Forecast-error variance of an AR(1)

Question

A stationary AR(1) has persistence \phi = 0.7 and white-noise shocks with variance \sigma^2 = 1. What is the variance of the 2-step-ahead forecast error?

Accepted Answer

Write the forecast error 2 steps ahead by iterating the AR(1). The error accumulates the shocks that arrive between now and the forecast date: X{t+2} - E[X{t+2}\mid Xt] = \varepsilon{t+2} + \phi\,\varepsilon{t+1}. These two shocks are uncorrelated, so their variances add (the second weighted by \phi^2): \operatorname{Var}(	ext{2-step error}) = \sigma^2 + \phi^2\sigma^2 = \sigma^2\,(1 + \phi^2). In general the k-step forecast-error variance is \sigma^2\dfrac{1-\phi^{2k}}{1-\phi^2}, which for k=2 reduces to \sigma^2(1+\phi^2). Plugging in \phi = 0.7 and \sigma^2 = 1: \operatorname{Var} = 1 	i…

Forecast-error variance of an AR(1)

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