case study

# case study - covariance can be any number and is not...

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for case study … use the last year as the testing data set use the first 14 years to come up with the model f(x1,x2,x3) = f(x5,52,53) ...only the distance matters, not the actual location cov(x2, x8) = gamma(6) = cov(x3, x9) only when stationary … if the lag doesn’t change, the cov doesn't either gamma(h) = gamma(-h) cov(xt, xt+h) = cov(h) = cov (xt+h, xt) = cov(-h) If you have a long enough sequence, you can calculate the true mean of the pop. accurately sample autocorrelation function -- normally distributed p(hat) (h) ~ N(0, 1/root(n)) in acf, top bar is +1.96[1/root(n)] lower bar is -1.96[1/root(n)] When they are IID, (because they are independent - it is useless) gamma X(t+h, h) = sigma^2 when h=0 sigma when h doe not = 0 IID has the same mean, same covariance [depends on h] → so it is stationary correlation coefficient is useful to normalize data b/c it is btwn -1 and 1

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Unformatted text preview: covariance can be any number and is not normalized have to compare it to the data to know if it is big or small white noise ~WN(0,sigma^2) = N(0,sigma^2) → normal distribution Random walk xt = w1 + w2 + … + wt E[Xt] = 0, Var[Xt] = t*sigma^2 s > t and cov(Ws, Xt) = 0 [what happens when t>s?] s = 10, t =5 cov(W10, w1+w2+. ..+w10). ..10th days news compared to 5 days news combined gamma (t+h, t) = 2*sigma^2 = t*sigma^2 where t is the # of things in common cov ( xt+h, xt) t = 2, h = 1 cov(x3, x2) = cov(w1+w2+w3, w1+w2) just count how many things are related → w1 and w2 in this case Is {Xt} stationary? NO, it depends on t (stationary should not depend on where you are) Var(Xt) = Var(phi Xt-1) + sigma^2 = phi^2 * Var(Xt-1) + sigma^2 If we have stationarity, Var(Xt) = Var(Xt-1) so they are constant Var(Xt) = sigma^2 / 1-phi^2...
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