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这是悉尼大学统计推断课程的代写成功案例。
STAT3023 课程简介
In today’s data-rich world more and more people from diverse fields are needing to perform statistical analyses and indeed more and more tools for doing so are becoming available; it is relatively easy to point and click and obtain some statistical analysis of your data. But how do you know if any particular analysis is indeed appropriate? Is there another procedure or workflow which would be more suitable? Is there such a thing as the best possible approach in a given situation? All of these questions (and more) are addressed in this unit. You will study the foundational core of modern statistical inference, including classical and cutting-edge theory and methods of mathematical statistics with a particular focus on various notions of optimality. The first part of the unit covers various aspects of distribution theory which are necessary for the second part which deals with optimal procedures in estimation and testing. The framework of statistical decision theory is used to unify many of the concepts. You will apply the methods learnt to real-world problems in laboratory sessions. By completing this unit you will develop the necessary skills to confidently choose the best statistical analysis to use in many situations.
Prerequisites
At the completion of this unit, you should be able to:
- LO1. deduce the (limiting) distribution of sums of random variables using moment-generating functions
- LO2. derive the distribution of a transformation of two (or more) continuous random variables
- LO3. derive marginal and conditional distributions associated with certain multivariate distributions
- LO4. classify many common distributions as belonging to an exponential family
- LO5. derive and implement maximum likelihood methods in various estimation and testing problems
- LO6. formulate and solve various inferential problems in a decision theory framework
- LO7. derive and apply optimal procedures in various problems, including Bayes rules, minimax rules, minimum variance unbiased estimators and most powerful tests.
STAT3023 Statistical Inference HELP（EXAM HELP， ONLINE TUTOR）
Let $X$ be a non-negative random variable with CDF $F$. Show that
$$
E[X]=\int_0^{\infty}(1-F(t)) d t .
$$
Hint: Argue that you can write $X=\int_0^{\infty} 1_{{t<X}} d t$, and then assume that you can switch the order of expectation and integral.
For a non-negative random variable $X$ we can write
$$
X=\int_0^X 1 d t=\int_0^{\infty} 1_{{t<X}} d t
$$
and therefore
$$
E[X]=E\left[\int_0^{\infty} 1_{{t<X}} d t\right]
$$
Reversing the order of integration yields
$$
E[X]=\int_0^{\infty} E\left[1_{{t<X}}\right] d t=\int_0^{\infty} P(t<X) d t=\int_0^{\infty}(1-F(t)) d t
$$
The interchange of expectation and integration for integrating a non-negative function is justified by a result known as Tonelli’s theorem.
For a positive value of $\alpha$ let $X$ be a continuous random variable with density
$$
f(x)= \begin{cases}\frac{\alpha}{x^{\alpha+1}} & \text { for } x \geq 1 \ 0 & \text { otherwise }\end{cases}
$$
Find $E[X]$ and $\operatorname{Var}(X)$.
The $n$-th moment of this density is
$$
\begin{aligned}
E\left[X^n\right] & =\int_1^{\infty} x^n \frac{\alpha}{x^{\alpha+1}} d x=\int_1^{\infty} \alpha x^{n-\alpha-1} d x \
& =\left{\begin{array}{ll}
{\left[\frac{\alpha}{n-\alpha} x^{n-\alpha}\right]_1^{\infty}} & \text { for } \alpha \neq n \
{[\alpha \log (x)]_1^{\infty}} & \text { for } \alpha=n
\end{array}= \begin{cases}\infty & \text { for } \alpha \leq n \
\frac{\alpha}{\alpha-n} & \text { for } \alpha>n\end{cases} \right.
\end{aligned}
$$
So
$$
E[X]= \begin{cases}\infty & \text { for } \alpha \leq 1 \ \frac{\alpha}{\alpha-1} & \text { for } \alpha>1\end{cases}
$$
and
$$
E\left[X^2\right]= \begin{cases}\infty & \text { for } \alpha \leq 2 \ \frac{\alpha}{\alpha-2} & \text { for } \alpha>2 .\end{cases}
$$
The variance of $X$ is therefore infinite if $\alpha \leq 2$, and is
$$
\operatorname{Var}(X)=E\left[X^2\right]-E[X]^2=\frac{\alpha}{\alpha-2}-\left(\frac{\alpha}{\alpha-1}\right)^2=\frac{\alpha}{(\alpha-2)(\alpha-1)^2}
$$
for $\alpha>2$.
Let $X_1, X_2$, and $V$ be independent random variables with
$$
\begin{aligned}
& E\left[X_1\right]=\mu \quad E\left[X_2\right]=\mu \quad E[V]=0 \
& \operatorname{Var}\left(X_1\right)=\sigma^2 \quad \operatorname{Var}\left(X_2\right)=\sigma^2 \quad \operatorname{Var}(V)=\tau^2 . \
&
\end{aligned}
$$
Let $Y_1=X_1+V$ and $Y_2=X_2+V$
(a) Find the means and variances of $Y_1$ and $Y_1$.
(b) Find $\operatorname{Cov}\left(Y_1, Y_2\right)$ and $\operatorname{Cov}\left(Y_1, V\right)$.
(a) The means are
$$
E\left[X_i\right]=E\left[Y_i+V\right]=E\left[Y_i\right]+E[V]=\mu
$$
Since the $X_i$ are independent of $V$ the variances are
$$
\operatorname{Var}\left(Y_i\right)=\operatorname{Var}\left(X_i+V\right)=\operatorname{Var}\left(X_i\right)+\operatorname{Var}(V)=\sigma^2+\tau^2 .
$$
(b) The covariance of $Y_1$ and $Y_2$ is
$$
\begin{aligned}
\operatorname{Cov}\left(Y_1, Y_2\right) & =\operatorname{Cov}\left(X_1+V, X_2+V\right) \
& =\operatorname{Cov}\left(X_1, X_2+V\right)+\operatorname{Cov}\left(V, X_2+V\right) \
& =\operatorname{Cov}\left(X_1, X_2\right)+\operatorname{Cov}\left(X_1, V\right)+\operatorname{Cov}\left(V, X_2\right)+\operatorname{Cov}(V, V) \
& =0+0+0+\operatorname{Var}(V)=\tau^2 .
\end{aligned}
$$
The covariance of $Y_i$ and $V$ is
$$
\begin{aligned}
\operatorname{Cov}\left(Y_i, V\right) & =\operatorname{Cov}\left(X_i+V, V\right) \
& =\operatorname{Cov}\left(X_i, V\right)+\operatorname{Cov}(V, V) \
& =0+\operatorname{Var}(V)=\tau^2 .
\end{aligned}
$$
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