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So, I have this function: $ y_t=v_t+\rho v_{t-1}+\rho^2 v_{t-2}+\dots+\rho^{t-1}v_1+\rho^ty_0 $

And I want to find the variance (and after that the covariance, but I should be able to do that..). I know that the variance is: $ Var[y_t]=\frac{var[v]}{1-\rho^2}=\frac{\sigma^2_v}{1-\rho^2} $

But somewhere I mess everything up : (

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    Did you get something from the solution below?2011-04-11

1 Answers 1

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You probably assume that the sequence $(v_t)$ is i.i.d. and that $(v_t)$ and $y_0$ are independent. Write $s_t$ for the variance of $y_t$ and $u$ for the variance of $v_t$. The variance of an independent sum is the sum of the variances, hence $ s_t=u+\rho^2u+\cdots+\rho^{2t-2}u+\rho^{2t}s_0=\frac{1-\rho^{2t}}{1-\rho^2}u+\rho^{2t}s_0, $ where the second formula assumes that $|\rho|<1$. To get your formula, one can either assume that $t\to+\infty$, or that $(y_t)$ is stationary. In the first case, $\rho^{2t}\to0$ hence $s_t\to u/(1-\rho^2)$. In the second case, $s_t=s_0$ hence plugging this value of $s_0$ in the formula above, one gets $s_t=u/(1-\rho^2)$ for every $t$.