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\title{Stat 311-Lecture 12 Other Continuous Random Variables}
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%\author{\Large Probability}
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Oct 27, 2005

\begin{enumerate}

\item Exponential random variable. $X \sim exp(\lambda)$. Density function: $f(x) = \lambda e^{-\lambda x}, x \geq
0$. c.d.f. $F(x) = 1- \exp(-\lambda x), x \geq 0$. $\mathbb{E} X^n =
\frac{n}{\lambda}\mathbb{E} X^{n-1}$.

$\mathbb{E} X = \frac{1}{\lambda}$, $\mathbb{V} X =
\frac{1}{\lambda^2}$.




\item Problem. The density function of an exponential random variable $X$ is
$f(x) = \frac{1}{\beta}e^{-y/\beta}$ for $x>0$. Assume that the mean
value of $X$ is 1. (a) Find the cumulative distribution function of
$X$ and plot its graph. (b) Derive the probability density function
of $Y=2X$ using the cumulative distribution function of $X$.



\item Problem. Let $X$ have the probability density function $f(x)=ce^{-x}$
for $ 1 < x < \infty$ and $f(x)=0$ otherwise.\\
 (a) Find the value $c$.\\
 {\em Solution.} $\int_1^\infty ce^{-x} \;dx = -ce^{-x} |_1^\infty = ce^{-1}=1$. So
 $c=e$.\\
  (b) Find the cumulative distribution function of $X$ and plot its graph. \\
{\em Solution.} Integrating $f$, we get $F(x)=0$ for $x \leq 1$ and
$F(x)= 1-e^{1-x}$ for $x>1$. $F$ is a nondecreasing function and
$F(\infty)=1$.\\
  (c) Find $P(-1< X \leq 3| X\geq 3)$.\\
{\em Solution.}  $P(-1< X \leq 3|X \geq 3) = \frac{P(X=3)}{P(X\geq
3)} =0$ since the probability that a continuous random variable
taking a single point is zero.

\item Gamma random variable. $X \sim Gamma(\alpha,\lambda)$.
$$f(x) = \frac{\lambda e^{-\lambda x}(\lambda x)^{\alpha
-1}}{\Gamma(\alpha)}, x\geq 0.$$ Note the gamma function
$$\Gamma(\alpha) = \int_0^{\infty} \lambda e^{-\lambda x}(\lambda
x)^{\alpha -1} \;dx = (\alpha-1)\Gamma(\alpha-1).$$ For integer
values, $\Gamma(\alpha) = (\alpha-1)!$. $$\mathbb{E} X =
\frac{\alpha}{\lambda}, \mathbb{V} X = \frac{\alpha}{\lambda^2}.$$

{\em Special cases.} $Gamma(1,\lambda) \sim exp(\lambda)$ and
$Gamma(\frac{n}{2},\frac{1}{2}) \sim \chi_n^2$, chi-squared
distribution with $n$ degrees of freedom. If $Z_i \sim N(0,1)$,
$Z_1^2 + \cdots, Z_n^2 \sim \chi_n^2$.

\item Problem. The lifetime (in hours) of an electronic component is a
random variable with density function given by
$$f(x)= e^{-x}, \; x>0.$$
Four of these components operate independently in a piece of
equipment. The equipment fails if at least two of the components
fail. Find the probability that the equipment will operate for at
least 2 hours without failure.

{\em Solution.} The probability of single component lasting more
than 2 hours is $p = \int_2^{\infty} e^{-x} \; dx = e^{-2}$. Then we
are seeking $${4 \choose 2}p^2(1-p)^2 + {4 \choose 3}p^3(1-p) + {4
\choose 4}p^4.$$



\item If $X$ and $Y$ are continuous random variables with joint c.d.f. $F(x,y)$,
express $P(a \leq X \leq b, c \leq Y \leq d)$ using $F$.\\
{\em Solution.} 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 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.
\end{enumerate}


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