1. exactly one prize,
2. at least one prize,
3. 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:

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

2. $$1-\left(\genfrac{}{}{0.0pt}{}{50}{0}\right){\left(\frac{1}{100}\right)}^0{\left(\frac{99}{100}\right)}^{50}\approx 0.3949939329$$

3. $$1-\left(\genfrac{}{}{0.0pt}{}{50}{0}\right){\left(\frac{1}{100}\right)}^0{\left(\frac{99}{100}\right)}^{50}-\left(\genfrac{}{}{0.0pt}{}{50}{1}\right){\left(\frac{1}{100}\right)}^2{\left(\frac{99}{100}\right)}^{49}\approx 0.08943531310$$

The normal approximations (from the Central Limit Theorem using the half-integer, or histogram area corrections) are

1. $\mathbb{P}\left[\frac{1∕2-50\cdot \left(1∕100\right)}{\sqrt{99\cdot 50}∕100}<Z<\frac{3∕2-50\cdot \left(1∕100\right)}{\sqrt{99\cdot 50}∕100}\right]\approx 0.4223907551$,
2. $\mathbb{P}\left[\frac{1∕2-50\cdot \left(1∕100\right)}{\sqrt{99\cdot 50}∕100}<Z\right]=1∕2=0.5$,
3. $\mathbb{P}\left[\frac{3∕2-50\cdot \left(1∕100\right)}{\sqrt{99\cdot 50}∕100}<Z\right]\approx 0.0776092449$.

The normal approximations are not good since the rule of thumb $npq>18$ is not satisfied, in fact $npq\approx 1∕2$.

Solution: Let $X_1,X_2,X_3\dots$ 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 $S_N=X_1+\dot{+}{X}_n$. We seek $\mathbb{P}\left[S_n>1000000\right]\le 0.01$. This is approximately the probability $\mathbb{P}\left[Z>\frac{1000000-12500n}{4330.127020\sqrt{n}}\right]$. Note (for example from tables) this requires

or $n<73.10947801$ so the bank can expect to make about 73 loans.

1. Find the probability that $0<W\left(1\right)<1$.
2. Find the probability that $0<W\left(1\right)<1$ and $1<W\left(2\right)-W\left(1\right)<3$.
3. Find the probability that $0<W\left(1\right)<1$ and $1<W\left(2\right)-W\left(1\right)<3$ and $0<W\left(3\right)-W\left(2\right)<1∕2$.

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 $\mathbb{P}\left[0<W\left(3\right)-W\left(2\right)<1∕2\right]=0.1915$. Then

1. $0.3413$
2. By independence of increments: $0.3413\cdot 0.1574=0.05372$.
3. By independence of increments: $0.3413\cdot 0.1574\cdot 0.1915=0.01029$.

1. Evaluate the probability that $W\left(5\right)\le 3$ given that $W\left(1\right)=1$.
2. Find the number $c$ such that $\mathbb{P}\left[W\left(9\right)>c|W\left(1\right)=1\right]=0.10$.

Solution:

1. Since $W\left(5\right)-W\left(1\right)\sim N\left(0,4\right)$, the required probability is $$\begin{array}{rcll}\mathbb{P}\left[W\left(5\right)>3|W\left(1\right)=1\right]& =& \mathbb{P}\left[W\left(5\right)-W\left(1\right)>3-1\right]& \\ & =& \mathbb{P}\left[\left(W\left(5\right)-W\left(1\right)\right)∕2>1\right]& \\ & =& 0.1586552539& \end{array}$$

2. Since $W\left(9\right)-W\left(1\right)\sim N\left(0,8\right)$, the required value can be deduced from $$\begin{array}{rcll}\mathbb{P}\left[W\left(9\right)>c|W\left(1\right)=1\right]& =& \mathbb{P}\left[W\left(9\right)-W\left(1\right)>c-1\right]& \\ & =& \mathbb{P}\left[\left(W\left(9\right)-W\left(1\right)\right)∕\left(2\sqrt{2}\right)>\left(c-1\right)∕\left(2\sqrt{2}\right)\right]& \\ & =& 0.10& \end{array}$$

   Then $\left(c-1\right)∕\left(2\sqrt{2}\right)=1.281551566$ and $c=4.624775211$.

1. Using a normal random variable generator (from Excel, Maple, Mathematica, Octave, MATLAB, R etc. , all have one and probably the TI-89 or equivalent has one too), find sample values of $X\left(1\right)$, $X\left(2\right)$, $X\left(4\right)$ and $X\left(9\right)$.
2. Explain why the distribution of $X\left(t\right)$ is normal with mean $0$ with variance $t$.
3. 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 $t=0$.