5-6 Joint Hypothesis Tests We can also test the joint hypothesis that all m of the correlation coefficients are simultaneously equal to zero using the @-statistic developed by box and pierce: Q=7x2 where T= sample size, m= maximum lag length The O-statistic is asymptotically distributed as a x m However, the Box Pierce test has poor small sample properties, so a variant has been developed, called the ljung-Box statistic O*=T(T+2 This statistic is very useful apfa portmanteau(general)test of linear dependence in time series
5-6 • We can also test the joint hypothesis that all m of the k correlation coefficients are simultaneously equal to zero using the Q-statistic developed by Box and Pierce: where T = sample size, m = maximum lag length • The Q-statistic is asymptotically distributed as a . • However, the Box Pierce test has poor small sample properties, so a variant has been developed, called the Ljung-Box statistic: • This statistic is very useful as a portmanteau (general) test of linear dependence in time series. Joint Hypothesis Tests m 2 = = m k Q T k 1 2 ( ) 2 1 2 2 ~ m m k k T k Q T T = − = +
5-7 An ACF Example (p234 · Question: Suppose that we had estimated the first 5 autocorrelation coefficients using a series of length 100 observations, and found them to be( from 1 to5):0.207,-0.013,0.086,0.005,-0.022 Test each of the individual coefficient for significance, and use both the box-Pierce and ljung-Box tests to establish whether they are jointly significant · Solution: a coefficient would be significant if it lies outside(0. 196, +0.196) at the 5% level, so only the first autocorrelation coefficient is significant Q5.09andQ*=5.26 Compared with a tabulated x(5=1l.1 at the 5%level, so the 5 coefficients are jointly insignificant
5-7 • Question: Suppose that we had estimated the first 5 autocorrelation coefficients using a series of length 100 observations, and found them to be (from 1 to 5): 0.207, -0.013, 0.086, 0.005, -0.022. Test each of the individual coefficient for significance, and use both the Box-Pierce and Ljung-Box tests to establish whether they are jointly significant. • Solution: A coefficient would be significant if it lies outside (-0.196,+0.196) at the 5% level, so only the first autocorrelation coefficient is significant. Q=5.09 and Q*=5.26 Compared with a tabulated 2 (5)=11.1 at the 5% level, so the 5 coefficients are jointly insignificant. An ACF Example (p234)
5-8 3 Moving Average processes Let u,(t1, 2,3, ..) be a sequence of independently and identically distributed (iid )random variables with E(u)=0 and Var(u =0, then y=H+l+日1n1+0u…+e4 is a gth order moving average model ma(q Or using the lag operator notation: Ly,=y L y=+∑0Ln+t1=H+(L)m (L)=1+61L+2L2+…+b 通常,可以将常数项从方程中去掉,而并不失一般性
5-8 • Let ut (t=1,2,3,...) be a sequence of independently and identically distributed (iid) random variables with E(ut )=0 and Var(ut )= 2 , then yt = + ut + 1 ut-1 + 2 ut-2 + ... + q ut-q is a q th order moving average model MA(q). • Or using the lag operator notation: Lyt = yt-1 L iyt = yt-i 通常，可以将常数项从方程中去掉，而并不失一般性。 3 Moving Average Processes t q i t t i yt i L u u (L)u 1 = + + = + = q L = + L + L ++ q L 2 1 1 2 ( )
5-9 移动平均过程的性质 Its properties are E(=u var(y)=%=(1+G2+02+、+02)a2 Covariances (3+661+6。22 O or S= 0 for s>q 自相关函数
5-9 移动平均过程的性质 • Its properties are E( yt )= Var( yt ) = 0 = (1+ )2 Covariances 自相关函数 1 2 2 2 2 + +...+ q + + + + = = + + − for s q for s q s s s q q s s 0 ( ... ) 1,2,..., 2 1 1 2 2
5-10 Example of an Ma Process Consider the following MA(2) process A1=l1+1l1-1+62l12 where u, is a zero mean white noise process with variance o (1) Calculate the mean and variance ofX, (i) Derive the autocorrelation function for this process (i.e express the autocorrelations, Ti, t,,... as functions of the parameters 8, and 82) (iii)If 81=-0.5 and 02=0.25, sketch the acf of X
5-10 Consider the following MA(2) process: where ut is a zero mean white noise process with variance . (i) Calculate the mean and variance of Xt (ii) Derive the autocorrelation function for this process (i.e. express the autocorrelations, 1 , 2 , ... as functions of the parameters 1 and 2 ). (iii) If 1 = -0.5 and 2 = 0.25, sketch the acf of Xt . Example of an MA Process Xt = ut + 1 ut−1 + 2 ut−2 2