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\date{March 3, 2005}

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\title{Stat 324: Lecture 11\\
Bivariate Distributions}

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\author{Moo K. Chung\\
mchung@stat.wisc.edu}

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\begin{enumerate}

\item {\em Motivation.} There are many practical situations where we
need to deal with more than two measurement simultaneously. For
instance we might be interested in the relationship between heights
and weights of a population.

\item {\em Definition.} For discrete random variables $X$ and $Y$,
the {\em joint probability distribution} is
$$p(x,y) = P(X=x,Y=y).$$
For continuous random variables $X$ and $Y$, the {\em joint density
function} $f(x,y)$ is defined as a function satisfying
$$P((X,Y) \in A) = \int\int_A f(x,y) \;dx \;dy.$$


\item {\em Definition.} Marginal probability function of $X$ is
given by $p_X(x) = \sum_y p(x,y)$. This is the probability function
of $X$. The marginal density function of $X$ is given by $f_X(x) =
\int_{-\infty}^{\infty} f(x,y) \;dy$.

\item Random variables $X$ and $Y$ are independent if
$p(x,y)=p_X(x)p_Y(y)$ for discrete case and $f(x,y) = f_X(x)f_Y(y)$
for continuous case  for all $x$ and $y$.

\item The conditional probability density of $Y$ given $X=x$ is
defined as $f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)}$. The conditional
expectation of $Y$ given $X=x$ is $$\mathbb{E}
(Y|X=x)=\int_{-\infty}^{\infty} yf_{Y|X}(y|x)\;dy.$$

\item {\em Problem.} If $X$ and $Y$ are jointly continuous random variables with joint (cumulative) distribution function $F(x,y)$, express $P(a \leq X \leq b, c \leq Y \leq d)$ using only $F$.\\
{\em Solution.} This is a fun problem testing whether you really
understand the concept of a joint cumulative distribution function.
Draw  $\{a \leq X \leq b, c \leq Y \leq d\} \subset \mathbb{R}^2$.
Then $P(a \leq X \leq b, c \leq Y \leq d) = F(b,d) - F(a,d) - F(b,c)
+ F(a,c)$ using the fact that $P(A \cup B) = P(A) +P(B) - P(A \cap
B)$.

\item {\em Problem.} Suppose that the joint probability density function $f$ of
$X$ and $Y$ is
$$f(x,y)=\left\{
\begin{array}{cc}
  k(x^2+y) & \mbox{ for } 0\leq x \leq 1, 0 \leq y \leq 2 \\
  0 & \mbox{ otherwise. }
\end{array} \right$$
Find the value $k$ and the marginal density functions.
Are $X$ and $Y$ independent random variables? Justify your answer.\\

{\it Solution.} $\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)
\; dxdy = \int_{0}^2 \int_{0}^1 k(x^2+y)\; dx dy =\int_{0}^2
k(\frac{1}{3} +y) \; dy = \frac{8}{3}k=1.$ So $k=\frac{3}{8}$.
$$f_X(x) = \int_{0}^{2} \frac{3}{8}(x^2+y)dy = \left\{\begin{array}{cc}
  \frac{3}{4}(x^2+1)  & \mbox{ for } 0 \leq x \leq 1, \\
  0 & \mbox {otherwise.}
\end{array} \right$$
$$f_Y(y) = \int_{0}^{1} \frac{3}{8}(x^2+y)dx = \left\{\begin{array}{cc}
  \frac{1}{8}(1+3y)  & \mbox{ for } 0 \leq y \leq 2, \\
  0 & \mbox {otherwise.}
\end{array} \right$$
Since $f(0,1) \neq f_X(0)f_Y(1)$, they are not independent.


{\bf Assigned Problems.} Exercise 5.10, Exercise 5.12. Your second
midterm is tentatively scheduled to be on April 5th.

\end{enumerate}
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