20240507 updated3

5_mathematical_statistics_note/chapter/1_Probability_and_Distributions.tex

@@ -184,7 +184,7 @@ \subsection{The Moment Generating Function}
 
 \section{Homework}
 
-\begin{exercise}{}{}
+\begin{exercise}{1.9.7}{}
     Show that the moment generating function of the random variable $X$
     having the pdf $f(x)=\frac{1}{3}$, $-1<x<2$, zero elsewhere, is
     \begin{align*}

5_mathematical_statistics_note/chapter/2_Multivariate_Distributions.tex

@@ -78,6 +78,60 @@ \section{Distributions of Two Random Variables}
 \section{Transformations: Bivariate Random Variables}
 
 \section{Conditional Distributions and Expectations}
+We now introduce a parameter $\rho$ of the joint distribution of $(X,Y)$ 
+which quantifies the dependence between $X$ and $Y$ (so that $\rho=0$ when $X$ and $Y$ are independent).
+We assume the existence of all expectation under discussion.
+
+\begin{definition}{}{}
+    Let $(X,Y)$ have a joint distribution. Denote the means of $X$ and $Y$
+    respectively by $\mu_1$ and $\mu_2$ and their respective variances by $\sigma_1^2$ and $\sigma_2^2$.
+    The covariance of $(X,Y)$ is 
+    \begin{align*}
+        \text{cov}(X,Y) = E[(X-\mu_1)(Y-\mu_2)].
+    \end{align*}
+\end{definition}
+
+\begin{remark}
+    Since the expectation operator is linear, then
+    \begin{align*}
+        \text{cov}(X,Y) &= E[XY-\mu_2X-\mu_1Y+\mu_1\mu_2] = E[XY] - \mu_2E[X] - \mu_1 E[Y] +\mu_1\mu_2\\
+                        &= E[XY] - \mu_1\mu_2-\mu_1\mu_2+\mu_1\mu_2 = E[XY] - \mu_1\mu_2. 
+    \end{align*}
+\end{remark}
+
+\begin{definition}{}{}
+    If each of $\sigma_1$ and $\sigma_2$ is positive then the correlation coefficient between $X$ and $Y$ is 
+    \begin{align*}
+        \rho = \frac{E[(X-\mu_1)(Y-\mu_2)]}{\sigma_1\sigma_2} = \frac{\text{cov}(X,Y)}{\sigma_1\sigma_2}.
+    \end{align*}
+\end{definition}
+
+\begin{remark}
+    We can relate these parameters as
+    \begin{align*}
+        E[XY]  &=\mu_1\mu_2 +\text{cov}(X,Y)\\
+            &= \mu_1\mu_2 +\rho\sigma_1\sigma_2.
+    \end{align*}
+\end{remark}
+
+\begin{theorem}{}{}
+    For all jointly distributed random variables $(X,Y)$ whose correlation coefficient $\rho$ exists
+    (so that $\sigma_1>0$ and $\sigma_2>0$ by the definition of $\rho$), we have $-1\leqs \rho \leqs 1$.
+\end{theorem}
+
+\begin{theorem}{}{}
+    If $X$ and $Y$ are independent random variables then $\text{cov}(X,Y)=0$ and hence $\rho=0$.
+\end{theorem}
+
+\begin{theorem}{}{}
+    Suppose $(X,Y)$ have a joint distribution with the variances of $X$ and $Y$ finite and positive.
+    Denote the means and variances of $X$ and $Y$ by $\mu_1,\mu_2$ and $\sigma_1^2,\sigma_2^2$, respectively,
+    and let $\rho$ be the correlation coefficient between $X$ and $Y$. If $E[Y|X]$ is linear in $X$ then
+    \begin{align*}
+        E[Y|X] = \mu_2 + \rho \frac{\sigma_2}{\sigma_1}(X-\mu_1) \text{ and }\\
+        E[Var(Y|X)] = \sigma_2^2(1-\rho^2).
+    \end{align*}
+\end{theorem}
 
 
 
@@ -87,7 +141,7 @@ \section{The Correlation Coefficient}
 
 \section{Homework}
 
-\begin{exercise}{}{}
+\begin{exercise}{2.3.6}{}
     Let the joint pdf of $X$ and $Y$ be given by
     \begin{align*}
         f(x,y) = \left\{\begin{matrix}
@@ -138,7 +192,7 @@ \section{Homework}
 
 
 
-\begin{exercise}{}{}
+\begin{exercise}{2.6.9}{}
     Let $X_1,X_2,X_3$ be iid with common pdf $f(x)=\exp(-x)$, $0<x<\infty$,
     zero elsewhere. Evaluste:\\
     (a) $P(X_1<X_2|X_1<2X_2)$.\\

5_mathematical_statistics_note/chapter/3_Some_Special_Distributions.tex

@@ -0,0 +1,54 @@
+\chapter{Some Special Distributions}
+
+\section{The Binomial and Related Distributions}
+
+
+\section{The Poisson Distribution}
+
+\section{The $\Gamma$,$\chi^2$, and $\beta$ Distributions}
+
+\section{The Normal Distribution}
+
+\section{The Multivariate Normal Distribution}
+
+\section{$t-$ and $F-$Distributions}
+
+\section{Mixture Distributions}
+
+
+\section{Homework}
+
+\begin{exercise}{3.2.17}{}
+    Let $X_1$ and $X_2$ be two independent random variables.
+    Suppose that $X_1$ and $Y=X_1+X_2$ have Possion Distributions
+    with means $\mu_1$ and $\mu>\mu_1$, respectively.
+    Find the distribution of $X_2$.
+\end{exercise}
+
+
+\begin{exercise}{3.4.21}{}
+    Let $f(x)$ and $F(x)$ be the pdf and the cdf, respectively, of a distribution of 
+    the continuous type such that $f'(x)$ exists for all $x$. Let the mean of the truncated distribution
+    that has pdf $g(y)=f(y)/F(b)$, $-\infty<y<b$, zero elsewhere, be equal to $-f(b)/F(b)$ for all real $b$.
+    Prove that $f(x)$ is a pdf of a standard normal distribution.
+\end{exercise}
+
+
+\begin{exercise}{3.5.9}{}
+    Say the correlation coefficient between the heights of husbands and wives is 
+    $0.70$ and the mean male height is $5$ feet $10$ inches with standard deviation $2$ inches,
+    and the mean female height is $5$ feet $4$ inches with standard deviation $1\frac{1}{2}$ inches.
+    Assuming a bivariate normal distribution, what is the best guess of the height of 
+    a woman whose husband's height is $6$ feet? Find a $95\%$ prediction interval for her height.
+\end{exercise}
+
+\section{Reference}
+
+\begin{itemize}
+    \item \href{https://tomoki-okuno.com/files/math/Ch3_sol.pdf}{ex3.2.17}
+    \item \href{https://cs.du.edu/~paulhorn/361/assn11-solns.pdf}{ex3.4.21}
+    \item \href{https://faculty.etsu.edu/gardnerr/4047/Beamer-Hogg-McKean-Craig/Proofs-HMC-3-5.pdf}{ex3.5.9}
+\end{itemize}
+
+
+

5_mathematical_statistics_note/chapter/4_Some_Elementary_Statistical_Inferences.tex

@@ -1,4 +1,222 @@
+\chapter{Some Elementary
+Statistical Inferences}
+
+\section{Introduction}
+Statistics is a branch of Mathematics, that deals with the collection, analysis, interpretation, and the presentation of the numerical data. 
+In other words, it is defined as the collection of quantitative data. 
+The main purpose of Statistics is to make an accurate conclusion using a limited sample about a greater population.
+\par
+
+\subsection{Types of statistics}
+Statistics can be classified into two different categories. The two different types of Statistics are:
+\begin{itemize}
+    \item Descriptive Statistics
+    \item Inferential Statistics
+\end{itemize}
+
+In Statistics, descriptive statistics describe the data, 
+whereas inferential statistics help you make predictions from the data. 
+In inferential statistics, the data are taken from the sample and allows you to generalize the population. 
+In general, inference means “guess”, which means making inference about something. 
+So, statistical inference means, making inference about the population. 
+To take a conclusion about the population, it uses various statistical analysis techniques. 
+In this article, one of the types of statistics called inferential statistics is explained in detail. 
+Now, you are going to learn the proper definition of statistical inference, types, solutions, and examples.
+
+\subsection{Statistical inference definition}
+Statistical inference is the process of analysing the result and making conclusions from data subject to random variation. 
+It is also called inferential statistics. 
+Hypothesis testing and confidence intervals are the applications of the statistical inference. 
+Statistical inference is a method of making decisions about the parameters of a population, based on random sampling. 
+It helps to assess the relationship between the dependent and independent variables. 
+The purpose of statistical inference to estimate the uncertainty or sample to sample variation. 
+It allows us to provide a probable range of values for the true values of something in the population. 
+The components used for making statistical inference are:
+\begin{itemize}
+    \item Sample Size
+    \item Variability in the sample
+    \item Size of the observed differences
+\end{itemize}
+
+\subsection{Types of statistical inference}
+There are different types of statistical inferences that are extensively used for making conclusions. 
+They are:
+\begin{itemize}
+    \item One sample hypothesis testing
+    \item Confidence Interval
+    \item Pearson Correlation
+    \item Bi-variate regression
+    \item Multi-variate regression
+    \item Chi-square statistics and contingency table
+    \item ANOVA or T-test
+\end{itemize}
+
+\section{Estimate
+Distribution Parameters}
+
+Until now we have studied Probability, proceeding as follows: we assumed parameters of all distributions to be known and, based on this, computed probabilities
+of various outcomes (in a random experiment). In this chapter we make the essential transition to Statistics, which is concerned with the exact opposite: the
+random experiment is performed (usually many times) and the individual outcomes
+recorded; based on these, we want to estimate values of the distribution parameters
+(one or more). 
+
+\begin{definition}{}{}
+    If the sample $\vect{X}_1,\vect{X}_2,..., \vect{X}_n$ are iid, then they constitute
+a random independent sample (RIS) of size $n$ from the population $\bold{X}$.
+\end{definition}
+
+\begin{definition}{}{}
+    Let $T = T(\vect{X}_1,\vect{X}_2,..., \vect{X}_n)$ be a function of the sample 
+    $\vect{X}_1,\vect{X}_2,..., \vect{X}_n$. Then $T$ is called a statistic.
+\end{definition}
+\begin{remark}
+    Once the sample is drawn, then $t=T(\vect{x}_1,\vect{x}_2,..., \vect{x}_n)$ is called the
+    realization of $T$ , where $\vect{x}_1,\vect{x}_2,..., \vect{x}_n$ is the value of the sample.
+\end{remark}
+
+\begin{example}{}{}
+    How should we estimate the mean $\mu$ of a Normal distribution
+    $N (\mu, \sigma)$, based on a RIS of size $n$? We would probably take $\overline{X}$ 
+    (the sample mean) to be a 'reasonable' estimator of $\mu$ 
+    [note that this name applies to the random variable $\overline{X}$, 
+    with all its potential (would-be) values; as soon as
+    the experiment is completed and a particular value of $\overline{X}$ 
+    recorded, this value (i.e. a specific number) is called an estimate of $\mu$].
+\end{example}
+
+There is a few related issues we have to sort out:
+\begin{itemize}
+    \item How do we know that $\overline{X}$ is a 'good' estimator of $\mu$, i.e. 
+    is there some sensible set of criteria which would enable us to judge the quality of individual
+    estimators?
+    \item Using these criteria, can we then find the best estimator of a parameter, at
+    least in some restricted sense?
+    \item Would not it be better to use, instead of a single number [the so called
+    {\it point estimate}, which can never precisely agree with the exact value of
+    the unknown parameter, and is thus in this sense always wrong], an interval
+    of values which may have a good chance of containing the correct answer?
+\end{itemize}
+The rest of this section tackles the first two issues. We start with
+
+\section{Confidence intervals}
+The last section considered the issue of so called {\it point estimates} (good, better
+and best), but one can easily see that, even for the best of these, a statement which
+claims a parameter, say $\mu$, to be close to $8.3$, is not very informative, unless we
+can specify what 'close' means. This is the purpose of a confidence interval,
+which requires quoting the estimate together with specific limits, e.g. $8.3\pm 0.1$ (or
+$8.2 \leftrightarrow  8.4$, using an interval form). 
+\par
+The limits are established to meet a certain (usually 95\%) level of confidence 
+(not a probability, since the statement does not involve any randomness -- we are either 100\% right, or 100\% wrong!).
+The level of confidence ($1-\alpha$ in general) corresponds to the original, a-priori
+probability (i.e. before the sample is even taken) of the procedure to get it right
+(the probability is, as always, in the random sampling). To be able to calculate
+this probability exactly, we must know what distribution we are sampling from.
+So, until further notice, we will assume that the distribution is Normal.
+
+\subsection{Confidence interval for mean $\mu$}
+
+\begin{theorem}{Sums of Independent Normal Random Variables}{}
+    If $\vect{X}_1,\vect{X}_2,...,\vect{X}_n$ are mutually independent normal random variables with means 
+    $\mu_1,\mu_2,...,\mu_n$ and variances $\sigma_1^2,\sigma_2^2,...,\sigma_n^2$,
+    then the linear combination:
+    \begin{align*}
+        \vect{Y} = \sum\limits_{i=1}^{n} c_i\vect{X}_i
+    \end{align*}
+    follows the normal distribution:
+    \begin{align*}
+        N(\sum\limits_{i=1}^{n} c_i\mu_i,\sum\limits_{i=1}^{n} c_i^2\sigma_i^2)
+    \end{align*}
+\end{theorem}
+
+\begin{corollary}{}{}
+    If $\vect{X}_1,\vect{X}_2,...,\vect{X}_n$ are observations of a random sample of size $n$ from a 
+    $N(\mu,\sigma^2)$ population.
+    \begin{itemize}
+        \item $\overline{X}=\frac{1}{n} \sum\limits_{i=1}^{n} \vect{X}_i$ is the sample mean of the $n$ observations, and
+        \item $S^2=\frac{1}{n-1}\sum\limits_{i=1}^{n}(\vect{X}_i-\overline{\vect{X}})^2$ is the sample variance of the $n$ observations.
+    \end{itemize}
+    Then: \\
+    (1) $\overline{X}\sim N(\mu,\frac{\sigma^2}{n})$;\\
+    (2) $\frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)$\\
+    (3) $\overline{X}$ and $S^2$ are independent
+\end{corollary}
+
+We first assume that, even though $\mu$ is to be estimated (being unknown), we still
+know the exact (population) value of $\sigma$ (based on past experience).
+We know that
+
+\section{Testing hypotheses}
+Suppose now that, instead of trying to estimate
+
+
+\section{Homework}
+\begin{exercise}{4.5.8}{}
+    Let us say the life of a tire in miles, say $X$, 
+    is normally distributed with mean
+    $\theta$ and standard deviation $5000$.
+    Past experience indicates that $\theta = 30,000$.
+    The manufacturer claims that the tires made by a new process have mean $\theta>30,000$.
+    It is possible that $\theta=35,000$.
+    Check his claim by testing $H_0:\theta=30,000$ against
+    $H_1:\theta>30,000$. We observe $n$ independent values of $X$, 
+    say $x_1,...,x_n$, and we reject $H_0$ (thus accept $H_1$)
+    if and only if $\overline{x}\geqs c$.
+    Determine $n$ and $c$ so that the power function $\gamma(\theta)$ of the test
+    has the values $\gamma(30,000)=0.01$ and $\gamma(35,000)=0.98$.
+\end{exercise}
+
+\begin{exercise}{4.5.11}{}
+    Let $Y_1<Y_2<Y_3<Y_4$ be the order statistics of a random sample of size 
+    $n=4$ from a distribution with pdf $f(x;\theta)=1/\theta,0<x<\theta$, zero elsewhere,
+    where $0<\theta$. The hypothesis $H_0:\theta=1$ is rejected and $H_1:\theta>1$ is accepted if 
+    the observed $Y_4\geqs c$.\\
+    (a) Find the constant $c$ so that the significance level is $\alpha=0.05$.\\
+    (b) Determine the power function of the test.
+\end{exercise}
+
+\begin{exercise}{4.6.5}{}
+    On page 373 Rasmussen (1992) discussed a paired design. A baseball coach 
+    paired $20$ members of his team by their speed; i.e., each member of the pair has 
+    about the same speed. Then for each, he randomly chose one member of the
+    pair and told him that if could beat his best time in circling the bases he would
+    give him an award (call this response the time of the "self" member). For the other
+    member of the pair the coach's instruction was an award if he could beat the time 
+    of the other member of the pair (call this response the time of the "rival" member). 
+    Each member of the pair knew who his rival was. The data are given below, but are
+    also in the file {\tt selfrival.rda}. Let $\mu_d$ be the true difference in times (rival minus 
+    self) for a pair. The hypotheses of interest are $H_0:\mu_d=0$ versus $H_1:\mu_d<0$. The
+    data are in order by pairs, so do not mix the order.\\
+    {\tt 
+    \quad self: 16.20 16.78 17.38 17.59 17.37 17.49 18.18 18.16 18.36 18.53 
+    15.92 16.58 17.57 16.75 17.28 17.32 17.51 17.58 18.26 17.87\\
+    \quad rival: 15.95 16.15 17.05 16.99 17.34 17.53 17.34 17.51 18.10 18.19
+    16.04 16.80 17.24 16.81 17.11 17.22 17.33 17.82 18.19 17.88} \\
+    (a) Obtain comparison boxplots of the data. Comment on the comparison plots.
+    Are there any outliers?\\
+    (b) Compute the paired $t$-test and obtain the $p$-value. Are the data significant at the 
+    $5\%$ level of significance?\\
+    (c) Obtain a point estimate of $\mu_d$ and a $95\%$ confidence interval for it.\\
+    (d) Conclude in terms of the problem. 
+
+\end{exercise}
+
+
+
+
 \section{Reference}
 \begin{itemize}
+    \item \href{https://byjus.com/maths/statistical-inference/}{Statistical Inference}
     \item \href{https://faculty.etsu.edu/gardnerr/4047/notes-Hogg-McKean-Craig/Hogg-McKean-Craig-4-1.pdf}{Sampling and Statistics}
+    \item \href{https://spartan.ac.brocku.ca/~jvrbik/MATH2P82/Statistics.PDF}{Chapter 3 RANDOM SAMPLING}
+    \item \href{https://spartan.ac.brocku.ca/~jvrbik/MATH2P82/Statistics.PDF}{Chapter 5 ESTIMATING DISTRIBUTION PARAMETERS}
+    \item \href{https://spartan.ac.brocku.ca/~jvrbik/MATH2P82/Statistics.PDF}{Chapter 6 CONFIDENCE INTERVALS}
+    \item \href{https://spartan.ac.brocku.ca/~jvrbik/MATH2P82/Statistics.PDF}{Chapter 7 TESTING HYPOTHESES}
+    \item \href{https://cloud.moezx.cc/Document/mooc/%E6%B5%99%E5%A4%A7%E6%A6%82%E7%8E%87%E8%AE%BA/%E7%AC%AC42%E8%AE%B2%20.pdf}{Sampling distribution of a single normal population}
+    \item \href{https://cloud.moezx.cc/Document/mooc/%E6%B5%99%E5%A4%A7%E6%A6%82%E7%8E%87%E8%AE%BA/%E7%AC%AC43%E8%AE%B2%20.pdf}{Sampling distribution of two normal populations}
+    \item \href{https://online.stat.psu.edu/stat414/lesson/26/26.2}{Sampling Distribution of Sample Mean}
+    \item \href{https://online.stat.psu.edu/stat415/lesson/25}{Power of a Statistical Test}
+    \item \href{https://tomoki-okuno.com/files/math/Ch4_sol.pdf}{ex4.5.8}
+    \item \href{https://tomoki-okuno.com/files/math/Ch4_sol.pdf}{ex4.5.11}
+    \item \href{https://zhuanlan.zhihu.com/p/570096188}{ex4.6.5}
 \end{itemize}

5_mathematical_statistics_note/chapter/preface.tex

@@ -5,5 +5,6 @@ \chapter*{Preface}
     \item \href{}{Introduction to Mathematical Statistics 8th Edition}
     \item \href{https://faculty.etsu.edu/gardnerr/4047/notes-Hogg-McKean-Craig.htm}{lecture note}
     \item \href{https://faculty.etsu.edu/gardnerr/4047/notes-Hogg-McKean-Craig/StudyGuide1.pdf}{Study Guide}
+    \item \href{https://zhuanlan.zhihu.com/p/570096188}{Introduction to mathematical statistics exercise solution}
 \end{itemize}
 

5_mathematical_statistics_note/note.ist

@@ -1,5 +1,5 @@
 % makeindex style file created by the glossaries package
-% for document 'note' on 2024-3-24
+% for document 'note' on 2024-4-28
 actual '?'
 encap '|'
 level '!'

5_mathematical_statistics_note/note.tex

@@ -54,6 +54,8 @@
 
 \input{chapter/1_Probability_and_Distributions.tex}
 \input{chapter/2_Multivariate_Distributions.tex}
+\input{chapter/3_Some_Special_Distributions.tex}
+\input{chapter/4_Some_Elementary_Statistical_Inferences.tex}
 
 
 

5_mathematical_statistics_note/notes_template.tex

@@ -360,4 +360,5 @@
 \newcommand{\ci}{\mathrm{i}}
 \newcommand{\hH}{\mathscr{H}}
 \newcommand{\hK}{\mathscr{K}}
-\newcommand{\inner}[2]{\langle#1,#2\rangle}
+\newcommand{\inner}[2]{\langle#1,#2\rangle}
+\newcommand{\vect}[1]{\boldsymbol{#1}}