Equilibrium Valuation with Growth Rate Uncertainty

Equilibrium Valuation with Growth Rate Uncertainty Lars Peter Hansen University of Chicago Toulouse p. 1/26

Portfolio dividends relative to consumption. 2 2.5 Port. 1 Port. 2 Port. 3 3 3.5 4 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2 2.5 3 Port. 4 Port. 5 Market 3.5 4 4.5 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 Toulouse p. 2/26

Dominant Eigenvalues: Matrix digression Growth state of the economy at time t is d x t C t /C t 1, where x t {x n : n = 1, 2,...,N}. The probabilities of transiting from one state to another are given by: a m,n = Prob(x t+1 = x n x t = x m ) One-period stochastic discount factor is assumed to be S t+1,t = β ( Ct+1 C t ) α = β(x t+1) α. Consider a cash flow growth process: D t = (C t ) λ Construct the matrix P where the (m, n) entry of this matrix is given by: p m,n = βa m,n (x n ) λ α. Toulouse p. 3/26

Cash flow valuation Let D t equal: D t = (C t ) λ φ(x t ) and write φ(x t ) as an N dimensional vector Φ. PΦ is the vector of date t prices of a payoff D t+1 multiplied by Dt = (C t ) λ. The date t value of an infinite cash flow is: P j Φ j=0 multiplied by Dt. Long run value dictated by the behavior of P j. Toulouse p. 4/26

Dominant eigenvalue Suppose that the matrix P has distinct eigenvalues, and write the eigenvalue decomposition as: P = T T 1 where is a diagonal matrix of eigenvalues. Then (P) j = T j T 1. Typically one eigenvalue will be positive with an associated positive eigenvector. Largest (in absolute value) eigenvalue. Write as exp( ν). Then lim exp(jν)(p j )Φ. j proportional to the dominant eigenvector, but not on Φ. Toulouse p. 5/26

Value Decomposition Suppose that Φ is strictly positive. 1 j log(p j )Φ is the yield on a j period security with cash flow D t+j once we adjust initial payout for the initial payout 1 j D t. Observations 1. Decomposition of value P j Φ j=0 by horizon. 2. lim j 1 j log(p j )Φ = ν1 N Toulouse p. 6/26

Illustration 2 x 10 3 4 Portfolio 1 2 x 10 3 Portfolio 5 4 log values 6 8 10 12 0 50 100 6 8 10 12 0 50 100 0.015 0.015 growth rate 0.01 0.01 0.005 0 50 100 quarters 0.005 0 50 100 quarters Toulouse p. 7/26

Avoid Markov chain approximation Introduce valuation operators that map functions into functions. Steps: 1. Construct a reference growth process - statistical decomposition 2. Build a family of valuation operators - economic model 3. Build a family of growth operators 4. Compute dominant eigenvalue and function 5. Construct tail returns, expected returns and expected excess returns after adjusting for expected growth Toulouse p. 8/26

Abstract operator formulation Ingredients: 1. {x t } be a stationary Markov process. 2. s t+1,t an economic model of a stochastic discount factor. Price of a payoff ψ(x t+1 is: E [exp(s t+1,t )ψ(x t+1 ) x t ] and depends only on x t. Markov pricing. 3. Martingale M t with increments that depend on x t and shocks that influence the evolution of x t used to build a stochastic growth process: exp(m t + ζt) Toulouse p. 9/26

Three operators 1. One-period valuation-growth operator: Pψ(x) = E [exp (s t+1,t + ζ + M t+1 M t )ψ(x t+1 ) x t = x]. 2. One-period growth operator (abstracts from valuation) Gψ(x) = E [exp (ζ + M t+1 M t )ψ(x t+1 ) x t = x]. 3. One-period valuation operator (abstracts from growth) P f ψ(x) = E [exp (s t+1,t ) ψ(x t+1 ) x t = x]. Toulouse p. 10/26

Long run value accounting Three eigenvalue problems: 1. Solve Pφ = exp( ν)φ. ν asymptotic rate of decay in value. 2. Gφ + = exp(ɛ)φ +. ɛ asymptotic growth rate of cash flow. ν + ɛ is an expected rate of return. 3. Solve P f φ f = exp( ν f )φ f. ν + ɛ ν f expected excess rate of return. Change martingales trace out long run risk-return relation. Toulouse p. 11/26

Dominant Eigenfunctions Use the dominant eigenfunction to construct a valuation process and the corresponding return. A valuation process {J t : t = 1, 2,...} is one for which the date t price of the security with liquidation value J t+1 is J t. Form: J t+1 = exp [(ν + ζ)(t + 1) + M t+1 M 0 ] φ (x t+1 ). Since φ is a positive eigenfunction, the date t value of the payoff J t+1 is indeed J t, verifying that J t+1 is indeed a valuation process. Toulouse p. 12/26

Observations The k-period return is: Rt+k k = J t+k = exp [(ν + ζ)k + M t+k M t ] φ (x t+k ) J t φ (x t ) 1. Riskiness of constructed one-period return (k=1) depends on riskiness of the stochastic component to the growth process M t+1 and of the logarithm of φ (x t+1 ). 2. Take expectations and logarithms: lim k 1 k log E ( R k t+k X t ) = ν + ɛ Toulouse p. 13/26

Constructed equity Build a security with same returns by valuing equity with a dividend ˆD t+1 = exp [ζ(t + 1) + M t+1 M 0 ] φ (x t+1 ). Using the eigenvalue property the price dividend ratio is given by: ˆP t ˆD t = exp ( ν) 1 exp ( ν) Includes both a pure discount factor (adjusted for risk) and a dividend growth factor. The implied discount rate is ν + ɛ since the asymptotic dividend growth factor for dividends is: exp(ɛ). Toulouse p. 14/26

Cash flow growth. 2 2.5 Port. 1 Port. 2 Port. 3 3 3.5 4 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2 2.5 3 Port. 4 Port. 5 Market 3.5 4 4.5 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 Toulouse p. 15/26

Useful time series martingale decomposition Suppose x t+1 = Gx t + Hw t+1 where G has stable eigenvalues and {w t+1 } is iid vector process of standard normally distributed random variables. Stationary process. Positive cash flow process expressed in logarithms: d t+1 d t = µ d + U d x t + ι 0 w t+1. Stationary increments. Alternatively, we may specify the process in moving-average form: d t+1 d t = µ d + ι(l)w t+1. where ι(z) = j=0 ι jz j Toulouse p. 16/26

Martingale decomposition Commonly used in establishing central limit approximations (e.g. see Hall and Heyde (1980)) and it is not limited to linear processes (e.g. see Hansen and Scheinkman (2003) for a nonlinear Markov version.) Write d t+1 d t = µ d + ι(1)w t+1 + Udx t+1 Udx t Thus 1. {d t } has growth rate µ d 2. {d t } has a martingale component has increment ι(1)w t+1. where ι(1) = ι 0 + U d G(I G) 1 H 3. {D t } asymptotic growth rate for cash flow is µ d + (1/2) ι(1) 2. Toulouse p. 17/26

Model: Investor Preferences Relax discounted expected utility theory model, but maintain recursivity and dynamic programming. Consider a Kreps and Porteus (1978) specification with a CES recursion: V t = [(1 β) (C t ) 1 ρ + βr t (V t+1 ) 1 ρ] 1 1 ρ. where C t is date t consumption and V t the date t continuation value of the consumption profile. R t adjusts the continuation value for risk via: R t (V t+1 ) = [E (V t+1 ) 1 α X t ] 1 1 α where X t is the current period information set. Special case: Cobb-Douglas specification (ρ = 1). The recursion becomes: V t = (C t ) (1 β) R t (V t+1 ) β. Toulouse p. 18/26

Investor Preferences Continued Recursion again V t = [(1 β) (C t ) 1 ρ + βr t (V t+1 ) 1 ρ] 1 1 ρ. R t (V t+1 ) = [E (V t+1 ) 1 α X t ] 1 1 α Observations: 1. 1 ρ is a measure of intertemporal substitution. 2. No reduction of intertemporal lotteries - the intertemporal allocation of risk matters!! 3. α is a measure of risk aversion for simple wealth gambles. Toulouse p. 19/26

Stochastic consumption growth An alternative recursion: V t C t = [ (1 β) + βr t ( Vt+1 C t+1 C t+1 C t ) 1 ρ ] 1 1 ρ Let v t denote the logarithm of the continuation value relative to the logarithm of consumption, and let c t denote the logarithm of consumption. Rewrite recursion as v t = 1 1 ρ log ((1 β) + β exp [(1 ρ) Q t(v t+1 + c t+1 c t )]), where Q t is the risk-sensitive recursion: Q t (v t+1 ) = 1 1 α log E (exp [(1 α)v t+1] X t ). Solve when ρ = 1 using consumption dynamics; compute derivatives. Toulouse p. 20/26

ρ = 1 limit where v t = βq t (v t+1 + c t+1 c t ). Q t (v t+1 ) = 1 1 α log E (exp [(1 α)v t+1] X t ), log linear consumption dynamics c t c t 1 = γ(l)w t + µ c, γ(z) = γ j z j. j=0 Toulouse p. 21/26

ρ = 1 solution Solve a linear expectational difference equation forward: v t = β j E (c t+j c t+j 1 µ c F t ) + µ v j=1 where µ v = β 1 β [µ c + (1 α) γ(β) γ(β)] 2 and γ(β) is the discounted response. The term γ(β) will be an important ingredient in our calculations. Easy to solve for v t when ρ = 1. Could extend to accommodate conditional volatility as in Tauchen and Lettau-Ludvigson-Wachter. Toulouse p. 22/26

Stochastic Discount Factor Valuation of one-period securities: S t+1,t = MV t+1mc t+1 MC t = β ( Ct+1 C t ) ρ ( Vt+1 ) ρ α R t (V t+1 ) The stochastic discount factor in the ρ = 1 case is: S t+1,t β ( Ct C t+1 ) [ ] (V t+1 ) 1 α R t (V t+1 ) 1 α. Depends on continuation values - much of the literature finds clever ways to avoid this dependence. Valuation of multi period securities multiplies up stochastic discount factors. Toulouse p. 23/26

Stochastic discount factor for ρ = 1. The logarithm of the stochastic discount factor can now be depicted as: s t+1,t log S t+1,t = δ γ(l)w t+1 µ c +(1 α)γ(β)w t+1 (1 α)2 γ(β) γ(β) 2 where β = exp( δ). The term γ(β)w t+1 is the is the solution to (1 β) β j [E(c t+j X t+1 ) E(c t+j X t )]. j=0 Illustrates the role of consumption predictability as featured by Bansal and Yaron (2004). Toulouse p. 24/26

Extraction of macro risk consumption 15 x consumption shock 10 3 10 5 0 15 x 10 3 earnings shock 10 5 0 without cointegration with cointegration 5 20 40 60 80 5 20 40 60 80 0.05 0.05 0.04 0.04 earnings 0.03 0.02 0.01 0 0.03 0.02 0.01 0 20 40 60 80 20 40 60 80 Toulouse p. 25/26

Questions Implied long run risk return tradeoff? Relation to robustness θ versus 1 α 1? Learning and sensitivity to model specification? Toulouse p. 26/26