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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 Yule-Walker 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+φ1yt-1) =μ +φ1E(yt-1) But also So E(yt)= μ +φ1 (μ +φ1E(yt-2)) = μ +φ1 μ +φ12 E(yt-2)) E(yt) = μ +φ1 μ +φ12 E(yt-2)) = μ +φ1 μ +φ12 (μ +φ1E(yt-3)) = μ +φ1 μ +φ12 μ +φ13 E(yt-3) 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[yt-E(yt)][yt-E(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, yt-1) = E[yt-E(yt)][yt-1-E(yt-1)] Since μ has been set to zero, E(yt) = 0 and E(yt-1) = 0, so γ1 = E[ytyt-1] γ1 = E[ = E[ = = 22 Solution (cont’d) For the second autocorrelation coefficient, γ2 = Cov(yt, yt-2) = E[yt-E(yt)][yt-2-E(yt-2)] Using the same rules as applied above for the lag 1 covariance γ2 = E[ytyt-2] = 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 yt-k after removing the effects of yt-k+1 , yt-k+2 , …, yt-1 . • At lag 1, the ACF = PACF always • At lag 2, τ22 = (τ2-τ12) / (1-τ12) • 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.

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