If I have a process of the form $Y_t=\mu+\phi Y _{t-1} + \epsilon_t$, how is the population variance derived? Assuming that $\epsilon_t$ has a zero mean and a variance of 2.
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1Go down recursively to find an expression of Yt. – Somabha Mukherjee Oct 07 '14 at 12:27
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Provided that $|\phi|<1$, you can do backwards substitution to arrive at $$ Y_t=\mu(1+\phi+\phi^2+\cdots)+\epsilon_t+\phi\epsilon_{t-1}+\cdots=\frac{\mu}{1-\phi}+\sum_{j=0}^\infty\phi^j\epsilon_{t-j}. $$ From here, you can compute $$ \text{Var}(Y_t)=E\left(\sum_{j=0}^\infty\phi^j\epsilon_{t-j}\sum_{j=0}^\infty\phi^j\epsilon_{t-j}\right)=\sum_{j=0}^\infty\phi^{2j}E(\epsilon_{t-j}^2)=\frac{\text{Var}{\epsilon}}{1-\phi^2}=\frac{2}{1-\phi^2}. $$ The second equality uses the fact that when the indices don't match, the expectation vanishes. Finally, if your process doesn't start at infinity in the past, you need to define $\text{Var}(Y_0)$ as $\frac{\text{Var}{\epsilon}}{1-\phi^2}$.
Kim Jong Un
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