# 统计代写|统计推断代写Statistical inference代考|Mean and variance of the sample mean

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## 统计代写|统计推断代写Statistical inference代考|Mean and variance of the sample mean

The following proposition establishes the mean and variance of the sample mean. Note that neither of these properties requires us to make any assumption about the parametric form of the population distribution.
Proposition 7.1.2 (Properties of the sample mean)
Suppose that $Y_1, \ldots, Y_n$ is a random sample from a population distribution with mean $\mu$, then
i. the mean of the sample mean is the population mean, $\mathbb{E}(\bar{Y})=\mu$.
If, in addition, the population distribution has finite variance, $\sigma^2<\infty$, then
ii. the variance of the sample mean is given by $\operatorname{Var}(\bar{Y})=\frac{\sigma^2}{n}$.
Proof.
i. The proof comes directly from the definition of the sample mean.
$$\begin{array}{rlr} \mathbb{E}(\bar{Y}) & =\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^n Y_i\right)=\frac{1}{n} \sum_{i=1}^n \mathbb{E}\left(Y_i\right) & \text { properties of expectation } \ & =\frac{1}{n} \sum_{i=1}^n \mu & \text { identically distributed } \ & =\mu . & \end{array}$$
Note that we are only using the fact that the elements of a random sample are identically distributed with the population distribution. There is no need to assume independence to get this result.
ii. The result for the variance of the sample mean can also be proved directly.
$$\begin{array}{rlr} \operatorname{Var}(\bar{Y}) & =\operatorname{Var}\left(\frac{1}{n} \sum_{i=1}^n Y_i\right)=\frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}\left(Y_i\right) & \text { by independence } \ & =\frac{1}{n^2} \sum_{i=1}^n \sigma^2 & \text { identically distributed } \ & =\frac{1}{n} \sigma^2 . & \end{array}$$

## 统计代写|统计推断代写Statistical inference代考|Central limit theorem

The central limit theorem is a remarkable result. One of its consequences is that the sample mean of a large random sample is approximately normally distributed, (almost) regardless of the distribution that we sample from. The mysterious appearance of the normal distribution in this context is widely used in inference. The proof we give relies on the existence of the moment-generating function of the population distribution. This is not a necessary condition for the central limit theorem to hold, but the proof is much more straightforward in this case.
Theorem 7.1.3 (Central limit theorem)
Suppose that $\boldsymbol{Y}=\left(Y_1, \ldots, Y_n\right)^T$ is a random sample with $\mathbb{E}(Y)=\mu$ and $\operatorname{Var}(Y)=\sigma^2$.
Provided $0<\sigma^2<\infty$, then
$$\frac{\bar{Y}n-\mu}{\sqrt{\sigma^2 / n}} \stackrel{d}{\longrightarrow} \mathrm{N}(0,1) \quad \text { as } \quad n \rightarrow \infty .$$ where $\bar{Y}_n=\frac{1}{n} \sum{j=1}^n Y_j$ is the sample mean.
Proof of special case.
As mentioned above, we only prove the case in which the moment-generating function of the $Y_j$ exists. If we define
$$Z_n=\frac{\bar{Y}n-\mu}{\sqrt{\sigma^2 / n}}$$ then we can readily show that $$Z_n=\frac{S_n-n \mu}{\sqrt{n \sigma^2}}$$ where $S_n=\sum{j=1}^n Y_j$ is the partial sum. We can get an expression for the momentgenerating function of $Z$ in terms of the moment-generating function of $Y$ by using the fact that $m_{S_n}(t)=\left[m_Y(t)\right]^n$, so
\begin{aligned} m_{Z_n}(t) & =\mathbb{E}\left(e^{t Z_n}\right)=\mathbb{E}\left{\exp \left[t\left(\frac{S_n-n \mu}{\sqrt{n \sigma^2}}\right)\right]\right} \ & =\exp \left(-\frac{\sqrt{n} \mu t}{\sigma}\right) \mathbb{E}\left[\exp \left(\frac{t}{\sqrt{n \sigma^2}} S_n\right)\right] \ & =\exp \left(-\frac{\sqrt{n} \mu t}{\sigma}\right) m_{S_n}\left(\frac{t}{\sqrt{n \sigma^2}}\right) \ & =\exp \left(-\frac{\sqrt{n} \mu t}{\sigma}\right)\left[m_Y\left(\frac{t}{\sqrt{n \sigma^2}}\right)\right]^n . \end{aligned}
Taking logs yields the cumulant-generating function,
$$K_{Z_n}(t)=-\frac{\sqrt{n} \mu t}{\sigma}+n K_Y\left(\frac{t}{\sqrt{n \sigma^2}}\right)$$

# 统计推断代考

## 统计代写|统计推断代写Statistical inference代考|Mean and variance of the sample mean

$$\mathbb{E}(\bar{Y})=\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^n Y_i\right)=\frac{1}{n} \sum_{i=1}^n \mathbb{E}\left(Y_i\right) \quad \text { prope }$$

$\operatorname{Var}(\bar{Y})=\operatorname{Var}\left(\frac{1}{n} \sum_{i=1}^n Y_i\right)=\frac{1}{n^2} \sum_{i=1}^n \operatorname{Var}\left(Y_i\right)$

## 统计代写|统计推断代写Statistical inference代考|Central limit theorem

$\frac{\bar{Y} n-\mu}{\sqrt{\sigma^2 / n}} \stackrel{{ }^d}{\longrightarrow} \mathrm{N}(0,1) \quad$ as $\quad n \rightarrow \infty$.

\begin } { \text { aligned } } \text { m_{Z_n } } ( t ) \& = \backslash m a t h b b { E } \backslash l e f t ( e ^ { \wedge } { t \quad Z _ { – } n } \backslash

$$K_{Z_n}(t)=-\frac{\sqrt{n} \mu t}{\sigma}+n K_Y\left(\frac{t}{\sqrt{n \sigma^2}}\right)$$

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