统计代写|统计推断代写Statistical inference代考|Pruperties of the Sample Mean and Variance

Doug I. Jones

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统计代写|统计推断代写Statistical inference代考|Pruperties of the Sample Mean and Variance

We have already seen how to calculate the means and variances of $\bar{X}$ and $S^{\prime 2}$ in general. Now; under the additional assumption of normality, we can derive their full distributions, and more. The properties of $\bar{X}$ and $S^2$ are summarized in the following theorm.

Theorem 5.3.1 Let $X_1, \ldots, X_n$ be a rantom sample from a $\mathrm{n}\left(\mu, \sigma^2\right)$ distribution and let $\dot{X}=(1 / n) \sum_{i=1}^n X_i$ and $S^2=[1 /(n-1)] \sum_{i=1}^n\left(X_i-\bar{X}\right)^2$. Then
a. $\dot{X}$ and $S^2$ are independent random variables.
b. $\bar{X}$ hos $\mathrm{n}\left(\mu, \sigma^2 / n\right)$ distribution,
c. $(n-1) S^{12} / \sigma^2$ has a chi squared distribution with $n-1$ degrees of freedom.
Proof: First note that, from Section 3.5 on location-scale families, we can assume, without loss of grnerality, that $\mu=0$ and $\sigma=1$. (Also see the discussion preceding Theorem 5.2 .11 .) Furthermore, part (b) has already been established in Example 5.2.8, leaving us tu prove parts (a) and (c).

To prove part (a) we will apply Theorem 4.6 .12 , and show that $\bar{X}$ and $S^2$ are functions of independent random vectors. Note that we can write $S^2$ as a function of $n-1$ deviations as follows:
\begin{aligned} S^{\prime 2} & =\frac{1}{n-1} \sum_{i=1}^n\left(X_i-\dot{X}\right)^2 \ & =\frac{1}{n-1}\left(\left(X_1-\bar{X}\right)^2+\sum_{i=2}^n\left(X_i-\bar{X}\right)^2\right) \ & =\frac{1}{n-1}\left(\left[\sum_{i=2}^n\left(X_i-\bar{X}\right)\right]^2+\sum_{i=2}^n\left(X_i-\dot{X}\right)^2\right) \cdot\left(\sum_{i=1}^n\left(X_i-\bar{X}\right)=0\right) \end{aligned}

统计代写|统计推断代写Statistical inference代考|The Derived Distributions: Student’s $t$ and S’nedecor’s $F$

The distributions derived in Section 5.3 .1 are, in a sense, the first step in a statistical analysis that assumes normalitv. In particular, in most practical cases the variance, $\sigma^2$, is unknown. Thus, to get any idea of the variability of $\bar{X}$ (as an estimate of $\mu$ ), it is necessary to estimate this variance. This topic was first addressed by $\mathrm{W}$. S. Gossct (who published under the pscudonym of Student) in the early 1900s. The landmark work of Student re’sulted in Student’s $t$ distribution or, more simply, the $t$ distribution. If $X_1, \ldots, X_n$ are a random sample from a $\mathrm{n}\left(\mu, \sigma^2\right)$, we know that the quantity
$$\frac{\dot{X}-\mu}{\sigma / \sqrt{n}}$$
is distributed as a $\mathrm{n}(0,1)$ random variable. If we knew the value of $\sigma$ and we measured $\bar{X}$, then we could use ( $J 3.3)$ as a basis for inference about $\mu$, since $\mu$ would then be the only unknown quantity. Most of the time, however, $\sigma$ is unknown. St.udent did the obvious thing-he looked at the distribution of
$$\frac{X-\mu}{S / \sqrt{n}}$$
a quantity that could be us’d as a basis for inference about $\mu$ when $\sigma$ was unknown.

The distribution of (5.3.4) is easy to derive, provided that we first notice a few simplifying maneuvers. Multiply (5.3.4) by $\sigma / \sigma$ and rearrange slightly to obtain
$$\frac{\bar{X}-\mu}{S / \sqrt{n}}=\frac{(\bar{X}-\mu) /(\sigma / \sqrt{n})}{\sqrt{S^2 / \sigma^2}} .$$
The numerator of $(5.3 .5)$ is a $\mathrm{n}(0,1)$ random variable, and the denominator is $\sqrt{\chi_{n-1}^2 /(n-1)}$, independcnt of the numerator. Thus, the distribution of (5.3.4) can be found by solving the simplitied problem of finding the distribution of $U / \sqrt{V / p}$, where $U$ is $n(0,1), V$ is $\chi_p^2$, and $U$ and $V$ are independent. This gives us Student’s $t$ distribution.

统计推断代考

统计代写|统计推断代写Statistical inference代考|Pruperties of the Sample Mean and Variance

B. $\bar{X}$ hos $\mathrm{n}\left(\mu, \sigma^2 / n\right)$分布;
C. $(n-1) S^{12} / \sigma^2$有一个卡方分布，其自由度为$n-1$。

\begin{aligned} S^{\prime 2} & =\frac{1}{n-1} \sum_{i=1}^n\left(X_i-\dot{X}\right)^2 \ & =\frac{1}{n-1}\left(\left(X_1-\bar{X}\right)^2+\sum_{i=2}^n\left(X_i-\bar{X}\right)^2\right) \ & =\frac{1}{n-1}\left(\left[\sum_{i=2}^n\left(X_i-\bar{X}\right)\right]^2+\sum_{i=2}^n\left(X_i-\dot{X}\right)^2\right) \cdot\left(\sum_{i=1}^n\left(X_i-\bar{X}\right)=0\right) \end{aligned}

统计代写|统计推断代写Statistical inference代考|The Derived Distributions: Student’s $t$ and S’nedecor’s $F$

$$\frac{\dot{X}-\mu}{\sigma / \sqrt{n}}$$

$$\frac{X-\mu}{S / \sqrt{n}}$$

(5.3.4)的分布很容易推导，只要我们首先注意到一些简化的操作。(5.3.4)乘以$\sigma / \sigma$，稍微重新排列得到
$$\frac{\bar{X}-\mu}{S / \sqrt{n}}=\frac{(\bar{X}-\mu) /(\sigma / \sqrt{n})}{\sqrt{S^2 / \sigma^2}} .$$
$(5.3 .5)$的分子是一个$\mathrm{n}(0,1)$随机变量，分母是$\sqrt{\chi_{n-1}^2 /(n-1)}$，与分子无关。因此，可以通过求解求解$U / \sqrt{V / p}$分布的简化问题来找到(5.3.4)的分布，其中$U$是$n(0,1), V$是$\chi_p^2$, $U$和$V$是独立的。这给了我们Student的$t$分布。

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