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这是南佛罗里达州大学统计推断课程的代写成功案例。
EEL6029 课程简介
Credit Hours: 3
This first-tier graduate course aims at establishing the background in mathematical statistics, and abstract modeling of complex cyber systems, data analytics, and Bayesian intelligence for graduate students in electrical and systems engineering.
Statistical inference is the process of using data analysis to infer properties of an underlying distribution of probability.Inferential statistical analysis infers properties of a population, for example by testing hypotheses and deriving estimates. It is assumed that the observed data set is sampled from a larger population.
Prerequisites
Statistical inference makes propositions about a population, using data drawn from the population with some form of sampling. Given a hypothesis about a population, for which we wish to draw inferences, statistical inference consists of (first) selecting a statistical model of the process that generates the data and (second) deducing propositions from the model.
Konishi & Kitagawa state, “The majority of the problems in statistical inference can be considered to be problems related to statistical modeling”. Relatedly, Sir David Cox has said, “How the translation from subject-matter problem to statistical model is done is often the most critical part of an analysis”
EEL6029 Statistical Inference HELP（EXAM HELP， ONLINE TUTOR）
(15 points) Let $\left(X_1, \cdots, X_n\right)$ be a sample from $N\left(\theta, \sigma^2\right)$, where $\sigma^2$ is known. For each of the following hypotheses, write out the acceptance region of a level $\alpha$ test and the $1-\alpha$ confidence interval that results from inverting the test.
(a). (5 points) $H_0: \theta=\theta_0$ versus $H_1: \theta \neq \theta_0$.
(b). (5 points) $H_0: \theta \geq \theta_0$ versus $H_1: \theta<\theta_0$. (a). (5 points) $H_0: \theta \leq \theta_0$ versus $H_1: \theta>\theta_0$.
(a) For the hypothesis $\$ H_{-} 0$ : $\backslash$ theta= $\backslash$ theta_ $0 \$$ versus $\$ H_{-} 1$ : $\backslash$ theta $\backslash$ neq $\backslash$ theta_0 $\$$, we can use a two-sided test. The acceptance region of a level \$lalpha\$ test is given by:
$$
\left{x \in \mathbb{R}^n: \frac{\sqrt{n}\left(\bar{x}-\theta_0\right)}{\sigma} \in\left(-z_{1-\alpha / 2}, z_{1-\alpha / 2}\right)\right}
$$
where $\$ \backslash \operatorname{bar}{x} \$$ is the sample mean and $\$ z_{-}{1-$-lalpha/2} $\$$ is the $\$(1-$ lalpha/2)\$-th quantile of the standard normal distribution.
To obtain the \$1-\alpha\$ confidence interval, we can use the following formula:
$$
\left(\bar{x}-z_{1-\alpha / 2} \frac{\sigma}{\sqrt{n}}, \bar{x}+z_{1-\alpha / 2} \frac{\sigma}{\sqrt{n}}\right)
$$
(b) For the hypothesis $\$ H_{-} 0$ : \theta $\backslash$ geq $\backslash$ theta_0 $\$$ versus $\$ H_{-} 1$ : \theta<\theta_0 , we can use a one-sided test. The acceptance region of a level $\$ \backslash$ alpha\$ test is given by:
$$
\left{x \in \mathbb{R}^n: \frac{\sqrt{n}\left(\bar{x}-\theta_0\right)}{\sigma} \leq-z_\alpha\right}
$$
where $\$ z _{a l p h a} \$$ is the $\$ \backslash a l p h a \$$-th quantile of the standard normal distribution.
To obtain the $\$ 1$-\alpha\$ confidence interval, we can use the following formula:
$$
\left(-\infty, \bar{x}+z_\alpha \frac{\sigma}{\sqrt{n}}\right)
$$
(c) For the hypothesis $\$ H_{-} 0$ : $\backslash$ theta $\backslash$ leq $\backslash$ theta_ $0 \$$ versus $\$ H_{-} 1$ : $\backslash$ theta $>$ theta_0 $\$$, we can again use a one-sided test. The acceptance region of a level $\$ \backslash a l p h a \$$ test is given by:
$$
\left{x \in \mathbb{R}^n: \frac{\sqrt{n}\left(\bar{x}-\theta_0\right)}{\sigma} \geq z_\alpha\right}
$$
where $\$ z$ z_alpha} $\$$ is the $\$ \backslash a l p h a \$$-th quantile of the standard normal distribution.
To obtain the $\$ 1$-\alpha\$ confidence interval, we can use the following formula:
$$
\left(\bar{x}-z_\alpha \frac{\sigma}{\sqrt{n}}, \infty\right)
$$
(15 points) Find a $1-\alpha$ confidence interval of minimum length for $\theta$, given $X_1, \cdots, X_n$ i.i.d. with probability density function (pdf)
(a). (5 points) $f(x \mid \theta)=1, \theta-1 / 2<x<\theta+1 / 2$.
(b). (10 points) $f(x \mid \theta)=2 x / \theta^2, 0<x<\theta, \theta>0$.
(a) Since $\$ X _1, \backslash$ dots, $X _n \$$ are i.i.d. with pdf $\$ f(x \backslash m i d \backslash$ theta) $=1 \$$ for $\$ \backslash$ theta$1 / 2<x<\mid$ theta $+1 / 2 \$$, the likelihood function is given by:
$$
L\left(\theta \mid x_1, \ldots, x_n\right)=\prod_{i=1}^n f\left(x_i \mid \theta\right)=\prod_{i=1}^n 1=1
$$
for $\$ \mid$ theta-|frac ${1}{2}<x_{-}{(1)}<\backslash$ cdots $<x_{-}{(n)}<\backslash$ theta $+\backslash$ frac ${1}{2} \$$, where $\$ x _{(1)}, \backslash$ dots,$x_{-}{(n)} \$$ denote the order statistics.
The maximum likelihood estimate of $\$$ theta $\$$ is the midpoint of the sample range, i.e., $\$ \backslash$ hat ${$ theta $}=\backslash f r a c\left{x\left{((1)}+x _{(n)}\right}{2} \$\right.$.
To construct a $\$ 1$-lalpha $\$$ confidence interval of minimum length for $\$$ theta\$, we need to find the endpoints $\$ \backslash$ ell $\$$ and $\$$ u that satisfy:
$$
P\left(\ell<\theta<u \mid X_1=x_1, \ldots, X_n=x_n\right) \geq 1-\alpha
$$
and minimize the length of the interval \$u-lell\$. Since the likelihood function is constant, any interval with endpoints between $\$ x_{-}{(1)}-\mid \operatorname{frac}{1}{2} \$$ and $\$ x{(n)}+\mid f r a c{1}{2} \$$ will have the same probability.
Therefore, we can choose the endpoints to be:
$$
\ell=x_{(1)}-\frac{1}{2} \quad \text { and } \quad u=x_{(n)}+\frac{1}{2}
$$
so that the interval has minimum length $\$ u-\backslash e l l=x _{(n)}-x _{(1)}+1 \$$.
The $\$ 1$-\alpha $\$$ confidence interval of minimum length for $\$ \backslash$ theta $\$$ is then given by:
$$
\left(x_{(1)}-\frac{1}{2}, x_{(n)}+\frac{1}{2}\right)
$$
(b) Since $\$ X_{-} 1, \backslash$ dots, $X _n \$$ are i.i.d. with pdf $\$ f(x \backslash m i d \backslash$ theta $)=2 x \wedge$ theta^ $2 \$$ for $\$ 0<x<\backslash$ theta $\$$, the likelihood function is given by:
$$
L\left(\theta \mid x_1, \ldots, x_n\right)=\prod_{i=1}^n f\left(x_i \mid \theta\right)=\frac{2^n}{\theta^{2 n}} \prod_{i=1}^n x_i
$$
for $\$ 0<x _{(1)}, \backslash$ Idots, $x _{(\mathrm{n})}<\backslash$ theta $\$$.
The log-likelihood function is:
$$
\ell\left(\theta \mid x_1, \ldots, x_n\right)=-n \ln (\theta)-2 \sum_{i=1}^n \ln x_i
$$
Differentiating with respect to $\$ \backslash$ theta $\$$ and setting the derivative to zero, we obtain:
$$
\hat{\theta}=\frac{2}{n} \sum_{i=1}^n x_i
$$
To check that this is a maximum, we need to check that the second derivative is negative:
$$
\frac{\partial^2 \ell\left(\theta \mid x_1, \ldots, x_n\right)}{\partial \theta^2}=\frac{2 n}{\theta^2}<0
$$
Thus, the maximum likelihood estimate of $\$ \backslash$ theta $\$$ is $\$ \backslash$ hat $\bigcap$
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