6.4 MA Model
MA (Moving Average，移動平均) Model:
1. MA( q)
yt = t + θ1t−1 + θ2t−2 + ··· + θqt−q,
which is rewritten as:
yt = θ(L)t,
where 2 q θ(L) = 1 + θ1L + θ2L + ··· + θqL .
111 2. Invertibility (反転可能性):
2 q The q solutions of x from θ(x) = 1 + θ1 x + θ2 x + ··· + θq x = 0 are outside the unit circle.
=⇒ MA(q) model is rewritten as AR( ∞) model.
Example: MA(1) Model: yt = t + θ1t−1
1. Mean of MA(1) Process:
E(yt) = E(t + θ1t−1) = E(t) + θ1E(t−1) = 0
2. Autocovariance Function of MA(1) Process:
γ = 2 = + θ 2 = 2 + θ + θ22 (0) E(yt ) E( t 1 t−1) E( t 2 1 t t−1 1 t−1) = 2 + θ + θ2 2 = + θ2 σ2 E( t ) 2 1E( t t−1) 1E( t−1) (1 1)
112 2 γ(1) = E(ytyt−1) = E((t + θ1t−1)(t−1 + θ1t−2)) = θ1σ
γ(2) = E(ytyt−2) = E((t + θ1t−1)(t−2 + θ1t−3)) = 0
3. Autocorrelation Function of MA(1) Process: θ 1 , for τ = 1, γ(τ) + θ2 ρ(τ) = = 1 1 γ(0) 0, for τ = 2, 3, ···.
Let x be ρ(1). θ 1 = x, i.e., xθ2 − θ + x = 0. + θ2 1 1 1
θ1 should be a real number. 1 1 1 − 4x2 > 0, i.e., − ≤ ρ(1) ≤ . 2 2
113 4. Invertibility Condition of MA(1) Process:
t = −θ1t−1 + yt
2 = (−θ1) t−2 + yt + (−θ1)yt−1
3 2 = (−θ1) t−3 + yt + (−θ1)yt−1 + (−θ1) yt−2 . .
s 2 t−s+1 = (−θ1) t−s + yt + (−θ1)yt−1 + (−θ1) yt−2 + ··· + (−θ1) yt−s+1
s When (−θ1) t−s −→ 0, the MA(1) model is written as the AR(∞) model, i.e.,
2 t−s+1 yt = −(−θ1)yt−1 − (−θ1) yt−2 − · · · − (−θ1) yt−s+1 − · · · + t
5. Likelihood Function of MA(1) Process:
γ = +θ2 σ2 γ = θ σ2 γ τ = The autocovariance functions are: (0) (1 1) , (1) 1 , and ( ) 0 for τ = 2, 3, ···.
114 The joint distribution of y1, y2, ··· , yT is: ( ) 1 1 f (y , y , ··· , y ) = |V|−1/2 exp − Y0V−1Y 1 2 T (2π)T/2 2 where 1 + θ2 θ 0 ··· 0 1 1 y1 .. . θ 1 + θ2 θ . . 1 1 1 y2 . . 2 .. .. Y = . , V = σ 0 θ 0 . . 1 . .. .. . . . 1 + θ2 θ 1 1 yT ··· θ + θ2 0 0 1 1 1
115 6. MA(1) +drift: yt = µ + t + θ1t−1
Mean of MA(1) Process:
yt = µ + θ(L)t,
where θ(L) = 1 + θ1L.
Taking the expectation,
E(yt) = µ + θ(L)E(t) = µ.
116 Example: MA(2) Model: yt = t + θ1t−1 + θ2t−2
1. Autocovariance Function of MA(2) Process: (1 + θ2 + θ2)σ2, for τ = 0, 1 2 (θ + θ θ )σ2, for τ = 1, γ τ = 1 1 2 ( ) θ σ2, τ = 2 for 2, 0, otherwise.
2. let −1/β1 and −1/β2 be two solutions of x from θ(x) = 0.
For invertibility condition, both β1 and β2 should be less than one in absolute value.
Then, the MA(2) model is represented as:
yt = t + θ1t−1 + θ2t−2
117 2 = (1 + θ1L + θ2L )t
= (1 + β1L)(1 + β2L)t
AR( ∞) representation of the MA(2) model is given by:
= 1 t + β + β yt ((1 1L)(1 2L) ) β1/(β1 − β2) −β2/(β1 − β2) = + yt 1 + β1L 1 + β2L
118 3. Likelihood Function: ( ) 1 1 f (y , y , ··· , y ) = |V|−1/2 exp − Y0V−1Y 1 2 T (2π)T/2 2
where 1 + θ2 + θ2 θ + θ θ θ 0 1 2 1 1 2 2 y1 .. θ + θ θ 1 + θ2 + θ2 θ + θ θ . 1 1 2 1 2 1 1 2 y2 . . 2 .. .. Y = . , V = σ θ θ + θ θ θ . 2 1 1 2 2 .. .. . . 1 + θ2 + θ2 θ + θ θ 1 2 1 1 2 yT θ θ + θ θ + θ2 + θ2 0 2 1 1 2 1 1 2
119 4. MA(2) +drift: yt = µ + t + θ1t−1 + θ2t−2
Mean:
yt = µ + θ(L)t,
2 where θ(L) = 1 + θ1L + θ2L .
Therefore,
E(yt) = µ + θ(L)E(t) = µ
120 Example: MA(q) Model: yt = t + θ1t−1 + θ2t−2 + ··· + θqt−q
1. Mean of MA(q) Process:
E(yt) = E(t + θ1t−1 + θ2t−2 + ··· + θqt−q) = 0
2. Autocovariance Function of MA(q) Process: −τ ∑q 2 2 σ (θ0θτ + θ1θτ+1 + ··· + θq−τθq) = σ θiθτ+i, τ = 1, 2, ··· , q, γ τ = ( ) i=0 0, τ = q + 1, q + 2, ··· ,
where θ0 = 1.
3. MA( q) process is stationary.
4. MA(q) +drift: yt = µ + t + θ1t−1 + θ2t−2 + ··· + θqt−q
121 Mean:
yt = µ + θ(L)t,
2 q where θ(L) = 1 + θ1L + θ2L + ··· + θqL .
Therefore, we have:
E(yt) = µ + θ(L)E(t) = µ.
122 6.5 ARMA Model
ARMA (Autoregressive Moving Average，自己回帰移動平均) Process
1. ARMA(p, q)
yt = φ1yt−1 + φ2yt−2 + ··· + φpyt−p + t + θ1t−1 + θ2t−2 + ··· + θqt−q,
which is rewritten as:
φ(L)yt = θ(L)t,
2 p 2 q where φ(L) = 1−φ1L−φ2L − · · · −φpL and θ(L) = 1+θ1L+θ2L + ··· +θqL .
2. Likelihood Function:
The variance-covariance matrix of Y, denoted by V, has to be computed.
123 Example: ARMA(1,1) Process: yt = φ1yt−1 + t + θ1t−1 Obtain the autocorrelation coefficient.
The mean of yt is to take the expectation on both sides.
E(yt) = φ1E(yt−1) + E(t) + θ1E(t−1), where the second and third terms are zeros. Therefore, we obtain:
E(yt) = 0.
The autocovariance of yt is to take the expectation, multiplying yt−τ on both sides.
E(ytyt−τ) = φ1E(yt−1yt−τ) + E(tyt−τ) + θ1E(t−1yt−τ).
Each term is given by:
E(ytyt−τ) = γ(τ), E(yt−1yt−τ) = γ(τ − 1),
124 2 (φ1 + θ1)σ , τ = 0, 2 σ , τ = 0, E( y −τ) = E( − y −τ) = σ2, τ = , t t t 1 t 1 0, τ = 1, 2, ··· , 0, τ = 2, 3, ··· . Therefore, we obtain;
γ = φ γ + + φ θ + θ2 σ2, (0) 1 (1) (1 1 1 1)
2 γ(1) = φ1γ(0) + θ1σ ,
γ(τ) = φ1γ(τ − 1), τ = 2, 3, ··· .
From the first two equations, γ(0) and γ(1) are computed by: ( )( ) ( 2 ) 1 −φ1 γ(0) 1 + φ1θ1 + θ 2 1 = σ −φ1 1 γ(1) θ1
( ) ( )−1 ( 2 ) γ(0) 1 −φ1 1 + φ1θ1 + θ 2 1 = σ γ(1) −φ1 1 θ1
125 ( )( ) ( ) σ2 1 φ 1 + φ θ + θ2 σ2 1 + 2φ θ + θ2 = 1 1 1 1 = 1 1 1 . − φ2 − φ2 1 1 φ1 1 θ1 1 1 (1 + φ1θ1)(φ1 + θ1)
Thus, the initial value of the autocorrelation coefficient is given by:
(1 + φ θ )(φ + θ ) ρ(1) = 1 1 1 1 . + φ θ + θ2 1 2 1 1 1 We have:
ρ(τ) = φ1ρ(τ − 1).
126 ARMA(p, q) +drift: yt = µ + φ1yt−1 + φ2yt−2 + ··· φpyt−p + t + θ1t−1 + θ2t−2 + ··· + θqt−q.
Mean of ARMA(p, q) Process: φ(L)yt = µ + θ(L)t, 2 p 2 q where φ(L) = 1 − φ1L − φ2L − · · · − φpL and θ(L) = 1 + θ1L + θ2L + ··· + θqL .
−1 −1 yt = φ(L) µ + φ(L) θ(L)t.
Therefore, µ −1 −1 −1 E(yt) = φ(L) µ + φ(L) θ(L)E(t) = φ(1) µ = . 1 − φ1 − φ2 − · · · − φp
127