New answers tagged asymptotics
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Is this a typo on P.75, Theorem 5.52 of the book "Asymptotic Statistics" by Van der Vaart?
First point: Since $d(\theta, \theta_0)<\delta$ it should be possible to select $\theta = \theta_0$ which gives $\sup=0>-C\cdot0^\alpha$.
Without being specific for this problem, if $x^*=\arg\...
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5
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Prove that the Deviance and the Generalised Pearson Statistic are asymptotically equivalent
In $\rm [I] ~p.68,$ Jørgensen notes
The saddlepoint approximation may be viewed as refinement of the normal approximation. The deviance $D(y, \mu) $ is approximately a quadratic form in $y$ for $\...
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Why is the asymptotic bias of the maximum likelihood estimate $b(\theta) = \frac{b_1(\theta)}{n}+\frac{b_2(\theta)}{n^2}+...$?
Thanks to the comment from User1865345, I found the answer in Quenoullie's jackknife paper. Here's a slightly modified explanation:
Given a series of observations $y_1, y_2, ..., y_n$, if the ...
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5
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Finding the limiting distribution of $\sqrt{n} (\hat{\tau} - \tau)$ as $n \rightarrow \infty$ for $N(\mu, \mu^2 \tau)$
Whuber's comment is to the point, since it matters whether you estimate two parameters or one.
Algebraically, you are correct up to and including the computation of the expected value of the negative ...
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2
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Why is the asymptotic bias of the maximum likelihood estimate $b(\theta) = \frac{b_1(\theta)}{n}+\frac{b_2(\theta)}{n^2}+...$?
This is not exclusive to the maximum likelihood estimators.
In general, it is often a valid expansion of the expected value of an estimator, say $\hat \theta, $
$$\mathbb E\hat{\theta}=\theta +\frac{...
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