Problem 1: ARMA basics
Empirical Methods in Finance Homework 3: Solution
Prof. . Lochstoer TA:
January 23, 2022
Copyright By PowCoder代写 加微信 powcoder
Consider the ARMA(1,1):
yt = 0.95×yt−1 −0.9×εt−1 +εt, (1)
where εt is i.i.d. Normal with mean zero and variance σ2 = 0.052. In the below you need to show your work in order to get full credit.
1. What is the first-order autocorrelation of yt?
2. What is the second-order autocorrelation of yt? Also, what is the ratio of the second-
order to first-order autocorrelation equal to? Give some intuition for this result.
3. If yt = 0.6 and εt = 0.1, what is (i) Et [yt+1], (ii) Et [yt+2] given the ARMA model?
4. Let xˆt = Et [yt+1] where the expectation is taken using the ARMA model. What is the unconditional mean, standard deviation, and first-order auto-correlation of xˆt? Suggested Solution:
V(yt)=0.952V(yt−1)+0.92σ2 +σ2 −2×0.95×0.9σ2 V (yt) = 2.56e − 3
C(yt, yt−1) = C(0.95yt−1 − 0.9εt−1 + εt, yt−1)
= 0.95V (yt−1) − 0.9σ2
ρ1 = C(yt, yt−1) = 0.071 V(yt)
C(yt, yt−2) = C(0.95yt−1 − 0.9εt−1 + εt, yt−2)
= 0.95C(yt−1, yt−2)
ρ2 = C(yt, yt−2) = 0.068 V(yt)
If we calculate the ratio of the second-order to first-order auto-correlation, the ratio is equal to 0.95, which is exactly the coefficient of yt−1. For a ARMA(1,1),the auto- correlation function decays like the corresponding AR(1) model.
Et(yt+1) = 0.95yt − 0.9εt = 0.48 Et(yt+2) = 0.95Et(yt+1) = 0.46
xˆt = Et [yt+1]
xˆt = 0.95yt − 0.9εt
E(xˆ ) = 0.95E(y ) − 0.9E(ε ) = 0 ttt
V(xˆ) = 0.952V(y )+0.92V(ε )−2×0.95×0.9σ2 = 6.41e−05 ttt
std(xˆ ) = 8.00e − 3 t
C(xˆ,xˆ ) C(0.95y −0.9ε,0.95y −0.9ε ) ρ(xˆ)= t t−1 = t t t−1 t−1
= 0.952C(yt, yt−1) − 0.95 × 0.9C(yt, εt−1) = 0.95
V(xˆ) V(xˆ) tt
The final answer might be a little different if you put the numbers into the equations at different steps of algebra, but it is better to use the number at the final step.
Problem 2: Year-on-year quarterly data and ARMA dynamics
Assume the true quarterly log market earnings follow: et = et−1 + xt,
xt = φ xt−1 + εt, where V(εt) = σε2 = 1 and εt is i.i.d. over time t.
The earnings data you are given is year-on-year earnings growth, with in logs is:
yt ≡ et − et−4.
1. Assume φ = 0. Derive auto-covariances of order 0 through 5 for yt.
2. Assume φ = 0. Determine the number of AR lags and MA lags you need in the ARMA(p,q) process for yt. Give the associated AR and MA coefficients.
3. Optional: Assume 0 < φ < 1. Repeat 1 and 2 under this assumption. Suggested Solution: In the general case of 0 < φ < 1, we have:
and yt can be written as
V ( x t ) = φ 2 V ( x t ) + σ ε2 V(xt)= σε2
yt = et − et−4
=xt−3 +xt−2 +xt−1 +xt
Recalling that the autocovariances of the AR(1) process are given by
C ( x t , x t − j ) = 1 − φ 2 σ ε2
We can now calculate the autocovariances:
C(yt,yt) = C(xt−i,xt−l) i=0 l=0
1 φ φ2 φ32 = 41−φ2 +61−φ2 +41−φ2 +21−φ2 σε
2+φ+φ2 =2· 1−φ σε2
C (yt, yt−1) = C (xt−i, xt−l) i=0 l=1
= (1+φ)(3+φ2)σε2 1−φ
C (yt, yt−2) = C (xt−i, xt−l) i=0 l=2
2+2φ+2φ2 +φ3 +φ4
C (yt, yt−3) = C (xt−i, xt−l) i=0 l=3
(1+φ)(1+φ2)2
C (yt, yt−4) = C (xt−i, xt−l) i=0 l=4
φ(1+φ)(1+φ2)2
C (yt, yt−5) = C (xt−i, xt−l) i=0 l=5
φ2 (1 + φ) (1 + φ2)2
From the above, we see that for j > 3, the autocovariances behave like those of an AR(1) process with parameter φ. Thus, we know that the ARMA representation is of the form
(1−φB)yt =θ(B)εt Expanding the left hand side, we have
(1−φB)yt =xt +(1−φ)xt−1 +(1−φ)xt−2 +(1−φ)xt−3 −φxt−4 =εt +εt−1 +εt−2 +εt−3
Thus, we see that the process is captured by an ARMA(1, 3) with coefficients φ (B) = 1 − φ B
θ(B)=1+B+B2 +B3
Plugging in φ = 0 and σε2 = 1 give the results in the simplified case.
Although not required, it is helpful to note the Wold Decomposition of this process is given by
yt =ψjεt−j
jl=0 φj j ≤ 3 ψj =
φ ψj−1 j > 3
which allows us to verify the autocovariances using the relation
γ (h) = ψj+|h|ψj j=0
Problem 3: ARMAs and expected and realized returns
Realized returns can be written:
rt+1 = Et (rt+1) + εt+1,
where Et (εt+1) = E (Et (rt+1) εt+1) = 0. That is, a random variable can be decomposed into its (conditional) expected value plus an innovation term (a shock). To simplify notation, denote conditional expected returns as xt ≡ Et (rt+1). Also, assume that xt follows a mean- zero AR(1), where I am making the mean-zero assumption just to simplify the math to come. Thus, we have the predictive system:
rt+1 = xt+εt+1, xt+1 = φ1xt+ηt+1,
where Et (ηt+1) = E (xtηt+1) = 0.
1. Assume that ηt+1 = −σηεt+1, where ση > 0. That is, shocks to expected returns are negatively perfectly correlated with shocks to realized returns. Now, derive the ARMA process for rt+1. Hint: use the above equations at time t and t − 1 to substitute out xt from the first equation. That way you will end up with rt+1 on the left hand side (as now), but only lags of rt as well as lags of the shock εt and the current short εt+1 on the right hand side.
2. Now, assume ηt+1 is a shock term that has arbitrary correlation ρ with εt+1. How would you now determine what the appropriate ARMA representation for rt+1 is? Hint: derive ACF for rt+1 in this more general case and determine what type of ARMA process matches this ACF.
Suggested Solution:
The first equation minus φ1 times second equation, which gives us, rt+1 −φ1rt =ηt +εt+1 −φ1εt
rt+1 = φ1rt + (−ση − φ1)εt + εt+1 The rt+1 follows an ARMA(1,1) process.
ηt+1 is a shock term that has arbitrary correlation ρ with εt+1
V ( η t ) = σ η2 σ ε2
C (ηt, εt) = ρV (ηt)V (εt) = ρσησε2
0<φ1 <1,Thevarianceofxt isgivenby
V ( x t ) = φ 21 V ( x t ) + σ ε2 V(xt)= σε2
The autocovariances of the AR(1) process are given by
C ( x t , x t − j ) = φ j1 σ ε2 1 − φ 21
We can now calculate the variance and autocovariances of rt: C(rt,rt)=C(xt−1 +εt,xt−1 +εt)
rt+1 = xt + εt+1 rt = xt−1 + εt
= C ( x t , x t ) + σ ε2 σ2
= ε +σ2 1−φ2 ε
= 2 − φ 21 σ 2 1−φ2 ε
1 C(rt+1,rt)=C(xt +εt+1,xt−1 +εt)
= C (xt, xt−1) + C (xt, εt) = C (xt, xt−1) + C (ηt, εt) = C (xt, xt−1) + ρσησε2
= φ1 σ2+ρσσ2 1−φ2 ε η ε
C (rt+2, rt) = C (xt+1 + εt+2, xt−1 + εt)
= C (xt+1, xt−1) + C (xt+1, εt)
= C (xt+1, xt−1) + C (φ1xt + ηt+1, εt) = C (xt+1, xt−1) + φ1C (xt, εt)
C (rt+3, rt) = C (xt+2 + εt+3, xt−1 + εt)
= C (xt+2, xt−1) + φ1C (xt+1, εt)
= C (xt+2, xt−1) + φ21C (xt, εt) Consider a general ARMA(1,1) model:
The ACF follows,
rt =φ0 +φ1rt−1 +εt −θ1εt−1
γ 1 = φ 1 γ 0 − θ 1 σ ε2 γj =φ1γj−1 j>1
The ACF we derived for rt is of the same form and the parameters of the ARMA(1,1) are,
θ1 = φ1 − ρση
In the general case, the process is still an ARMA(1,1) process when ηt+1 is a shock term that has arbitrary correlation ρ with εt+1.
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com