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\title{Stat 311-Lecture 08 Discrete Random Variables}
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%\author{\Large Probability}
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Oct 4, 2005


\begin{enumerate}

\item {\em Negative binomial distribution with parameter $r$ and $p$.} Let $X$ be the number of
coin tossing until $r$ number of heads are accumulated. Then $$p(x)
={x-1 \choose r-1} p^r (1-p)^{x-r}, x =r, r+1, \cdots.$$ The
geometric distribution is a special case of a negative binomial
distribution with $r=1$.

$\mathbb{E} X^k = \frac{r}{p} \mathbb{E} (Y-1)^{k-1}$ where $Y$ is a
negative binomial with parameter $r+1$ and $p$. By letting $k=1$,
$\mathbb{E} X = \frac{r}{p}$. By letting $k=2$, $\mathbb{E}X^k =
\frac{r}{p}(\frac{r+1}{p} -1)$. So $\mathbb{V} X =
\frac{r(1-p)}{p^2}$.

\item The Banach match problem. A mathematician carries two match
boxes each containing $n$ matches. The probability of using each
match box is 1/2. If he found one of the box is empty, what is the
probability that the one box contains $k$ matches?

{\em Solution.} Let $A_K$ be the above event.  Compute
$P(A_n),P(A_{n-1}),\cdots$ and you will see the pattern. $P(A_k) =
2{2n-k \choose n}(\frac{1}{2})^{2n-k}, k=0,1,\cdots,n$.


\item {\em Hypergeometric distribution.} A box contains $N$ balls,
of which $m$ are white and $N-m$ are black. Let $X$ be the number of
white balls in $n$ draws without replacement.
$$p(x) = \frac{ {m \choose x}{N-m \choose n-x}}{{N \choose n}},
x=0,1,\cdots,\min(n,m).$$

\item Two people toss a fair coin $n$ times each. Find the
probability that they throw equal number of heads.\\
{\it solution.} Let $E_i$ be the event the both throws $i$ heads.
Then we are computing $$P( \cup_{i=0}^n E_i) = \sum_{i=0}^n
P(E_i).$$ Note $P(E_i) = {n \choose i}(\frac{1}{2})^n {n \choose
i}(\frac{1}{2})^n$.

This problem can be also solved using random variables. Let $X$ be
the number of heads for the first person and $Y$ be the number of
heads for the second person. Then $X,Y \sim^{i.i.d.}
\mbox{Binomial}(n,\frac{1}{2})$.
\bq P(X=Y) &=& \sum_{i=0}^n P(X=i,Y=i)\\&=&\sum_{i=0}^n P(X=i)P(Y=i) \\
&=& \sum_{i=0}^n {n \choose i}(\frac{1}{2})^i (\frac{1}{2})^{n-i}
{n \choose i}(\frac{1}{2})^i (\frac{1}{2})^{n-i} \\
&=& \frac{1}{2^{2n}}\sum_{i=0}^n {n \choose i}{n \choose i} \eq From
the hypergeometric distribution, $${2n \choose n} =\sum_{i=0}^n {n
\choose i}{n \choose i}.$$ Hence, the probability is $\frac{{2n
\choose n}}{2^{2n}}$.

\end{enumerate}

{\bf NOTE:} The first midterm exam will cover lectures 1-9 (upto
Chapter 4). {\bf HW 3} due Oct 20. Solve the following 9 problems.
Chapter 4 Problems 42, 44, 48, 60, 64, 75, 76. Theoretical Exercises
31, 32.


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