Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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 $s\geqslant0$, $$ E[\sigma^2_{t+s}\mid F_{t}]=sw+\sigma^2_{t}. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.