Lag operator computations

Question

given $$h_t=\alpha_0+(1-\beta_1)\epsilon^2_{t-1}+\beta_1Lh_t$$ where $$Lh_t=h_{t-1}$$ how can I obtain $$h_t=\alpha_0/(1-\beta_1)+(1-\beta_1)\sum_{1=0}^\inf\beta_1^i\epsilon^2_{t-1-i}$$?

Answer 1

Start with

\begin{eqnarray} h_t &=& \alpha_0 + (1 - \beta_1)\epsilon_{t - 1}^2 + \beta_1 \color{blue}{h_{t - 1}} \\ &=& \alpha_0 + (1 - \beta_1)\epsilon_{t - 1}^2 + \beta_1 \left[\color{blue}{\alpha_0 + (1 - \beta_1)\epsilon_{t - 2}^2 + \beta_1 h_{t - 2}}\right] \\ &=& \alpha_0 (1 + \beta_1) + (1 - \beta_1)[\epsilon_{t - 1}^2 + \beta_1 \epsilon_{t - 2}^2] + \beta_1^2 \color{red}{h_{t-2}} \\ &=& \alpha_0 (1 + \beta_1) + (1 - \beta_1)[\epsilon_{t - 1}^2 + \beta_1 \epsilon_{t - 2}^2] + \beta_1^2 \left[\color{red}{\alpha_0 + (1 - \beta_1)\epsilon_{t - 3}^2 + \beta_1 h_{t - 3}}\right] \\ &=& \alpha_0 (1 + \beta_1 + \beta_1^2) + (1 - \beta_1) [\epsilon_{t - 1}^2 + \beta_1 \epsilon_{t - 2}^2 + \beta_1^2 \epsilon_{t - 3}^2] + \beta_1^2 h_{t - 3} \end{eqnarray}

you can add a few more terms, but it is already fairly straightforward to see the pattern. If you consider $n$-lags you will get

$$ h_t = \alpha_0 \sum_{k=0}^n \beta_1^k + (1 - \beta_1)\sum_{k = 0}^n \beta_1^k \epsilon_{t - k - 1}^2 + \beta_1^n h_{t - n} \tag{1} $$

And now it is a matter of taking the limit $n\to\infty$. If $|\beta_1| < 1$ then the first term in Eqn. (1) is just a convergent geometric series and the last term goes to zero, so you end up with

$$ h_t = \frac{\alpha_0}{1 - \beta_1} + (1 - \beta_1)\sum_{k = 0}^\infty \beta_1^k \epsilon_{t - k - 1}^2 $$

Answer 2

Use generating functions. Call:

$\begin{align*} H(z) &= \sum_{n \ge 0} h_n z^n \\ E(z) &= \sum_{n \ge 0} \epsilon^2_n z^n \end{align*}$

Write your recurrence as:

$\begin{align*} h_{t + 1} &= a_0 + (1 - \beta_1) \epsilon^2_t + \beta_1 h_t \end{align*}$

Multiply by $z^t$, sum over $t \ge 0$, recognize the resulting sums:

$\begin{align*} \sum_{t \ge 0} h_{t + 1} z^t &= a_0 \sum_{t \ge 0} z^t + (1 - \beta_1) \sum_{t \ge 0} \epsilon^2_t z^t + \beta_1 \sum_{t \ge 0} h_t \\ \frac{H(z) - h_0}{z} &= a_0 \frac{1}{1 - z} + (1 - \beta_1) E(z) + \beta_1 H(z) \end{align*}$

Solve for $H(z)$, split into partial fractions:

$\begin{align*} H(z) &= \frac{h_0 + (a_0 - h_0 + E(z) (1 - \beta_1)) z - E(z) (1 - \beta_1) z^2} {1 - (\beta_1 + 1) z + \beta_1 z^2} \\ &= - \frac{E(z) (1 - \beta_1)}{\beta_1} + \frac{a_0 }{1 - \beta_1} \cdot \frac{1}{1 - z} + \frac{E(z) - \beta_1 (a_0 + 2 E(z)) + E(z) \beta_1^2 + \beta_1 (1 - \beta_1)} {\beta_1 (1 - \beta_1) (1 - \beta_1 z)} \end{align*}$

We want the coefficient of $z^n$ in this, written $[z^n]$. This is a handful. but the pieces are easy to handle:

$\begin{align*} [z^n] E(z) &= \epsilon^2_n \\ [z^n] \frac{1}{1 - \beta_1 z} &= \beta_1^n \\ [z^n] \frac{E(z)}{1 - \beta_1 z} &= \sum_{0 \le t \le n} \epsilon^2_t \beta_1^{n - t} \\ [z^n] \frac{1}{1 - z} &= 1 \end{align*}$

The rest is a bit of messy algebra.