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Let's say we have a GARCH($1,1$) process specified as follows:

$y_t = \epsilon_t \sqrt h_t, \quad \epsilon_t \sim N(0,1) \quad \text{i.i.d.}$

$h_t = a_0 + a_1 y^2_{t-1} + b_1 h_{t-1}.$

If we were trying to estimate the parameters $\Theta = (a_0, a_1, b_1)$, and we have a sample $\{y_k\}^n_{k=1}$, would it make a difference if we estimated the parameters from $\{y_k\}^n_{k=1}$ or $100\times\{y_k\}^n_{k=1}$? I basically want to know which (if any) parameters would also scale. I am rescaling the $y$ for numerical stability.

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    $\begingroup$ You need to take a square root of $h_t$ in the first equation. To answer your question, $a_0$ will scale quadratically with $y_t$, while $a_1$ and $b_1$ will stay the same. $\endgroup$ – Richard Hardy Dec 4 '18 at 6:54
  • $\begingroup$ Thanks @RichardHardy ! So if I'm estimating these parameters using MLE, and scaling returns by x100 makes the optimisation more stable, is it still okay to estimate the parameters like this and then rescale the estimates by the invariance property? What would be the rescaling factor? $\endgroup$ – Vykta Wakandigara Dec 5 '18 at 8:03
  • $\begingroup$ I think all is fine. As I said, $a_0$ scales quadratically with $y_t$. So if you multiply $y_t$ by a positive constant $c$, the resulting $a_0$ will be $c^2$ times larger than the one on the original scale. $\endgroup$ – Richard Hardy Dec 5 '18 at 9:45
  • $\begingroup$ Also, how would one estimate the proper standard errors? Is there a text that I can read to access practical garch estimation issues and other mathematical expositions? $\endgroup$ – Vykta Wakandigara Dec 5 '18 at 13:42
  • $\begingroup$ I think that decent software for GARCH estimation will do all those things for you, including scaling for numerical purposes and outputting standard errors. There are many sources you could read. Perhaps the most solid is Francq and Zakoian's book "GARCH Models: Structure, Statistical Inference and Financial Applications" (2010). $\endgroup$ – Richard Hardy Dec 5 '18 at 14:43
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$a_0$ will scale quadratically with the scaling of $y_t$, while $a_1$ and $b_1$ will stay the same. Here is why. Take the original model

\begin{aligned} y_t &= \sqrt h_t \epsilon_t, \\ h_t &= a_0 + a_1 y^2_{t-1} + b_1 h_{t-1}, \\ \epsilon_t &\sim i.i.N(0,1) \\ \end{aligned}

and scale $y_t$ with a positive constant $c$. This turns the model into

\begin{aligned} cy_t &= \sqrt{c^2 h_t} \epsilon_t, \\ c^2 h_t &= c^2 a_0 + a_1 (cy_{t-1})^2 + b_1 (c^2 h_{t-1}), \\ \epsilon_t &\sim i.i.N(0,1) \\ \end{aligned}

which can be expressed as

\begin{aligned} \tilde y_t &= \sqrt{\tilde h_t} \epsilon_t, \\ \tilde h_t &= \tilde a_0 + a_1\tilde y^2_{t-1} + b_1 (\tilde h_{t-1}), \\ \epsilon_t &\sim i.i.N(0,1) \\ \end{aligned}

where $\tilde y_t = c y_t$ and $\tilde a_0 = c^2 a_0$.

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Define $\tilde{y}_t = cy_t$ with $c > 0$ and you get $$ \tilde{y}_t = \sqrt{\tilde{h}_t}\epsilon_t $$ where $\tilde{h}_t = c^2 h_t$ is the new volatility process. The dynamics for this scaled up volatility process can be obtained by multiplying both sides of the original equation by $c^2$: \begin{align*} \tilde{h}_t &= c^2 a_0 + c^2 a_1 y^2_{t-1} + c^2 b_1 h_{t-1} \\ &= \tilde{a}_0 + a_1 \tilde{y}^2_{t-1} + b_1 \tilde{h}_{t-1}. \end{align*} The volatility is a linear function of its past value, and the past squared observation. You can see that $a_1$ and $b_1$ stay the same, but $a_0$ turns into $\tilde{a}_0 = c^2 a_0$.

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  • $\begingroup$ You are right, I missed the effect on $a_1$. $\endgroup$ – Richard Hardy Dec 9 '18 at 20:27
  • $\begingroup$ I actually think I misspecified the GARCH process in the question originally. The volatility equation should have : $h_t = a_0 + a_1 y^2_{t-1} + b_1 h_{t-1}$. I suppose the standard formulation writes $y_t$ as $\epsilon_t$. You may review the edits that I have made. Therefore I think the effect that @RichardHardy sighted still hold. $\endgroup$ – Vykta Wakandigara Dec 9 '18 at 21:59
  • $\begingroup$ @VyktaWakandigara, indeed. I did not notice this so I wrote the answer for the classical form of the GARCH model. Now that you have fixed the model, my original answer is correct again. $\endgroup$ – Richard Hardy Dec 10 '18 at 6:15

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