時間序列與財務計量第1次作業(碩二暑假)
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Teks penuh
(2) Problem1：Rewrite the following expressions without using the lag operator. 答： 1、由定義ℒ 𝑘 𝑦𝑡 ≡ 𝑦𝑡−𝑘 知： (1 − 𝐿5 )𝒴𝑡 = ℰ𝑡 可得：𝑦𝑡 − 𝑦𝑡−5 = 𝜀𝑡 2、由性質(𝐿𝑘 + 𝐿𝑗 )𝑦𝑡 = 𝐿𝑘 𝑦𝑡 + 𝐿𝑗 𝑦𝑡 = 𝑦𝑡−𝑘 + 𝑦𝑡−𝑗 知： 𝒴𝑡 = (. 2 + 5𝐿 + 0.8𝐿2 ) ℰ𝑡 𝐿 − 0.6𝐿3. (𝐿 − 0.6𝐿3 )𝒴𝑡 = (2 + 5𝐿 + 0.8𝐿2 )ℰ𝑡. 1. 𝑦𝑡−1 − 0.6𝑦𝑡−3 = 2ℰ𝑡 + 5ℰ𝑡−1 + 0.8ℰ𝑡−2 3、由性質(𝐿𝑘 𝐿𝑗 )𝑦𝑡 = 𝐿𝑘+𝑗 𝑦𝑡 = 𝑦𝑡−𝑗−𝑘 知： 𝐿3. 𝒴𝑡 = 2(1 + ) ℰ𝑡 L. 𝐿3. 𝒴𝑡 = 2ℰ𝑡 + ( ) ℰ𝑡 L. 𝑦𝑡 = 2ℰ𝑡 + 2ℰ𝑡−2. Problem2：Rewrite the following expressions in lag operator form. 答： 1、𝒴𝑡 + 𝒴𝑡−1 + ⋯ + 𝒴𝑡−𝑛 = 𝛼 + ℰ𝑡 + ℰ𝑡−1 + ℰ𝑡−2 + ⋯ + ℰ𝑡−𝑚 ，whereαis a constant. (1 + L + 𝐿2 + ⋯ + 𝐿n )𝑦𝑡 = α + (1 + L + 𝐿2 + ⋯ + 𝐿m )ℰ𝑡 𝑗 ( ∑𝑛j=0 𝐿𝑗 )𝑦𝑡 = α + ( ∑𝑚 j=0 𝐿 ) ℰ𝑡 或者為：等比級數的公式為：𝑆𝑛 = 1−𝐿𝑛+1 1−𝐿. 𝑦𝑡 = α +. 1−𝐿𝑚+1 1−𝐿. ℰ𝑡. 𝑎1 (1−𝑟 𝑛 ) 1−𝑟. 同乘1 − L 後，. (1 − 𝐿𝑛+1 )𝑦𝑡 = (1 − 𝐿)𝛼 + (1 − 𝐿𝑚+1 ) ℰ𝑡 再由性質 Lc = c 知： (1 − 𝐿𝑛+1 )𝑦𝑡 = 𝛼 − α + (1 − 𝐿𝑚+1 ) ℰ𝑡 (1 − 𝐿𝑛+1 )𝑦𝑡 = (1 − 𝐿𝑚+1 ) ℰ𝑡. 2、𝒴𝑡 = ℰ𝑡−2 + ℰ𝑡−1 + ℰ𝑡 𝑦𝑡 = (𝐿2 + 𝐿 + 1)ℰ𝑡 = (1 + 𝐿 + 𝐿2 )ℰ𝑡 = ( ∑2j=0 𝐿𝑗 ) ℰ𝑡.
(3) Problem3：Define the autocovariance function as γ(t, s) = Cov(𝒴𝑡 , 𝒴𝑡−𝑠 ) Given a constant α，consider the following four autocovariance functions： 2、γ(t, s) = 𝑒 −αs. 1、γ(t, s) = α. 3、γ(t, s) = αs. 4、γ(t, s) =. α S. Which autocovariance function(s) are consistent with weak stationarity, and which are not？Why？ 答： 若一個時間序列的均數為常數，變異數為有限，又自我共變異數是 s 的函數且與 t 無關， 則為弱定態的時間序列 1、 γ(t, s) = α 2 γ(0) = γ(1) = γ(2) = ⋯ = γ(m) = α γ(0) = Var(𝑦𝑡 ) = α 雖變異數為有限，而自我共變異數是 s 的零次函數且與 t 無關，故為弱定態的時間序列 1. 1. α. 另外考量Var(𝑦̅𝑡 ) = Var (𝑛 ∑𝑛𝑖=1 𝑦𝑡 ) = 𝑛2 ∑𝑛𝑖=1 𝑉𝑎𝑟(𝑦𝑡 ) = 𝑛. 𝛂 = 𝟎 ，故此𝐀𝐂𝐅又符合的一致性的弱定態。 𝐧→∞ 𝒏. 𝐥𝐢𝐦 𝑽𝒂𝒓(𝒚̅𝒕 ) = 𝐥𝐢𝐦. 𝐧→∞. 2、 γ(t, s) = 𝑒 −αs γ(0) = Var(yt ) = e−α0 = 1 γ(1) = 𝑒 −α , γ(m) = 𝑒 −𝑚α 【備註：必須進一步符合γ(m) = γ(−m)才可成立】 變異數為有限，自我共變異數是 s 的函數且與 t 無關，故為弱定態的時間序列 1. 1. 1. 另外考量Var(𝑦̅𝑡 ) = Var (𝑛 ∑𝑛𝑖=1 𝑦𝑡 ) = 𝑛2 ∑𝑛𝑖=1 𝑉𝑎𝑟(𝑦𝑡 ) = 𝑛 1 = 0 ，故此 ACF 又符合的一致性的弱定態。 n→∞ 𝑛. lim 𝑉𝑎𝑟(𝑦̅𝑡 ) = lim. n→∞. 3、γ(t, s) = αs γ(0) = Var(yt )= 0 ,. γ(1) =α , 1. γ(m) = mα 1. , 倘若n → ∞，γ(m)為發散值 0. 另外考量Var(𝑦̅𝑡 ) = Var (𝑛 ∑𝑛𝑖=1 𝑦𝑡 ) = 𝑛2 ∑𝑛𝑖=1 𝑉𝑎𝑟(𝑦𝑡 ) = 𝑛 0 = 0 ，故此 ACF 看似又符合的一致性的弱定態。 n→∞ n→∞ 𝑛 然而考量在自然界的法則，𝛄(𝐧)不應為發散之狀態，是故本小題不為選項之 1 lim 𝑉𝑎𝑟(𝑦̅𝑡 ) = lim. 4、γ(t, s) =. α S. γ(0) = Var(yt ) =. 𝛼 0. γ(1) =α. γ(n) =. 變異數不存在，未符合𝐕𝐚𝐫(𝐲𝐭 ) < ∞，故不是弱定態的時間序列. 𝛼 𝑛.
(4) 𝑖𝑖𝑑. Problem4：Given e𝑡 ~ (0, 𝜎 2 ), find the autocovariance function for the following process. 𝑖𝑖𝑑. 答：由e𝑡 ~ (0, 𝜎 2 )可知： E(e𝑡 ) = 0, ∀ t , Var(e𝑡 ) = 𝜎 2 , ∀ 𝑡. 又Var(𝑒𝑡 ) = E(𝑒𝑡 2 ) − E(𝑒𝑡 )2 𝜎 2 = E(𝑒𝑡 2 ) − 02 ，則. E(𝑒𝑡 2 ) = 𝜎 2 ，∀ t E(e𝑡 , e𝑡−𝑘 ) = 0 ，∀ k, t 由定義 5 可知：γ(t, k) = Cov(𝒴𝑡 , 𝒴𝑡−𝑘 ) = E[(𝑦𝑡 − 𝜇𝑡 )(𝑦𝑡−𝑘 − 𝜇𝑡−𝑘 )] = γ(k) 1、 𝒴𝑡 = 𝑒𝑡 + 𝜃1 e𝑡−1 ，本題可得 3. γ(0) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡 ) = Var(𝑦𝑡 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡 − E(𝑦𝑡 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡 − 0)] = E(𝑦𝑡 2 ) = E[(𝑒𝑡 + 𝜃1 e𝑡−1 )(𝑒𝑡 + 𝜃1 e𝑡−1 )] = E(𝑒𝑡 2 ) + 𝜃1 2 E(𝑒𝑡−1 2 ) = 𝜎 2 + 𝜃1 2 𝜎 2 = (1 + 𝜃1 2 )𝜎 2 γ(1) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡−1 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡−1 − E(𝑦𝑡−1 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡−1 − 0)] = E[(𝑦𝑡 )(𝑦𝑡−1 )] = E[(𝑒𝑡 + 𝜃1 e𝑡−1 )(𝑒𝑡−1 + 𝜃1 e𝑡−2 )] = 𝜃1 E(𝑒𝑡−1 2 ) = 𝜃1 𝜎 2 γ(2) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡−2 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡−2 − E(𝑦𝑡−2 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡−2 − 0)] = E[(𝑦𝑡 )(𝑦𝑡−2 )] = E[(𝑒𝑡 + 𝜃1 e𝑡−1 )(𝑒𝑡−2 + 𝜃1 e𝑡−3 )]. =0. (1 + 𝜃1 2 )𝜎 2 ，k = 0 故可得自我共變異函數γ(𝑘) = { 𝜃1 𝜎 2 ,k = 1 0 ,k > 1 2、𝒴𝑡 = 𝑒𝑡 + 𝜃1 𝑒𝑡−1 + 𝜃2 𝑒𝑡−2 γ(0) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡 ) = Var(𝑦𝑡 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡 − E(𝑦𝑡 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡 − 0)] = E(𝑦𝑡 2 ) = E[(𝑒𝑡 + 𝜃1 𝑒𝑡−1 + 𝜃2 𝑒𝑡−2 )(𝑒𝑡 + 𝜃1 𝑒𝑡−1 + 𝜃2 𝑒𝑡−2 )] = E(𝑒𝑡 2 ) + 𝜃1 2 E(𝑒𝑡−1 2 ) + 𝜃2 2 E(𝑒𝑡−2 2 ). = 𝜎 2 + 𝜃1 2 𝜎 2 + 𝜃2 2 𝜎 2 = (1 + 𝜃1 2 + 𝜃2 2 )𝜎 2. γ(1) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡−1 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡−1 − E(𝑦𝑡−1 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡−1 − 0)] = E[(𝑦𝑡 )(𝑦𝑡−1 )] = E[(𝑒𝑡 + 𝜃1 𝑒𝑡−1 + 𝜃2 𝑒𝑡−2 )(𝑒𝑡−1 + 𝜃1 𝑒𝑡−2 + 𝜃2 𝑒𝑡−3 )] = 𝜃1 E(𝑒𝑡−1 2 ) + 𝜃1 𝜃2 E(𝑒𝑡−2 2 ) = 𝜃1 𝜎 2 + 𝜃1 𝜃2 𝜎 2 = (1．𝜃1 + 𝜃1 ．𝜃2 )𝜎 2 γ(2) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡−2 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡−2 − E(𝑦𝑡−2 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡−2 − 0)] = E[(𝑦𝑡 )(𝑦𝑡−2 )] = E[(𝑒𝑡 + 𝜃1 𝑒𝑡−1 + 𝜃2 𝑒𝑡−2 )(𝑒𝑡−2 + 𝜃1 𝑒𝑡−3 + 𝜃2 𝑒𝑡−4 )] = 𝜃2 E(𝑒𝑡−2 2 ) = 𝜃2 𝜎 2 γ(3) = 𝐶𝑜𝑣(𝑦𝑡 , 𝑦𝑡−3 ) = E[(𝑦𝑡 − E(𝑦𝑡 ))(𝑦𝑡−3 − E(𝑦𝑡−3 ))] = E[(𝑦𝑡 − 0)(𝑦𝑡−3 − 0)] = E[(𝑦𝑡 )(𝑦𝑡−3 )] = E[(𝑒𝑡 + 𝜃1 𝑒𝑡−1 + 𝜃2 𝑒𝑡−2 )(𝑒𝑡−3 + 𝜃1 𝑒𝑡−4 + 𝜃2 𝑒𝑡−5 )] = 0. 故可得自我共變異函數. (1 + 𝜃1 2 + 𝜃2 2 )𝜎 2 ，k = 0 2 ,k = 1 γ(𝑘) = (𝜃1 + 𝜃1 𝜃2 )𝜎 2 𝜃2 𝜎 ,k = 2 ,k > 2 {0.
(5)
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