# Tag Info

## New answers tagged asymptotics

1

You are missing a lot of context here (presumably taken from linear regression) so I will focus solely on the asymptotic distribution:$^\dagger$ $$\sqrt N (\sigma^2 - \hat{\sigma}^2) \overset{\text{Dist}}{\rightarrow} \text{N}(0, 2 \sigma^4). \quad \quad$$ The left-hand-side here is a scaled version of the estimation error, which is affected by the size $N$. ...

1

Note that this expectation does not exist for all $g$, because if $g$ is sufficiently fast-growing then $E[g(\bar X_n)]$ may not be finite for any $n$. For example, this happens if each $X_i$ follows a standard normal distribution, so that $\bar X_n \sim N(0, \frac 1 {\sqrt{n}})$, and $g(x) = e^{x^8}$. (Though the specific $g$ that you are most interested in ...

2

Let us begin with the example $g(x)=|x| 1_{|x|>c}$. Suppose that $X_1,\dots, X_n$ are centered and have finite second moment $\sigma^2$. Denote $F_n(x)=\mathbb{P}\left(\frac{1}{\sigma\sqrt{n}}\overline{X}_n\le x\right)$. We work with $\frac{1}{\sigma \sqrt{n}}\overline{X}_n$ instead of $\overline{X}_n$ as it simplifies the computations, please change $c$ ...

0

Start by showing that $X=O_p(1)$, then you can take advantage of the distribution of $X_n$ being close to that of $X$ to show that for large $n$, $X_n$ can't be much more likely than $X$ to exceed any specified bound $M$

3

I find the literature in MLE a bit fuzzy with nomenclature here, so I might have some stuff off, and I will try to stick to the nomenclature you introduced. We have the observed Fisher information: $$\left[\mathcal {J}(\theta)\right]_{ij} = -\left(\frac{\partial^2 \log f}{\partial \theta_i \partial \theta_j}\right)$$ And the empirical Fisher information: \...

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