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\begin{document}

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%\date{January 23, 2003}

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\title{Stat 312: Lecture 22\\
Correlation coefficient.}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}
\date{Dec 9, 2004}
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\maketitle \thispagestyle{empty}
\begin{enumerate}

\item {\em Correlation coefficient} between two random
variables $X$ and $Y$ is defined as $$\rho(X,Y) = \frac{{\bf
Cov}(X,Y)}{\sqrt{{\bf Var}(X){\bf Var}(Y)}}.$$ The {\em sample
correlation coefficient} $r$ between two samples $x_i$ and $y_j$ is
defined as $r = S_{xy}/\sqrt{S_{xx}S_{yy}}.$ {\em Example} Is there
any linear relationship between the mathematical achievement test
scores ($x_i$) and calculus grades
($y_i$) for 10 college freshmen?\\
x$\|39,43,21,64,57,47,28,75,34,52$\\
y$\|65,78,52,82,92,89,73,98,56,75$
\begin{verbatim}
x<-c(39,43,21,64,57,47,28,75,34,52)
y<-c(65,78,52,82,92,89,73,98,56,75)
> cor(x,y)
[1] 0.8397859
\end{verbatim}
We are interested in testing $H_0: \rho = 0 \mbox{ vs. } H_1: \rho
\neq 0.$ In order to do this, we need to know the distribution of
the sample correlation coefficient $r$ under the null assumption.
\item Suppose the joint probability density function $f$ of $X$ and $Y$ is the {\em
bivariate normal} such that $X \sim N(\mu_X,\sigma_X^2)$, $Y \sim
N(\mu_Y,\sigma_Y^2)$ and $\rho(X,Y)=\rho$. In this situation, it can
be shown that $$\mathbb{E}(Y|X=x) = \beta_0 + \beta_1 x,$$ where
$\beta_1 = \rho \sigma_Y/\sigma_X$. This is the regression model we
studied before. Note that $\rho =0$ if and only if $\beta_1=0$.
Hence testing $$H_0:\rho =0 \mbox{ vs. } H_1:\rho \neq 0$$ is
equivalent
 to $$H_0:\beta_1 =0 \mbox{ vs. } H_1:\beta_1 \neq 0.$$ So the appropriate test
 statistic is $$T= (\hat \beta_1-\beta_1)/S_{\hat
 \beta_1} = \beta_1/S_{\hat
 \beta_1} \sim t_{n-2}$$
under the null assumption. It can be shown that the test statistic
is equivalent to $T= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$.

{\bf Note.} In bivariate normal data, $\rho=0$ if and only if $X$
and $Y$ are independent. So testing for independence is equivalent
to testing $\rho=0$ in this situation.


\begin{verbatim}
> summary(lm(y~x))
Call: lm(formula = y ~ x)
Coefficients:
            ...  t value Pr(>|t|)
(Intercept) ...   4.794  0.00137
...
x 0.7656 0.1750   4.375 0.00236
... Multiple R-Squared: 0.7052
...
\end{verbatim}
\item {\em Fisher $z$-transformation}. Given sample correlation $r$,
$$z = \tanh^{-1} r = \frac{1}{2}\ln(\frac{1+r}{1-r})\sim N\Big(\rho,\frac{1}{\sqrt{n-3}}\Big).$$
This is asymptotic results for large sample.

\begin{verbatim}
>fisher.z<-function(r){atanh(r)}
>cor.test(x,y,"two.sided","pearson"
          ,conf.level=0.95)
Pearson's product-moment correlation data:
x and y t=4.375, df =8, p-value = 0.002365
...
95 percent confidence interval:
 0.4459632 0.9611846
sample estimates:
      cor
0.8397859
\end{verbatim}
{\tt R}-package {\tt cor.test} with {\tt pearson} option is based on
the exact distribution of a test statistic using $t_{n-2}$
distribution.


\end{enumerate}
\end{document}