Is it true that the asymptotic covariance matrix is equal to the covariance matrix of parameter estimates? If not, what is it? And what is the difference between the covariance matrix and the asymptotic covariance matrix in that case? Thanks in advance!
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3The asymptotic covariance matrix is an approximation to the covariance matrix of the sampling distribution of parameter estimates that gets better as the number of samples on which the parameter estimates are based increases. – tchakravarty Jan 10 '15 at 14:02
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Given an iid sample $(X_1,\ldots,X_N)$ from a parametric distribution with density $f_\theta(\cdot)$, $\theta$ being the unknown parameter, an estimator $\hat{\theta}(X_1,\ldots,X_N)$ has a distribution with mean $\mu_n(\theta)$ and variance-covariance matrix $\Sigma_n(\theta)$. So $\Sigma_n(\theta)$ is the variance-covariance matrix of $\hat{\theta}(X_1,\ldots,X_N)$ in the sense that $$\mathbb{E}_\theta\left[ \left\{\hat{\theta}(X_1,\ldots,X_N)-\mu_n(\theta)\right\} \left\{\hat{\theta}(X_1,\ldots,X_N)-\mu_n(\theta)\right\}^{\text{T}} \right] = \Sigma_n(\theta)\,.$$
Now, if $\hat{\theta}(X_1,\ldots,X_N)$ is a convergent estimator and if there exists a limiting distribution for $\hat{\theta}(X_1,\ldots,X_N)$, it means there exists a sequence $(\phi_n)$ increasing to $+\infty$, e.g., $\phi_n=\sqrt{n}$, such that $$\phi_n\left\{\hat{\theta}(X_1,\ldots,X_N)-\mu_n(\theta)\right\}\stackrel{\text{dist}}{\longrightarrow} G_\theta$$ where $G_\theta$ denotes a distribution indexed by $\theta$ and the limiting distribution of the l.h.s. This limiting distribution has a variance $\Xi_\theta$ that is called the asymptotic variance.
Xi'an
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Why do you say "e.g. $\phi_n=\sqrt n$"? Shouldn't it always be $\sqrt n$? – user56834 Jan 10 '18 at 14:52
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There are estimators that converge faster or more slowly than $\sqrt{n}$ as e.g. the Uniform one. – Xi'an Jan 10 '18 at 14:58