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{\Large \textbf{Sample Questions: Maximum Likelihood Part 1}}%\footnote{}
\vspace{1 mm}

STA312 Fall 2023. Copyright information is at the end of the last page.
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\vspace{5mm}

\noindent
Let $X_1, \ldots, X_n$ be independent Pareto random variables with density 
{ \large
$f(x|\theta) = \left\{ \begin{array}{ll} % ll means left left
   \frac{\theta}{x^{\theta+1}} & \mbox{for $ x \geq 1$}  \\
   0     & \mbox{for } x<1 
   \end{array}  \right. $ % $
} % End size
where $\theta>0$. The Pareto has a decreasing density with a heavy right tail, sometimes used as a model for the unequal distribution of wealth. 
\begin{enumerate} 

% \item What is the cumulative distribution function 

\item Derive a formula for the maximum likelihood estimate of $\theta$. Include the second derivative test. Show your work and \textbf{circle your final answer}. 

\pagebreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item Give a formula for $\widehat{v}_n$, the estimated asymptotic variance of $\widehat{\theta}_n$. Show a little work. \vspace{40mm}

\item The file \href{http://www.utstat.toronto.edu/~brunner/data/legal/pareto.data.txt}
                    {\texttt{http://www.utstat.toronto.edu/$\sim$brunner/data/legal/pareto.data.txt}}
has a set of raw data. Calculate
    \begin{enumerate}
        \item The maximum likelihood estimate $\widehat{\theta}_n$.
        \item A 95\% confidence interval for $\theta$.
    \end{enumerate}
Your answers are numbers. Circle and label them. Bring your printout to the quiz. 
{ \small
\begin{verbatim}
> rm(list=ls())
> x = scan("http://www.utstat.toronto.edu/~brunner/data/legal/pareto.data.txt")
Read 150 items
> x
  [1]    5.47    2.54    4.01    1.22    2.74    4.99    1.24    4.35  227.65
 [10]    3.20    4.35    1.02    1.17    3.49    1.61   10.43    9.04    1.07
 [19]    4.80    1.14    1.41   36.62    5.38    1.98    1.43    1.54    1.42
 [28]    1.06    1.68    1.44    1.52    2.25    1.62    1.01   53.79    1.11
 [37]    1.52   28.39   15.55    3.96    2.73    6.43    4.35    1.29    2.04
 [46]    1.04    1.68    1.89    1.78    2.57    1.39    5.49    1.07    1.74
 [55]    5.68    1.43    1.58   42.42    2.11    1.07    1.27    1.03    1.02
 [64]   10.92    1.43    2.18    6.28    1.81    4.42    1.93    2.39    3.75
 [73]    1.65    1.01    1.31    2.66    1.08    1.36    1.22    2.20    2.79
 [82]    1.11    2.01    3.11    2.02    2.21    1.05    5.69    1.16    5.47
 [91]    2.19    1.22    1.37    1.37    1.63    3.55    1.13    1.26    1.21
[100]    1.35    4.36    4.59    1.47    2.22    4.34    2.19    1.80    1.68
[109]   31.22    3.63    1.01    1.60    2.39    2.21    1.22    1.54    2.39
[118] 1008.06    3.37    1.30    4.46    1.01    7.51    1.14    1.02    1.12
[127]    2.83    1.79    1.67    1.45    1.74    1.66    1.01    1.34    1.12
[136]    1.30    1.18   10.96    2.28    2.33    1.51    1.41    1.19    1.40
[145]    1.83    1.33   31.98    5.86    3.28    1.53
> thetahat = 1/mean(log(x)); thetahat
[1] 1.100816
> n = length(x); vhat = thetahat^2/n; se = sqrt(vhat); se
[1] 0.08988125
> low95 = thetahat - 1.96*se; up95 = thetahat + 1.96*se
> c(low95,up95) # 95% CI
[1] 0.9246488 1.2769833
\end{verbatim}
} % End size
\pagebreak %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item For the Pareto distribution, the well-known 80-20 rule (80\% of the wealth is held by 20\% of the population) corresponds to a value of $\theta = 1.16$. Using a two-sided large-sample $Z$-test and the usual $\alpha=0.05$ significance level, test whether these data are compatible with $\theta = 1.16$.

    \begin{enumerate}
        \item There are two critical values, one for the lower tail and one for the upper tail. What are they? The answers are numbers.
        \item What is the value of the test statistic? The answer is a number. Circle it.
        \item Use R to calculate the 2-sided $p$-value. The answer is a number. %\vspace{120mm}

\begin{verbatim}



> # Critical value(s). Just say plus and minus 1.96, or ...
> c(qnorm(0.025),qnorm(0.975))
[1] -1.959964  1.959964
> 
> Z = (thetahat-1.16)/se; Z
[1] -0.6584683
> 
> pvalue = 2 * (1-pnorm(abs(Z))); pvalue
[1] 0.5102373



\end{verbatim}

        \item Do you reject the null hypothesis? Answer Yes or No.
        \item Are the results statistically significant? Answer Yes or No.
        \item Do these data contradict  claim that $\theta = 1.16$? Answer Yes or No.
    \end{enumerate}

\end{enumerate} % End of all the questions

\vspace{30mm}

\noindent
\begin{center}\begin{tabular}{l}
\hspace{6in} \\ \hline
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This assignment was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Mathematical and Computational Sciences, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website: 
     
\begin{center}
\href{http://www.utstat.toronto.edu/brunner/oldclass/312f23} {\small\texttt{http://www.utstat.toronto.edu/brunner/oldclass/312f23}}
\end{center}

\end{document}

# Simulating the Pareto with inverse cdf technique

rm(list=ls())
set.seed(9999)
n = 150; theta = 1.16 # See Wikipedia article
y = runif(n); x = (1-y)^(-1/theta); x = round(x,2)
x

# MOM 
xbar = mean(x); xbar/(1+xbar) # .9241136
1/mean(log(x)) # 1.100816


rm(list=ls())
x = scan("http://www.utstat.toronto.edu/~brunner/data/legal/pareto.data.txt")
x
thetahat = 1/mean(log(x)); thetahat
n = length(x); vhat = thetahat^2/n; se = sqrt(vhat); se
low95 = thetahat - 1.96*se; up95 = thetahat + 1.96*se
c(low95,up95) # 95% CI

# Critical value(s). Just say plus and minus 1.96, or ...
c(qnorm(0.025),qnorm(0.975))

Z = (thetahat-1.16)/se; Z

pvalue = 2 * (1-pnorm(abs(Z))); pvalue