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\title{Stat 311-Lecture 14 Correlation}
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%\author{\Large Probability}
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Nov 22, 2005


\begin{enumerate}


\item The covariance between $X$ and $Y$ is
$${\bf Cov} (X,Y) = \mathbb{E} (X-\mu_x)(Y-\mu_y),$$
where $\mu_x = \mathbb{E} X, \mu_y = \mathbb{E} Y$. It measures the
relationship between two random variables.


\item {\em Correlation coefficient} is given by $$\rho(X,Y) =
\frac{{\bf Cov}(X,Y)}{\sqrt{{\bf Var}(X){\bf Var}(Y)}}$$. It is the
normalized version of the covariance.

\item $${\bf Cov} (aX+b,cY+d) = ac {\bf Cov} (X,Y).$$
$${\bf Cov} (X,X)
= \mathbb{V} X.$$
$$\rho(aX+b,cY+d) = \rho(X,Y).$$


\item Suppose that the joint distribution of $X$ and $Y$ is a uniform distribution over $x^2 + y^2 \leq 1$. Determine the correlation coefficient of $X$ and $Y$.\\
{\em Solution.} The joint density is $f(x,y)= \frac{1}{\pi}$,for
$(x,y) \in \{(x,y)|x^2 + y^2 \leq 1\}$ and 0 otherwise. The marginal
density is
$$ f_X(x) =\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} \; dy
           = \frac{2\sqrt{1-x^2}}{\pi}, -1 \leq x \leq 1.$$
Since $f_X(x)$ is an even function, $xf_X(x)$ is odd so from
symmetry, $\mathbb{E}X = 0$. Similarly, $\mathbb{E}Y=0$. Also you
should be able to show $\mathbb{E}(XY) = 0$. Therefore, $\rho = 0$.

\item Suppose that $X$ and $Y$ are identically distributed Bernoulli random variables with parameter $0< p < 1$ such that ${\bf Cov}(X,Y) = 0$. Show that $X$ and $Y$ are independent.\\
{\em solution.} The condition ${\bf Cov}(X,Y)=0$ is equivalent to
$P(X=1,Y=1)=P(X=1)P(Y=1)=p^2$. Then it follows
$P(X=1,Y=0)=P(X=1)-P(X=1,Y=1)= p(1-p)=P(X=1)P(Y=0)$ and similarly
for other cases.
So we have $P(X=i,Y=j) = P(X=i)P(Y=j) \mbox{ for all } i,j =0,1.\\

\item For random variables $X$ and $Y$, prove or disprove that
$X$ and $Y-\mathbb{E}(Y|X)$ are correlated. \\
{\it Solution.} \bq &&{\bf cov}(X,Y-\mathbb{E}(Y|X))\\ &=&
\mathbb{E}\{X(Y-E(Y|X))\}
-\mathbb{E}(X)\mathbb{E}\{Y-\mathbb{E}(Y|X)\} \\
&=& \mathbb{E}(XY)-\mathbb{E}\{X\mathbb{E}(Y|X)\}-\mathbb{E}(X)\mathbb{E}(Y) + \mathbb{E}(X)\mathbb{E}\{\mathbb{E}(Y|X)\} \\
&=&\mathbb{E}(XY)-\mathbb{E}\{\mathbb{E}(XY|X)\}
-\mathbb{E}(X)\mathbb{E}(Y) + \mathbb{E}(X)\mathbb{E}(Y) =0\eq

\item Let $X_1,\dots,X_n \stackrel{i.i.d.}{\sim} N(\mu,\sigma^2)$. Then are $\bar X$ and $X_j - \bar X$ independent?

{\it solution.} $\bar X \sim N(\mu,\frac{\sigma^2}{n})$. Since $X_j
- \bar X$ is a linear combination of $X_i$'s, it is also distributed
as normal. Any uncorrelated normal distributions are independent. So
we only need to check if ${\bf Cov}( \bar X,X_j-\bar X)=0$. \bq &&
{\bf Cov}(\bar X, X_j - \bar X) \\&=& {\bf
Cov}(\frac{1}{n}\sum_{i=1}^n X_i, X_j -
\frac{1}{n}\sum_{k=1}^nX_k) \\
&=& \frac{1}{n}\sum_{i=1}^n {\bf Cov}(X_i,X_j)
-\frac{1}{n^2}\sum_{i,k=1}^n {\bf Cov}(X_i,X_k) \\
&=& \frac{1}{n}\mathbb{V}(X_j) - \frac{1}{n^2}\sum_{i=1}^n \mathbb{V}(X_i)\\
&=&  \frac{1}{n}\mathbb{V}(X_j) - \frac{1}{n}\mathbb{V}(X_j)=0\eq


\end{enumerate}

{\bf Note. 1.} The second midterm exam result: mean $25.2 \pm 8.6$.
max 40, min 5. 12 students above 30. 18 students between 20 and 30.
11 students below 20.


\begin{figure}[h]
\includegraphics[scale=0.55]{midterm2005.distribution.eps}
\centering \caption{Number of students (vertical) below the given
exam score (horizontal).}
\end{figure}






 {\bf Note. 2. } The final homework problem will be
posted in the TA's class website by this Thursday. Solve 9 problems
in Chapter 6 and 7. Due date: December 15 (Thursday).


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