TS Module 3: Trends HW

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TS Module 3: Trends HW

TS Module 3: Trends HW

Author	Message
nacho	nacho posted 13 Years Ago #11159 #
Forum Newbie Group: Forum Members Posts: 3, Visits: 1	LIAPP, I used the following 'givens' and I made a table like the one below to help me through the calculations. Hope it helps. Givens: N=50; Var(e)=1; Var(Y_t) = Var(mu)+Var(A)+Var(B); rho_1=Cov(Y_t,Y_t-1)/Var(Y_t); that last 'given' is explained in this forum and in the book (I think) Table: # Var Cov(Y_t,Y_t-1) rho_1 Formula from bottom of page 28 1 0+1+1=2; Var(e)=1; 1 / 2 = .5; 2/50 * [1 + 2(50-1)/50(.5)]=0.0792 2 0+1+1/4=1.25; .5Var(e)=.5 ; .5/1.25=.4 ; 1.25/50[1+1.96(.4)=0.0446 and so on and so forth. I'd write it al lout, but I don't have the time. 0
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NEAS	NEAS posted 10 Years Ago #13039 #
Supreme Being Group: Administrators Posts: 4.5K, Visits: 1.6K	Jacob: Formula 3.2.3 applied to the first time series gives y-bar = 0.0792. But mu + e_t + e_t-1 from 1 to 50 is (1/2500) × Var(50 mu + å ₍₅₀₎ + 2 × e₍₁₎ + …+ 2 × e₍₄₉₎ ) = (1/2500) × (0 + Var(e ₍₅₀₎ + 4 × (e₍₁₎ + …+ e₍₄₉₎ ) ) ) = (1 + 4 × 49) / 2500 = 0.0788 Rachel:The difference is subtle. The formula in the textbook assumes 50 observations are taken from this time series. Your second method assumes the time series starts at observation #1. The difference is one error term. Your second method doesn’t have an error term for period zero. Including this error term raises the variance of y-bar by 0.004. Intuition: The formula in the textbook assumes the time series process is infinite. We observe fifty values, such as 1.014 through 1.063, and compute the variance of y-bar. Your computation assumes the series starts at observation #1, so the error term e₀ is zero. Use the formula in the textbook, since its assumptions are closer to real life. We may observe only certain observations, but the time series may have existed for much longer. 0
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