stationarity - Joint distribution = f(x1 x2 … xt...

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stationarity - you only have one sample size in time series use the pattern over different times to do inferencing stability assumption...the behavior of the time series is the same at all time periods the distribution of x1 is the same as the dist of x2... E(Xt) = E(r1+r2+...+rt) = E(r1) + E(r2) + …. + E(rt) = 0 the expected value is 0 because it is r ~ N(0, sigma^2) if it was N(1, sigma^2) it would be 1 + 1 + 1….. E(Xs, Xt) = E(Xs Xt) - E(Xs) - E(Xt) = E[(Xs-(mu)s)(Xt-(mu)t)] If Xt is stationary, then E(Xt) = E(Xt-1) = E(Xt-2)... when you put in phi0 or theta0 ***For AR(p) E(Xt) = phi0 / 1 - phi1 - phi2 For MA(q) E(Xt) = theta(0) if Wt ~ N(0, sigma^2) autocovariance - the covariance for time series the same path at different time points cov(x,x) = var(x) autocorrelation = covariance(s,t) / sqrt[var(t) * var (s)]
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Unformatted text preview: Joint distribution = f(x1, x2, …, xt) everything about the random variables Check for weak stationary: mean doesn't change over time variance does not change covariance doesn’t change over time (does not depend on time) implies that the variance is going to be stable (the same over time) As long as the lag is the same, the difference in variance should be the same second order stationary (covariance stationary) only concentrate on mean, variance, and covariance to see if it is weakly stationary gamma = covariance gamma(t + h, t) = E(Xt+h - mu) * (Xt - mu) = E(Xh - mu) * (X0 - mu) gamma (h, 0) autocorrelation = gamma(h) / gamma(0) cov(1,2) = cov(2,1) gamma(0) = cov(xt, xt) = var(xt) = sigma^2...
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