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\title{Stat 471: Lecture 17\\
Bootstrap III: Regression}
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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
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\maketitle \thispagestyle{empty}
\begin{enumerate} 
\item Continuing lecture 16, if $f$ is not normal, $T$ will be still a pivot but we do not know its distribution. So we generate 1000 bootstrap replicates of $T^* = \frac{\bar X^* - \bar X}{S^*/\sqrt{n}}$ of $T$ and get the bootstrap CI
$$\Big(\bar x - t_{1-\alpha/2}^*S/\sqrt{n} , \bar x - t_{\alpha/2}^*S/\sqrt{n}\Big).$$
\begin{verbatim}
t=inline('(mean(x)-u)/(std(x)/
length(x))','x','u')
for i=1:1000;
  trep(i)=t(boot(arm,1),mean(arm))
end;
>> quantile(trep,[0.05 0.95])
  -19.8246   22.0677
\end{verbatim}


\item For given bivariate data $(x_i,y_i), i=1, \cots, n$ suppose we have model
$$Y_i = g(x_i; \beta) + \epsilon_i.$$
Let $g(x; \beta) = \beta_0 + \beta_1 x$. Fitting a somewhat crazy model like 
$g(x; \beta) = \frac{\beta_0}{\beta_1 - \beta_2 e^{-\beta_3 x}}$ is also possible. The least squares estimation (LSE) is given by
$$\hat \beta = \arg \min_{\beta} \sum_{i=1}^n \Big(y_i - g(x_i; \beta)\Big)^2$$

Based on LSE, the observed residual is
\bqn \hat \epsilon_i = y_i - g(x_i; \hat \beta). \label{eq:res}\eqn
A bootstrap dataset is then
\bqn Y_i^* = g(x_i; \hat \beta) + \epsilon_i^*. \label{eq:bs}\eqn
The bootstrap estimate of $\beta$ is then
\bqn \hat \beta^* = \arg \min_{\beta} \sum_{i=1}^n \Big(y_i^* - g(x_i; \beta)\Big)^2. \eqn

\item Let's find CI for $\beta_1$ for model 
$$\tt{grip} = \beta_0 + \beta_1 \tt{arm}.$$
 The pivot for $\beta_1$ is $T = \frac{\hat \beta_1 - \beta_1}{SE{\hat \beta_1}}$, where $SE \hat \beta_1$ = \sqrt{{\bf Var} \hat \beta_1}$. If you don't know the standard error $SE$, we estimate it, i.e. $T=(\hat \beta_1 - \beta_1)/S_{\hat \beta_1}$. Following eq.1,2 and 3 we generate 1000 bootstrap replicates of $T^*=\frac{\hat \beta_1^* - \hat \beta}{SE \hat \beta_1^*}$. We need to estimate $SE \hat \beta_1^*$ via bootstrap which requires bootstrapping from bootstrapped residual $e_i^*$ (there is a formula for computing the standard error but well let's forget that). Then $100(1-\alpha)\%$ CI for $\beta_1$ is
$$(\hat \beta_1 - t^*_{1-\alpha/2} S_{\hat \beta_1},\; \hat \beta_1 - t^*_{\alpha/2} S_{\hat \beta_1}). $$
\begin{verbatim}
beta=inline('pinv([ones(147,1) x])*y')
>>b=beta(arm,grip)
b=
   54.7081  
   0.7050
e=grip-b(1)-b(2)*arm;
for j=1:1000
  bse=boot(e,1);
  bsgrip = bse + b(1) + b(2)*arm;
  bsb=beta(arm,bsgrip);
  for i=1:50
     bs2e=boot(bse,1);
     bs2grip = bs2e + b(1) + b(2)*arm;
     bs2b=beta(arm,bs2grip);
     bsbeta(i)=bs2b(2);
  end;
  trep(j)=(bsb(2)-0.7050)/std(bsbeta);
end;
>>quantile(trep,[0.05 0.95])
-1.5308    1.6580
\end{verbatim}
\end{document}