 0.95 Rearranging the inequalities, we have � � Pr θ̂ − α97.5% < θ0 < θ̂ − α2.5% = 0.95 � � Hence, a 95% confidence interval is θ̂ − α97.5% , θ̂ − α2.5% , which illustrates why we � � � � �� ∗ ∗ ˆ ˆ ˆ ˆ have θ − θ97.5% − θ , θ − θ2.5% − θ as the bootstrapped confidence interval. θ∗ −θ̂ 27 has approximately the same distribution as θ̂ − θ0 . The 2.5 and 97.5 percentiles of the two distributions are�also approximately � � the same. � As a result, α2.5% and α97.5% can ∗ ∗ be approximated by θ2.5% − θˆ and θ97.5% − θˆ . 28