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Unformatted text preview: ws. The mean is
given by • The autocovariances and autocorrelation functions can be obtained by
solving what are known as the YuleWalker equations: • If the AR model is stationary, the autocorrelation function will decay
exponentially to zero.
17 Sample AR Problem
• Consider the following simple AR(1) model yt = μ + φ1 yt −1 + ut
(i) Calculate the (unconditional) mean of yt.
For the remainder of the question, set μ=0 for simplicity.
(ii) Calculate the (unconditional) variance of yt.
(iii) Derive the autocorrelation function for yt. 18 Solution (i) Unconditional mean:
E(yt) = E(μ+φ1yt1)
=μ +φ1E(yt1)
But also
So E(yt)= μ +φ1 (μ +φ1E(yt2))
= μ +φ1 μ +φ12 E(yt2))
E(yt) = μ +φ1 μ +φ12 E(yt2))
= μ +φ1 μ +φ12 (μ +φ1E(yt3))
= μ +φ1 μ +φ12 μ +φ13 E(yt3) 19 Solution (cont’d)
An infinite number of such substitutions would give
E(yt) = μ (1+φ1+φ12 +...) + φ1∞y0
So long as the model is stationary, i.e. , then φ1∞ = 0.
So E(yt) = μ (1+φ1+φ12 +...) = μ
1 − φ1 (ii) Calculating the variance of yt: yt = φ1 yt −1 + ut
From Wold’s decomposition theorem:
yt (1 − φ1 L) = ut
yt = (1 − φ1 L) −1 ut
yt = (1 + φ1 L + φ1 L2 + ...)ut
2 20 Solution (cont’d)
So long as φ1 < 1 , this will converge. yt = ut + φ1ut −1 + φ1 ut − 2 + ...
2 Var(yt) = E[ytE(yt)][ytE(yt)]
but E(yt) = 0, since we are setting μ = 0.
Var(yt) = E[(yt)(yt)]
= E[ u t + φ 1u t −1 + φ 1 2 u t − 2 + .. u t + φ 1u t −1 + φ 1 2 u t − 2 + ..
= E[(ut 2 + φ12ut −12 + φ14ut − 2 2 + ... + cross − products )]
2
2
2
4
2
= E[(ut + φ1 ut −1 + φ1 ut − 2 + ...)]
2
2
2
= σ u + φ12σ u + φ14σ u + ...
2
2
4
= σ u (1 + φ1 + φ1 + ...)
σ u2
=
(1 − φ12 ) ( )( ) 21 Solution (cont’d) (iii) Turning now to calculating the ACF, first calculate the autocovariances:
γ1 = Cov(yt, yt1) = E[ytE(yt)][yt1E(yt1)]
Since μ has been set to zero, E(yt) = 0 and E(yt1) = 0, so
γ1 = E[ytyt1]
γ1 = E[
= E[
=
= 22 Solution (cont’d) For the second autocorrelation coefficient,
γ2 = Cov(yt, yt2) = E[ytE(yt)][yt2E(yt2)]
Using the same rules as applied above for the lag 1 covariance
γ2 = E[ytyt2]
= E[
= E[
=
=
= 23 Solution (cont’d) • If these steps were repeated for γ3, the following expression would be
obtained
γ3 = and for any lag s, the autocovariance would be given by γs =
The ACF can now be obtained by dividing the covariances by the
variance: 24 Solution (cont’d) τ0 = τ1 = τ2 = τ3 =
… τs =
25 The Partial Autocorrelation Function (denoted τkk)
• Measures the correlation between an observation k periods ago and the
current observation, after controlling for observations at intermediate
lags (i.e. all lags < k). • So τkk measures the correlation between yt and ytk after removing the
effects of ytk+1 , ytk+2 , …, yt1 . • At lag 1, the ACF = PACF always • At lag 2, τ22 = (τ2τ12) / (1τ12) • For lags 3+, the formulae are more complex. 26 The Partial Autocorrelation F...
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This note was uploaded on 09/20/2013 for the course FINA 5170 taught by Professor Janebargers during the Summer '13 term at Greenwich School of Management.
 Summer '13
 JaneBargers
 Financial Markets

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