We consider the following GARCH(1, 1) model:

$y_t = h_t \epsilon_t$ where $(\epsilon_t)_{t \in \{1, \dots, n\}}$ are i.i.d. random variables with mean 0 and standard deviation 1.

$h_t = \omega + \alpha \epsilon_{t-1}^2 + \beta h_{t-1}^2$

A standard way of estimating the parameters of such a model is to use the quasi maximum likelihood method, i.e. maximising the log-likelihood of the model under the assumption that the $(\epsilon_t)_{t \in \{1, \dots, n\}}$ are gaussian, that is:

$\displaystyle\mathscr{l}(\omega, \alpha, \beta, y_1, \dots, y_n) = -\frac{1}{2}\log(2\pi) - \frac{1}{2}\sum_{t=1}^n\left(\log(h_t) + \frac{y_t^2}{h_t}\right)$

The estimator is shown to have good properties even if this assumption does no hold. The asymptotic distribution of that estimator is:

$\sqrt{n}\left(\hat{\theta}_n - \theta_0\right) \rightarrow \mathcal{N}\left(0, A(\theta_0)^{-1} B(\theta_0) A(\theta_0)^{-1}\right)$

where $\hat{\theta}_n$ is the vector of estimated parameters, $\theta_0$ the true vector of parameters, $B(\theta_0)$ the outer product of the gradient and $A(\theta_0)$ the Hessian matrix, both evaluated at the true value of the parameters.

We can get a good approximation of the Hessian asymptotically by replacing the true value of the parameter by its estimate. But when it comes to the gradient, $B(\hat{\theta}_n)$ is by definition equal to zero since $\hat{\theta}_n$ maximises $\mathscr{l}$. So it cannot be a good approximation of $B(\theta_0)$.

That leads me to my question: how do you estimate the covariance matrix of the asymptotic distribution of the parameters?


1 Answer 1


1) What is set equal to zero is the gradient of the log-likelihood of the whole sample, namely

$$\frac {\partial}{\partial \theta}\displaystyle\mathscr{l}(\theta, y_1, \dots, y_n) = \frac {\partial}{\partial \theta} \left(- \frac{1}{2}\sum_{t=1}^n\left(\log(h_t) + \frac{y_t^2}{h_t}\right)\right) = 0$$

This gives us a sum of derivatives (each derivative for each observation).

$$\frac {\partial}{\partial \theta}\displaystyle\mathscr{l}(\theta, y_1, \dots, y_n) = \frac {\partial}{\partial \theta} \left(- \frac{1}{2}\left(\log(h_1) + \frac{y_1^2}{h_1}\right)\right) +...+ \frac {\partial}{\partial \theta} \left(- \frac{1}{2}\left(\log(h_n) + \frac{y_n^2}{h_n}\right)\right) = 0$$

This does not mean that each derivative separately is set equal to zero

2) The (outer product of the) Gradient, and the Hessian used in the Asymptotic Variance are those of the likelihood for the typical observation $t$ namely,

$$B(\theta_0) = \frac {\partial}{\partial \theta} \left(- \frac{1}{2}\left(\log(h_t) + \frac{y_t^2}{h_t}\right)\right) \cdot \frac {\partial}{\partial \theta} \left(- \frac{1}{2}\left(\log(h_t) + \frac{y_t^2}{h_t}\right)\right)^T$$

$$A(\theta_0) = \frac {\partial^2}{\partial \theta^2} \left(- \frac{1}{2}\left(\log(h_t) + \frac{y_t^2}{h_t}\right)\right)$$

Their sample estimates are sample means of the above evaluated at $\hat \theta$, so, for the outer product of the Gradient,

$$B(\hat \theta) = \frac 1n \sum_{t=1}^n\left[\frac {\partial}{\partial \theta} \left(- \frac{1}{2}\left(\log(h_t) + \frac{y_t^2}{h_t}\right)\right)\cdot \frac {\partial}{\partial \theta} \left(- \frac{1}{2}\left(\log(h_t) + \frac{y_t^2}{h_t}\right)\right)^T\right]$$

which doesn't equal zero.


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