 If you buy a lottery ticket in 50 independent lotteries,
and in each lottery your chance of winning a prize is
$1\u2215100$,
write down and evaluate the probability of winning and also approximate
the probability using the Central Limit Theorem.
 exactly one prize,
 at least one prize,
 at least two prizes.
Explain with a reason whether or not you expect the approximation to be a
good approximation.
Solution: The exact probabilities are easy:

$$\left(\genfrac{}{}{0.0pt}{}{50}{1}\right){\left(\frac{1}{100}\right)}^{1}{\left(\frac{99}{100}\right)}^{49}\approx 0.3055586198$$

$$1\left(\genfrac{}{}{0.0pt}{}{50}{0}\right){\left(\frac{1}{100}\right)}^{0}{\left(\frac{99}{100}\right)}^{50}\approx 0.3949939329$$

$$1\left(\genfrac{}{}{0.0pt}{}{50}{0}\right){\left(\frac{1}{100}\right)}^{0}{\left(\frac{99}{100}\right)}^{50}\left(\genfrac{}{}{0.0pt}{}{50}{1}\right){\left(\frac{1}{100}\right)}^{2}{\left(\frac{99}{100}\right)}^{49}\approx 0.08943531310$$
The normal approximations (from the Central Limit Theorem using the
halfinteger, or histogram area corrections) are
 $\mathbb{P}\left[\frac{1\u2215250\cdot \left(1\u2215100\right)}{\sqrt{99\cdot 50}\u2215100}<Z<\frac{3\u2215250\cdot \left(1\u2215100\right)}{\sqrt{99\cdot 50}\u2215100}\right]\approx 0.4223907551$,
 $\mathbb{P}\left[\frac{1\u2215250\cdot \left(1\u2215100\right)}{\sqrt{99\cdot 50}\u2215100}<Z\right]=1\u22152=0.5$,
 $\mathbb{P}\left[\frac{3\u2215250\cdot \left(1\u2215100\right)}{\sqrt{99\cdot 50}\u2215100}<Z\right]\approx 0.0776092449$.
The normal approximations are not good since the rule of thumb
$npq>18$ is not satisfied,
in fact $npq\approx 1\u22152$.
 A bank has $1,000,000 available to make for car loans. The loans are in
random amounts uniformly distributed from $5,000 to $20,000. Make a
model for the total amount that the bank loans out. How many loans can
the bank make with 99% confidence that it will have enough money
available?
Solution: Let ${X}_{1},{X}_{2},{X}_{3}\dots \phantom{\rule{0em}{0ex}}$
be a sequence of random variables representing the individual loan amounts.
These random variables may reasonably be assumed to be independent, and
of course are identically distributed random variables on the interval
$\left[5000,20000\right]$. Then
$E\left[Xi\right]=12500$ and
$Var\left[{X}_{i}\right]=18,750,000$ so
$\sigma =4330.127020$. Then the total loan
amount is ${S}_{N}={X}_{1}+\stackrel{\u0307}{+}{X}_{n}$. We seek
$\mathbb{P}\left[{S}_{n}>1000000\right]\le 0.01$. This is approximately
the probability $\mathbb{P}\left[Z>\frac{100000012500n}{4330.127020\sqrt{n}}\right]$.
Note (for example from tables) this requires
$$\frac{100000012500n}{4330.127020\sqrt{n}}>2.326347874$$
or $n<73.10947801$
so the bank can expect to make about 73 loans.
 Let $W\left(t\right)$
be standard Brownian motion.
 Find the probability that $0<W\left(1\right)<1$.
 Find the probability that $0<W\left(1\right)<1$
and $1<W\left(2\right)W\left(1\right)<3$.
 Find the probability that $0<W\left(1\right)<1$
and $1<W\left(2\right)W\left(1\right)<3$
and $0<W\left(3\right)W\left(2\right)<1\u22152$.
Solution: First note that $\mathbb{P}\left[0<W\left(1\right)<1\right]=0.3413$,
$\mathbb{P}\left[1<W\left(2\right)W\left(1\right)<3\right]=0.1574$, and
$\mathbb{P}\left[0<W\left(3\right)W\left(2\right)<1\u22152\right]=0.1915$.
Then
 $0.3413$
 By independence of increments: $0.3413\cdot 0.1574=0.05372$.
 By independence of increments: $0.3413\cdot 0.1574\cdot 0.1915=0.01029$.
 Let $W\left(t\right)$
be standard Brownian motion.
 Evaluate the probability that $W\left(5\right)\le 3$
given that $W\left(1\right)=1$.
 Find the number $c$
such that $\mathbb{P}\left[W\left(9\right)>cW\left(1\right)=1\right]=0.10$.
Solution:
 Since $W\left(5\right)W\left(1\right)\sim N\left(0,4\right)$,
the required probability is
$$\begin{array}{rcll}\mathbb{P}\left[W\left(5\right)>3W\left(1\right)=1\right]& =& \mathbb{P}\left[W\left(5\right)W\left(1\right)>31\right]& \text{}\\ & =& \mathbb{P}\left[\left(W\left(5\right)W\left(1\right)\right)\u22152>1\right]& \text{}\\ & =& 0.1586552539& \text{}\end{array}$$
 Since $W\left(9\right)W\left(1\right)\sim N\left(0,8\right)$,
the required value can be deduced from
$$\begin{array}{rcll}\mathbb{P}\left[W\left(9\right)>cW\left(1\right)=1\right]& =& \mathbb{P}\left[W\left(9\right)W\left(1\right)>c1\right]& \text{}\\ & =& \mathbb{P}\left[\left(W\left(9\right)W\left(1\right)\right)\u2215\left(2\sqrt{2}\right)>\left(c1\right)\u2215\left(2\sqrt{2}\right)\right]& \text{}\\ & =& 0.10& \text{}\end{array}$$
Then $\left(c1\right)\u2215\left(2\sqrt{2}\right)=1.281551566$
and $c=4.624775211$.
 Use your Homework 5 record of a
$100$flip coin flip
sequence. Scoring ${Y}_{i}=+1$
for each Head and ${Y}_{i}=1$
for each Tail on each flip, keep track of the accumulated sum
${T}_{n}={\sum}_{i=1}^{n}{Y}_{i}$ for
$n=1,\dots ,100$
representing the net fortune at any time. Plot the resulting
${T}_{n}$ versus
$n$ on the interval
$\left[0,100\right]$. Finally, using
$N=10$ plot the rescaled
approximation ${W}_{10}\left(t\right)=\left(1\u2215\sqrt{10}\right)S\left(10t\right)$
on the interval $\left[0,10\right]$
on the same set of axes.
 Let $Z$
be a single normally distributed random variable, with mean
$0$ and
variance $1$,
i.e. $Z\sim N\left(0,1\right)$.
Then consider the continuous time stochastic process
$X\left(t\right)=\sqrt{t}Z$.
 Using a normal random variable generator (from Excel, Maple,
Mathematica, Octave, MATLAB, R etc. , all have one and probably
the TI89 or equivalent has one too), find sample values of $X\left(1\right)$,
$X\left(2\right)$,
$X\left(4\right)$
and $X\left(9\right)$.
 Explain why the distribution of $X\left(t\right)$
is normal with mean $0$
with variance $t$.
 Is $X\left(t\right)$
a Brownian motion? Explain why or why not.
Solution: No, $X\left(t\right)$
is not Brownian motion for two reasons in spite of the fact that
$\sqrt{t}Z\sim N\left(0,t\right)$
(which follows from being a scalar multiple by
$\sqrt{t}$ of the
$N\left(0,1\right)$ random
variable $Z$.)
First, for ${t}_{1}<{t}_{2}\le {t}_{3}<{t}_{4}$,
$X\left({t}_{2}\right)X\left({t}_{1}\right)=\left(\sqrt{{t}_{2}}\sqrt{{t}_{1}}\right)Z$ is not
independent of $X\left({t}_{4}\right)X\left({t}_{3}\right)=\left(\sqrt{{t}_{4}}\sqrt{{t}_{3}}\right)Z$
since both are multiples of the same sample value
$Z$ drawn
from the $N\left(0,1\right)$
population.
Second, the distribution of $\left(\sqrt{{t}_{2}}\sqrt{{t}_{1}}\right)Z$
is normal with variance ${\left(\sqrt{{t}_{2}}\sqrt{{t}_{1}}\right)}^{2}\ne {t}_{2}{t}_{1}$.
Note nevertheless that $X\left(0\right)=0$
and $X\left(t\right)$ is
continuous at $t=0$.