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Suppose $\sigma_{t}^{2} = w + \alpha_{1}y_{t-1}^{2} + \beta_{1}\sigma^{2}_{t-1}$ where $\alpha_{1} + \beta_{1} = 1$ and $y_{t} = \sigma_{t}e_{t}$ and $e_{t}$ is $ N(0,1)$. How do you show that

$E[\sigma_{t+m}^{2}|F_{t-1}] = mw + \sigma^{2}_{t}$? Here is my attempt:

$E[y_{t}^{2}] = \sigma_{t}^{2}$

so $E[\sigma_{t}^{2}] = w + \alpha_{1}E[y_{t-1}^{2}] + \beta_{1}E[\sigma_{t-1}^{2}]$. I am stuck at this part. I know I have to somehow utilize that $\alpha_{1} + \beta_{1} = 1$ but I am not sure how.

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What is $F_t $? – Stefan Hansen Apr 30 '13 at 11:43
Its the past history or filtration. – phil12 Apr 30 '13 at 11:47
up vote 2 down vote accepted

Assume that, for every $t$, $\sigma^2_t$ is $F_t$-measurable and $e_{t}$ is independent of $F_t$.

Then $\sigma^2_{t+1}=w+\gamma_t\sigma^2_{t}$ where $\gamma_t=\alpha_1e_{t}^2+\beta_1$ is independent of $F_{t}$ and has expectation $\alpha_1+\beta_1=1$, and $\sigma^2_{t}$ is measurable with respect to $F_{t}$. Hence, $$ E[\sigma^2_{t+1}\mid F_{t}]=w+E[\gamma_t]\sigma^2_{t}=w+\sigma^2_{t}. $$ Iterating this yields, for every $s\geqslant0$, $$ E[\sigma^2_{t+s}\mid F_{t}]=sw+\sigma^2_{t}. $$

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