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I am in the process of working through some problem sets. I have studied some time series, but my knowledge of ARCH models is pretty basic. I am given the following information:

$Y_t = a_0 + a_1 Y_{t-1} + \epsilon_t$

where $\epsilon_t | I_{t-1} \sim N(0,h_t)$ and $h_t = \alpha_0 + \alpha_1 \epsilon_{t-1}^2 + \alpha_2 \epsilon_{t-2}^2$

I solved for the following,

Conditional variance:

$E(Y_t | I_{t-1}) = a_0 + a_1 Y_{t-1}= \mu$

then

$Var(Y_t)=E[(Y_t- \mu)^2] = E(E_t^2) = ht = \alpha_0 + \alpha_1 \epsilon_{t-1}^2 + \alpha_2 \epsilon_{t-2}^2$

Unconditional Variance:

At this point I think we can create a new series for $Y_t$ since we are not conditioning, so I wrote,

$Y_t = a_0 + a_1 (a_0 + a_1 Y_{t-2} + \epsilon_{t-1}) + \epsilon_t$, repeat this infinitely many times and get

$Y_t = \frac{a_0}{1-a_1} + \sum_{j=0}^\infty a_1^j \epsilon_{t-j}$

$E(Y_t) = \frac{a_0}{1-a_1}$ and

$Var(Y_t) = E[(Y_t - \mu )^2] = \sum_{j=0}^\infty a_1^{2j} \epsilon_{t-j}^2 = (\alpha_0 + \alpha_1 \epsilon_{t-1}^2 + \alpha_2 \epsilon_{t-2}^2) + a_1^2 (\alpha_0 + \alpha_1 \epsilon_{t-2}^2 + \alpha_2 \epsilon_{t-3}^2) + a_1^4(\alpha_0 + \alpha_1 \epsilon_{t-3}^2 + \alpha_2 \epsilon_{t-4}^2) + \dots$

I am not sure if this cleans up nicely, but what does this say for the variance? it seems like the unconditional is just going to blow up infinitely (given I could have made a mistake in the derivation of this question). I am also not sure if I should state some stability conditions in this question similar to AR(1) processes (e.g., $\alpha_1 < 1$ and $\alpha_2 < 1$), since these are variances do those conditions apply?

If I was to conclude a statement, is it acceptable to stay the model captures volatility clustering via 2 period lags?

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I cannot directly answer your question, but I think I can shed some light. What I seem to show is that under some restrictions, the unconditional variance is finite. However, I am not sure how to relate my answer to the notion of volatility clustering.

Note that if we assume stationarity, then $\sigma^2=E(\epsilon_t^2)$ for all $t$ and it follows by the law of total expectations that $$\sigma^2=E(E(\epsilon_t^2|I_{t-1}))=E(h_t)=\alpha_0+\alpha_1\sigma^2+\alpha_2\sigma^2$$ using your expression for $h_t$. Thus, $\sigma^2=\frac{\alpha_0}{1-(\alpha_1+\alpha_2)}$ is a solution whenever $\alpha_1+\alpha_2<1$.

Now, by the law of total variance, \begin{align*}V(Y_t)&=E(V(Y_t|I_{t-1}))+V(E(Y_t|I_{t-1}))\\ &=E(\epsilon_t^2)+V(\mu_t)\\ &=\sigma^2+a_1^2V(Y_{t-1}) \end{align*} using your results, where $\mu_t=a_0+a_1 Y_{t-1}$. Assuming stationarity, and thus that the variance is constant for all $t$, it follows that $$V(Y_t)=\frac{\sigma^2}{1-a_1^2}$$ is a solution for $a_1\neq\pm 1$ and $\alpha_1+\alpha_2<1$.

I encourage you to clearly note my assumptions. I cannot further assess this answer as I know next to nothing about time series.

(Also note that we from the assumption of stationarity have that $E(Y_t)=a_0+a_1E(Y_{t-1})$ implies that $$E(Y_t)=\frac{a_0}{1-a_1}$$ is a solution.)

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