3
$\begingroup$

I'm currently estimating a DCC-type model by maximum likelihood. Im using the command solnp and it return an object where I can compute the Hessian H evaluated at the optimal values.

If I want to find the covariance matrix of the estimators, following Davidson and MacKinnon (2004), I can compute $$ \hat{Var} ( \hat{\theta} ) = -H^{-1} ( \hat{\theta} ) $$ However, there is a robust estimation for distribution miss-especification: the sandwich estimator or the quasi-maximum likelihood estimator (QMLE). It is as follows: $$ \hat{Var} ( \hat{\theta} ) = H^{-1} ( \hat{\theta} ) G^{T} ( \hat{\theta} ) G ( \hat{\theta} ) H^{-1} ( \hat{\theta} ) $$ where G is the gradient.

As solnp does not compute the gradient, I am using the command gHgenb to compute it at the optimal values of the parameters.

The problem is that when I compute the non-robust covariance matrix I get the following results:

                     Mean           SE          t        Pvalue
[Joint]dcca1 0.0087397376 9.267334e-03  0.9430692  3.461028e-01
[Joint]dccb1 0.9653333580 1.255250e-02 76.9036529 1.723509e-278
             0.0001513922 4.115874e-05  3.6782522  2.604074e-04

And when I compute the sandwich estimators I get the following results:

                     Mean           SE          t Pvalue
[Joint]dcca1 0.0087397376 6.622702e-09  1319663.4      0
[Joint]dccb1 0.9653333580 6.706874e-08 14393193.4      0
             0.0001513922 1.538495e-10   984028.1      0

where the standard errors are extremely low.

The questions is: what should I do? Should I multiplicate any of the matrix by the number of observations (which one?)?, as Bollerslev and Wooldridge (1992) state that the non-robust matrix is: $$ \hat{Var} ( \hat{\theta} ) = -H^{-1} ( \hat{\theta} ) / T $$

Thank you so much.

References:

Bollerslev, T., & Wooldridge, J. M. (1992). Quasi-maximum likelihood estimation and inference in dynamic models with time-varying covariances. Econometric reviews, 11(2), 143-172.

Davidson, R., & MacKinnon, J. G. (2004). Econometric theory and methods (Vol. 5). New York: Oxford University Press.

$\endgroup$

2 Answers 2

1
$\begingroup$

I just found the answer to my problem by comparing the manual method versus the QMLE robust standard errors for a GARCH using the package fGarch. The solution is as follows: Let hessian be the hessian computed by the optimization at the optimal values. Now, I computed a $N\times K$ matrix (where $N$ is the number of observations and $K$ the number of parameters). Using the command grad, from the package numDeriv, I computed the gradient for every observation in the optimal parameters and with those values I filled the matrix $N\times K$, which I called hessian. Then, the Information matrix and the QMLE robust standard errors are computed as follows:

meat<-t(gradient)%*%gradient
Info<-solve(hessian)%*%meat%*%solve(hessian)
SE_robust <- sqrt(diag(Info))
$\endgroup$
0
$\begingroup$

The sandwich matrix is $-H^{-1}VH^{-1}$ for $V$ an estimate of the expectation of $G(\theta_0)^TG(\theta_0)$, where $G$ is a row vector (following the poster's notation). Using $G(\hat{\theta})^TG(\hat{\theta})$ to approximate this expectation will not provide a good answer, because $G(\hat{\theta}) = 0$ by definition of $\hat{\theta}$!

To provide slightly more detail: if $\hat{\theta}$ solves $\sum_{i=1}^{n}\psi_i(\theta) = 0$ where $\psi_i$ are independent functions of the response $Y_i$, and $\theta_0$ solves $\Psi(\theta) = 0$ where $\Psi = E(\psi_1)$, then $\sqrt{n}(\hat{\theta} - \theta_0)$ is asymptotically $N(0,V)$ where $V = H^{-1}SH^{-1}$. These matrices are $H = E\nabla_\theta\psi_1(\theta_0)$ and $S = E\psi_1(\theta_0)\psi_1(\theta_0)^T$. We approximate $H$ by $\tilde{H} = \sum_{i=1}^{n}\psi^{\prime}_{i}(\hat{\theta})$ and $V$ by $\tilde{V} = \sum_{i=1}^{n}\psi_i(\hat{\theta})\psi_i(\hat{\theta})^T$. This last step is the part that's different from the question: $\sum_{i=1}^{n}\psi_i(\hat{\theta}) = 0$ so $\left(\sum_{i=1}^{n}\psi_i(\hat{\theta})\right)\left(\sum_{i=1}^{n}\psi_i(\hat{\theta})\right)^T = 0$, but $\sum_{i=1}^{n}\psi_i(\hat{\theta})\psi_i(\hat{\theta})^T \neq 0$.

Here is a useful reference: https://www.stat.berkeley.edu/~census/mlesan.pdf

My explanation is a poor recreation of material from van der Vaart's Asymptotic Statistics, Section 5.3, page 51/52; that's where to look for the good stuff!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.