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QUESTION

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\item[(a)]
Consider the exponential distribution $$f(x) = \lambda e^{-\lambda
x}, \quad x \geq 0.$$

\begin{description}

\item[(i)]
Prove that the mean of the exponential distribution is
$\lambda^{-1}$.

\item[(ii)]
Write down an expression which demonstrates that a probability
density function has the no-memory property.

\item[(iii)]
Prove that the exponential distribution has the no-memory
property.

\item[(iv)]
The inter-arrival time of the school buses is believed to be
exponentially distributed with a mean of $20$ minutes. You have
been waiting for the bus for $30$ minutes; what is the probability
that you have to wait for more than one hour at the end?

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\item[(b)]
Using the mixed congruential generator $$x_{n+1} = (7x_n+11){\rm
mod}\, 31$$ and seed $x_0=9$, generate a stream of five random
numbers in the interval $[0,30]$.  Use these to generate five
random numbers in the interval $[0,1)$, to three decimal place
accuracy.

\item[(c)]
By using the inverse transformation method, show that
$-{{1}\over{\lambda}} \ln (U)$ is exponentially distributed with
mean $\lambda^{-1}$. Here, $U$ is a continuous random variable
uniformly distributed over $(0,1)$.

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ANSWER

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\item[(a)]

\begin{description}

\item[(i)]

\begin{eqnarray*}
E(x)&=&\int_0^\infty t\lambda e^{-\lambda t}\,dt\\
&=&\int_0^\infty te^{-\lambda t}d(\lambda t)\\
&=&\int_0^\infty-td(e^{-\lambda t})\\ &=&\int_0^\infty e^{-\lambda
t}\,dt-te^{-\lambda t}|_0^\infty\\ &=&\int_0^\infty e^{-\lambda
t}\,dt-0\\ &=&\frac{1}{\lambda}
\end{eqnarray*}

\item[(ii)]
A probability distribution $f(x)$ is said to have the no-memory
property if

$$\textrm{Prob}(x>t+h|x\geq t)=\textrm{Prob}(x>h).$$

\item[(iii)]
For the exponential distribution function, it is clear that

$$\textrm{Prob}(x>h)=\int_h^\infty\lambda e^{-\lambda
t}\,dt=e^{-\lambda h}.$$

We note that

$$\textrm{Prob}(x>h+t|x\geq t)=\frac{\textrm{Prob}(x>h+t\textrm{
and }x\geq t)}{\textrm{Prob}(x\geq t)}.$$

Therefore we have

$$\textrm{Prob}(x>h+t\textrm{ and }x\geq
t)=e^{-\lambda(t+h)}\textrm{ and Prob}(x\geq t)=e^{-\lambda t}.$$

Hence

$$\textrm{Prob}(x>t+h|x\geq
t)=\frac{e^{-\lambda(t+h)}}{e^{-\lambda t}}=\textrm{Prob}(x>h).$$

\item[(iv)]
By the no memory property, the probability is given by

$$\int_30^\infty\frac{1}{20}e^{-\frac{t}{20}}\,dt=e^{-\frac{3}{2}}.$$

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\item[(b)]
$x_0=9,\ x_1=12,\ x_2=2,\ x_3=25,\ x_4=0,\ x_5=11$ therefore
$U_1=0.387,\ U_2=0.065,\ U_3=0.806,\ U_4=0,\ U_5=0.355$.

\item[(c)]
We note that cumulative probability distribution of the
Exponential distribution is

$$F(x)=\int_0^x\lambda e^{-\lambda t}\,dt=10e^{-\lambda x}.$$

By using the inverse transform method we have

$$x=F^{-1}(U)=-\frac{1}{\lambda}\log(1-U)$$

Here $U$ is the continuous random variable uniformly distributed
over $(0,1)$. Since $U$ and $1-U$ have the same probability
distribution,

$$-\frac{1}{\lambda}\log U$$

is also exponentially distributed with mean $\lambda^{-1}$.

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