Solution: Purchase a forward contract to change GBP to USD in one year’s time at rate $a$ to be determined. Assume that you borrow $X$ USD and then exchange it for $X∕1.5641$ GBP. You then invest it in banks in Great Britain for a year obtaining $Xexp\left(0.0026\cdot \right)∕1.5641$. Then you can exercise your forward contract and change the GBP to USD, obtaining $aXexp\left(0.0026\cdot 1\right)∕1.5641$. Then you have to pay back the loan in the amount $Xexp\left(0.002\cdot 1\right)$. These amounts should be the same or an arbitrage opportunity exists. Then

or $a=1.5632$.

1. What is the continuously compounded annual rate of increase of the price of gold over the period January 2006 to October 2009?
2. What is the continuously compounded annual rate of increase of the price of gold over the period October 2009 to September 2010?
3. In October 2009, one can borrow or lend money at 5% interest, again assume it compounded continuously. In view of this, describe a strategy that will make a profit in October 2010, involving borrowing or lending money, assuming that the rate of increase in the price gold stays constant over this time.
4. The article suggests that the rate of increase for gold will stay constant. Did it? In view of this, what do you expect to happen to interest rates and what principle allows you to conclude that? Did they?

Solution:

1. The continuously compounded annual rate of increase of gold over this period is the solution $r$ of the equation
   $$g\left(T\right)=g\left(0\right)exp\left(rT\right)$$

   where $T=46∕12$, $g\left(T\right)=1050$, $g\left(0\right)=510$, or

   $$r=ln\left(g\left(T\right)∕g\left(0\right)\right)∕T=ln\left(1050∕510\right)∕\left(46∕12\right)=0.1883829697,$$

   about $18.8\%$ annual rate.

2. Same calculation with $T=11∕12$, $g\left(0\right)=1300$, and $g\left(0\right)=1050$. Then
   $$r=ln\left(g\left(T\right)∕g\left(0\right)\right)∕T=ln\left(1300∕1050\right)∕\left(11∕12\right)=0.23299$$

   about $23.3\%$ annual rate.

3. One could borrow $1050 at $5\%$ continuously compounded interest and buy one ounce of gold at $1050. Holding the gold for one year, its value in October 2010 is $1050exp\left(0.188\right)=1267.66$. You could sell the gold, making $\$217.66$ per ounce. You then repay the loan which will require $1050\cdot exp\left(0.05\right)=1103.83$, or approximately $53.83$ in interest. The profit per ounce of gold is $217.66-53.83=163.83$.
4. The principle of arbitrage says that this opportunity will not stay for long, so if the rate of increase of gold stays constant (or even increases as it did over the last year!), then interest rates must rise. However, interest rates actually fell over the period. The conclusion is not that the principle of no-arbitrage opportunities is incorrect, but rather that the true market is more complex than a simple market consisting only of gold and bonds (loans). The true market has other factors keeping rates low.

Solution: With a decrease in base rate of default of 20% from $p=0.05$ to $p=0.04$, the default rate for the 10-tranche decreases from $0.028188$ to $0.0068445$, a decrease by a factor of $0.24281$.

With a decrease in base rate of default of 20% from $p=0.05$ to $p=0.04$, the default rate for the 10-CDO decreases from $5.4389\times 10^{-4}$ to $2.2300^{-9}$, a decrease by a factor of $4.1001\times 10^{-6}$.

Solution: Here is one way to get a table (or matrix) of default probabilities for the tranches:

Solution: To graph the default rate as a function of the tranche number, the following command in Octave will suffice:  plot( 1 - binocdf([0:100], 100, 0.05))  and the resulting plot is:

Figure 1: Default rate as a function of the tranche number, for $n=100$ and $p=0.05$.

To get a better look at the interesting part of the graph from tranche 0 to tranche 11:  plot( 1 - binocdf([0:11], 100, 0.05))  and the resulting close-up is:

Figure 2: Close-up of the default rate as a function of the tranche number, for $n=100$ and $p=0.05$.