Slides 4. Matthieu Gomez Fall 2017

Transcription

1 Slides 4 Matthieu Gomez Fall 2017

2 How to Compute Stationary Distribution of a Diffusion?

3 Kolmogorov Forward Take a diffusion process dx t = µ(x t )dt + σ(x t )dz t How does the density of x t evolves? Theorem (Kolmogorov Forward) Denote g t (x) the density of x t. We have: dg dt = x(µ(x)g(x)) xx(σ2 (x)g(x)) In particular, if a stationary density exists, it must satisfy 0 = x(µ(x)g(x)) xx(σ2 (x)g(x)) 1

4 Heuristic Proof For any function f, E[f (x t+dt )] can be written in two ways + + f (x)g t+dt (x)dx = [(f (x) + df (x))g t (x)dx Assume that f is a twice differentiable and use Ito s lemma on the RHS to obtain + + f (x)dg t (x)dx = (µ(x) xf (x) σ(x)2 xxf (x))g t (x)dx Assume that f decays fast enough as x + and use integration by parts to obtain + + f (x)dg t (x)dx = f (x)[( x(µ(x)g t (x)) x (σ(x) 2 g t ))dtdx This equality must hold for all f satisfying the conditions above. Therefore, we obtain dg t dt (x) = x(µ(x)g t(x)) x (σ2 (x)g t ) 2

5 Special Cases Take a Ornstein-Uhlenbeck ( AR(1) in discrete time) dx = θ(x µ)dt + σdz t Kolmogorov Forward gives 0 = x(θ(x µ)g(x)) + σ xx(g(x)) The solution is a normal distribution θ g(x) = πσ 2 e κ(x µ)2 /σ 2 Take a Cox-Ingersoll-Ross process dx = θ(x µ)dt + σ xdz t Kolmogorov Forward gives The solution is a gamma distribution with ω = 2κ/σ 2 and ν = 2κµ/σ 2 0 = x(κ(x µ)g(x)) + σ xx(xg(x)) g(x) = ων Γ(ν) xν 1 e ωx 3

6 General Case General case 0 = x(µ(x)g(x)) xx(σ2 (x)g(x)) Integrating wrt x 0 = µ(x)g(x) x(σ2 (x)g(x)) This is an ODE of degree one. We know that the solution has the following form g(x) = Cm(x) with m(z) 1 s(x)σ 2 (x) and C 1 and C 2 are constant chosen so that g(x) 0 and g(x)dx = 1 4

7 Solution Alternatively, we can solve with a computer the ODE 0 = x(µ(x)g(x)) xxσ2 (x)g(x) 1. Define a state space s = (s 1, s 2,..., s n). 2. The problem is to find g positive elementwise such that Ag = 0 5

8 Approximating First Derivative To approximate x(µ(x)g(x)), use the following matrix x(µ(x)g(x)) = D 1 g where D 1 is the matrix µ s µ 1 µ s s µ 0 2 µ s s D 1 = µ n 2 µ n 1 0 s s µ n 1 µn s s Formally µ n s x(µ(x)g(x)) i = µ+ g i i µ + i 1 g i 1 s g i+1µ i+1 g iµ i s where µ + i = max(µ i, 0) and µ i = min(µ i, 0) 6

9 Approximating Second Derivative To approximate xx(σ 2 (x)g(x)), use the following matrix where D 2 is the matrix D 2 = Formally σ2 1 σ 2 ( s) 2 2 xx(σ 2 (x)g(x)) = D 2 g ( s) σ1 2 ( s) 2 2 σ2 2 σ 2 ( s) 2 3 ( s) σ 2 2 ( s) 2 2 σ2 3. ( s) σ2 n 2 ( s) σ 2 n 2 ( s) 2 σ 2 n 1 ( s) σ2 n 1 ( s) 2 σn 1 2 ( s) 2 xx(σ 2 (x)g(x)) = σ2 i+1 g i+1 + σ 2 i 1 g i 1 2σ 2 i g i ( s) 2 σn 2 ( s) 2 σ2 n ( s) 2 7

10 Solve it We need to find a vector g such that (D D2 )g = 0 How can we be sure that there exists a solution g that is positive everywhere? 8

11 Theorem Suppose a square matrix A is such that 1. Sum of each column is 0 2. All elements off-diagonal are positive or null Then there exists a unique g 0 such that Ag = 0 9

12 Dicrete Time Version Proof. Take a positive real number δ. The matrix P = I + δa is such that 1. sum of each column is 1 2. all elements are positive or null (for δ small enough) Therefore, we know (existence stationary distribution for markov chain in discrete time) there is a unique g 0 such that Pg = g In particular, this means there exists a unique g 0 such that Ag = 0 10

13 By Force of Habit: Campbell Cochrane (1995)

14 Assumptions Utility is given by U(C t, X t ) = (C t X t ) 1 γ 1 γ In particular U C (C t, S t ) = (C t X t ) γ Define S t = C t X t C t We can write the marginal utility of consumption as U C (C t, S t ) = C γ t S γ t 11

15 Euler Equation. SDF is Λ t = e ρt U c(c t ) = e ρt C γ t S γ t = e ρt γ(c t+s t ) Assume that the process for c = log C and s = log S are given by dc = µ cdt + σ cdz t ds = µ sdt + σ sdz t Using Ito s lemma κ = σ Λ = γ(σ c + σ s) (1) r = µ Λ = ρ + γ(µ c + µ s) γ2 2 (σ2 c + σ2 s + 2σcσs) (2) 12

16 Evolution State Variable We assume the following law of motion for s t ds t = κ s(s t s)dt + λ(s t )σ cdw t with s and the function s λ(s) to be specified later. Plugging this expression for µ s and σ s in Equation (1) and Equation (2), we obtain κ = (1 + λ(s t ))γσ c (3) r = ρ + γ(µ c κ s(s t s)) γ2 2 (1 + λ(s t)) 2 σ 2 c (4) 13

17 Evolution State Variable Now choose s and λ to satisfy two things 1. Interest rate is constant: there exists K such that κs(s s) λ(s) = K + 2 γσc The volatility of X must be zero around the steady state s. Formally, we want We have σ X (s) = 0 and σ X (s) = 0 (5) X = C(S 1) Denote σ x the instantaneous volatility of ln X. Applying Ito s lemma The conditions Equation (5) mean Joining the three equations, we obtain σ X = (1 λ(s) e s 1 )σc λ(s) = e s 1 and λ (s) = e s λ(s) = e 1 s 2(s s) 1 γ e s = σ κ s 14

18 Euler Equation Plugging the value of s and λ in the Euler equation Equation (3) and Equation (4), we obtain κ = (1 + λ(s t ))γσ c r = ρ + γµ c γ 2 κs 15

19 Wealth-Consumption Ratio We can now solve for wealth-to-consumption ratio V using market pricing for wealth The law of motion of wealth is E[ dw ] = rdt + κσwdt ĉdt W One can interpret it as a market pricing equation for the wealth portfolio E[ C W dt + dw W ] = rdt + κσ Wdt Denote V the wealth-to-consumption ratio W/C We have dt d(vc) d(vc) + E[ ] = rdt + κσ[ V VC VC ]dt Denote µ V and σ V the geometric drift and volatility of V, i.e. dv V = µ Vdt + σ V dz t Applying Ito s lemma, we obtain 1 V + µc + σ2 c 2 + µ V + σ cσ V = r + κ(σ c + σ V ) (6) After substituting µ V and σ V using Ito s lemma, we obtain the following ODE: 1 V + µc + σ2 c 2 + V (s) V(s) µs(s) + 1 V (s) V (s) 2 V(s) σ2 s(s) + σ c V(s) σs(s) = r(s) + κ(s)(σc + V (s) V(s) σs) 16

20 Graphs Figure 1 17

21 Graphs Figure 2 18

22 Figure 3 19

23 Brunnemeir Nagel 2008: Do Wealth Fluctuations Generate Time-Varying Risk Aversion? Micro-Evidence on Individuals Asset Allocation?

24 Habit Model Remember the SDF κ = σ Λ = σ (C X) γ C = γ C X σc γ X C X σx Remember that at s = s, σ X = 0, that is C κ = γ C X σc Intuition: At s = s, it is as if the risk aversion of the representative agent is γ C C X We can inverse this formula σ c = (1 X C ) κ γ Intuition: the higher C, the higher the volatility of consumption 20

25 Habit Model Now suppose σ V 0 where V = W/C σ W = (1 XV W ) κ γ Dividing by σ R α = (1 XV W ) κ γσ R The habit model predicts that α increases in W is high 21

26 Omitted variables We want to hold γ constant = add household fixed effects (i.e. absorb any household specific determinant of portfolio share). We want to hold X, κ, σ R constant = add time fixed effects (i.e. absorb any year specific determinant of portfolio share). Life events may change wealth and portfolio shares at the same time,for reasons unrelated to habits. Think of getting fired, getting married, etc. = throw away households with big life events + add household life-cycle variables as controls We obtain the specification α = γ i + δ t + β log(w it ) + γx it + ɛ where X it is a set of household life-cycle controls Measurement error in W Measurement error in RHS biases down the estimates Because wealth is also used when computing the portfolio share α, it introductes a negative correlation between wealth and portfolio shares = instrument change in wealth by inheritance or change in income 22

27 Figure 4: The coefficient of in column 1 implies that 10 percent growth in real wealth leads to a tiny reduction in the share of risky liquid assets by , e.g., from 50 percent to percent. 23

28 Figure 5: The point estimate of in the first column implies that an increase in liquid wealth by 10 percent implies a roughly 1 percent increase in the probability or participating in the stock market. 24