经济代写|计量经济学代写Econometrics代考|Best27

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

经济代写|计量经济学代写Econometrics代考|Asymptotic Normality and Central Limit Theorems

There is the same sort of close connection between the property of asymptotic normality and central limit theorems as there is between consistency and laws of large numbers. The easiest way to demonstrate this close connection is by means of an example. Suppose that samples are generated by random drawings from distributions with an unknown mean $\mu$ and unknown and variable variances. For example, it might be that the variance of the distribution from which the $t^{\text {th }}$ observation is drawn is
$$\sigma_t^2 \equiv \omega^2\left(1+\frac{1}{2}(t(\bmod 3))\right) .$$
Then $\sigma_t^2$ will take on the values $\omega^2, 1.5 \omega^2$, and $2 \omega^2$ with equal probability. Thus $\sigma_t^2$ varies systematically with $t$ but always remains within certain limits, in this case $\omega^2$ and $2 \omega^2$.

We will suppose that the investigator does not know the exact relation (4.26) and is prepared to assume only that the variances $\sigma_t^2$ vary between two positive bounds and average out asymptotically to some value $\sigma_0^2$, which may or not be known, defined as
$$\sigma_0^2 \equiv \lim {n \rightarrow \infty}\left(\frac{1}{n} \sum{t=1}^n \sigma_t^2\right) .$$
The sample mean may still be used as an estimator of the population mean, since our law of large numbers, Theorem 4.1, is applicable. The investigator is also prepared to assume that the distributions from which the observations are drawn have absolute third moments that are bounded, and so we too will assume that this is so. The investigator wishes to perform asymptotic statistical inference on the estimate derived from a realized sample and is therefore interested in the nondegenerate asymptotic distribution of the sample mean as an estimator. We saw in Section $4.3$ that for this purpose we should look at the distribution of $n^{1 / 2}\left(m_1-\mu\right)$, where $m_1$ is the sample mean. Specifically, we wish to study
$$n^{1 / 2}\left(m_1-\mu\right)=n^{-1 / 2} \sum_{t=1}^n\left(y_t-\mu\right),$$
where $y_t-\mu$ has variance $\sigma_t^2$.

经济代写|计量经济学代写Econometrics代考|Asymptotic Normality and Central Limit Theorems

Since all the $y_t$ ‘s are mutually independent and have mean zero, no term in the quadruple sum of (4.27) can be nonzero unless the indices either are all the same or fall into pairs (with, for instance, $r=t$ and $s=u$ with $r \neq s$ ). If all the indices are the same, then the value of the corresponding term is just the fourth moment of the distribution of the $y_t$ ‘s. But there can only be $n$ such terms. With the factor of $n^{-2}$ in (4.27), we see that these terms contribute to (4.27) only to order $n^{-1}$. On the other hand, the number of terms for which the indices fall in pairs is $3 n(n-1),{ }^4$ which is $O\left(n^2\right)$. Thus the latter terms contribute to (4.27) to the order of unity. But, and this is the crux of the argument, the value of each of these terms is just the square of the variance of each $y_t$, or $\sigma^4$. Thus, to leading order, the fourth moment of $S_n$ depends only on the variance of the $y_t$ ‘s; it does not depend on the fourth moment of the distribution of the $y_t$ ‘s. ${ }^5$.

A similar argument applies to all the moments of $S_n$ of order higher than 2. Thus, to leading order, all these moments depend only on the variance $\sigma^2$ and not on any other property of the distribution of the $y_t$ ‘s. This being so, if it is legitimate to characterize a distribution by its moments, then the limiting distribution of the sequence $\left{S_n\right}_{n=1}^{\infty}$ depends only on $\sigma^2$. Consequently, the limiting distribution must be the same for all possible distributions with the variance of $y_t$ equal to $\sigma^2$, regardless of other properties of that distribution. This means that we may calculate the limiting distribution making use of whatever distribution we choose, provided it has mean 0 and variance $\sigma^2$, and the answer will be independent of our choice.

The simplest choice is the normal distribution, $N\left(0, \sigma^2\right)$. The calculation of the limiting distribution is very easy for this choice: $S_n$ is just a sum of $n$ independent normal variables, namely, the $n^{-1 / 2} y_t$ ‘s, all of which have mean 0 and variance $n^{-1} \sigma^2$. Consequently, $S_n$ itself is distributed as $N\left(0, \sigma^2\right)$ for all $n$. If the distribution is $N\left(0, \sigma^2\right)$ for all $n$ independent of $n$, then the limiting distribution is just the $N\left(0, \sigma^2\right)$ distribution as well. But if this is so for normal summands, we may conclude by our earlier argument that the limiting distribution of any sequence $S_n$ made up from independent mean-zero summands, all with variance $\sigma^2$, will be $N\left(0, \sigma^2\right)$

经济代写|计量经济学代写Econometrics代考|Asymptotic Normality and Central Limit Theorems

$$\sigma_t^2 \equiv \omega^2\left(1+\frac{1}{2}(t(\bmod 3))\right) .$$

$$\sigma_0^2 \equiv \lim n \rightarrow \infty\left(\frac{1}{n} \sum t=1^n \sigma_t^2\right)$$

$$n^{1 / 2}\left(m_1-\mu\right)=n^{-1 / 2} \sum_{t=1}^n\left(y_t-\mu\right),$$

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