# Reparametrization of the GJR-GARCH(1,1) model (Asymmetric GARCH models)

Xia et al. (2017) state that the power-transformed and threshold GARCH(1,1,1) model is equivalent to the GJR-GARCH(1,1) model (see page 355 of Xia et al. (2017)). It is not clear why the first model can be rewritten as the latter model.

MODEL 1: PTTGARCH(1,1,1)

The power-transformed and threshold GARCH(1,1,1) model takes the following form:

(1) $y_{t}$ = $\sigma_t$ $\epsilon_t$

(2) $\sigma_t^2$ = $\alpha_0$ + $\alpha_1$ $(y_{t-1}^{+})^2$ + $\alpha_2$ $(y_{t-1}^{-})^2$ + $\beta$ $\sigma_{t-1}^2$,

where $(y^{+})$ denotes positive values of random variable $y$, and where $(y^{-})$ denotes negative values of random variable $y$; moreover, $\alpha_0$, $\alpha_1$, $\alpha_2$, $\beta$ are (unknown) parameters. Essentially, this model splits the effect of past returns ($y_{t-1}$) on volatility into two distinct effects: the effect of past positive returns ($y_{t-1}^{+}$), and the effect of past negative returns ($y_{t-1}^{-}$).

MODEL 2: GJR-GARCH(1,1)

The GJR-GARCH(1,1) model takes the following form:

(a) $y_{t}$ = $\sigma_t$ $\epsilon_t$

(b) $\sigma_t^2$ = $\alpha_0$ + $\gamma_1$ $(y_{t-1})^2$ + $\gamma_2$ $(y_{t-1})^2$ $I_{(y_{t-1}>0)}$ + $\beta$ $\sigma_{t-1}^2$,

where $I_{(y_{t-1}>0)}$ is an indicator function which takes value 1 if $y_{t-1}$ is positive, and takes value 0 if $y_{t-1}$ is negative.

QUESTIONS

Is it true that the two models are equivalent? It seems that the two models are not equivalent in that I cannot transform equation (2) into equation (b), and I cannot transform equation (b) into equation (2). Am I missing something here? Is there a mathematical way to show that equation (2) is exactly the same as equation (b), and/or vice versa?

References

Xia et al. (2017). Bayesian Analysis of Power-Transformed and Threshold GARCH Models: A Griddy-Gibbs Sampler Approach. Computational Economics, Volume 50, Issue 3, pp. 353–372.

I think you can rewrite $y_{t-1} = y_{t-1}^{+} + y_{t-1}^{-}$.

Also $y_{t-1}^{+} = y_{t-1} I_{(y_{t-1}>0)}$ and $y_{t-1}^{-} = y_{t-1} I_{(y_{t-1} \le 0)}$

Since the intersection of the two sets $\{y_{t-1}>0\}$ and $\{y_{t-1} \le 0\}$ is zero it is easy to see that

$(y_{t-1})^2 = (y_{t-1})^2 I_{(y_{t-1}>0)} + (y_{t-1})^2 I_{(y_{t-1} \le 0)} = (y_{t-1}^{+})^2 + (y_{t-1}^{-})^2$

Hence $\gamma_1$ $(y_{t-1})^2 = \gamma_1(y_{t-1}^{+})^2 + \gamma_1(y_{t-1}^{-})^2$ and you can rewrite

(b) $\sigma_t^2$ = $\alpha_0$ + $\gamma_1$ $(y_{t-1})^2$ + $\gamma_2$ $(y_{t-1})^2$ $I_{(y_{t-1}>0)}$ + $\beta$ $\sigma_{t-1}^2$ as

(b) $\sigma_t^2 = \alpha_0 + \gamma_1(y_{t-1}^{+})^2 + \gamma_1(y_{t-1}^{-})^2 + \gamma_2(y_{t-1}^{+})^2 + \beta \sigma_{t-1}^2$

Now pose $\alpha_1 = \gamma_1 + \gamma_2$ and $\alpha_2 = \gamma_1$ and you get

(2) $\sigma_t^2$ = $\alpha_0$ + $\alpha_1$ $(y_{t-1}^{+})^2$ + $\alpha_2$ $(y_{t-1}^{-})^2$ + $\beta$ $\sigma_{t-1}^2$,