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#1
11-02-2007, 07:53 PM
 colby2152 Note Contributor SOA Join Date: Feb 2006 Location: Virginia Studying for FAP College: PSU '07 Favorite beer: Oskar Blues Old Chub Scotch Ale Posts: 4,171
Convexity

I had trouble with this question on a TIA practice exam (Exam #6 - Problem 14).

An investor sees the following table of call options prices in the newspaper one morning for a non-dividend-paying stock. The investor exploits an arbitrage opportunity, using (either buying or selling short) a block of 100 options expiring in 6 months and a block of 100 options expiring in 2 years. What is the present value of the minimum guaranteed profit for the investor if the continuous risk-free rate of interest is 6%?

$\begin{array}{|c|c|c|}\hline Strike & Time - to - Expiry & Call \\ \hline 100 & 6- months & 30 \\ \hline 105 & 6- months & 27 \\ \hline 110 & 6- months & 25 \\ \hline 100 & 2- years & 34 \\ \hline 105 & 2- years & 31 \\ \hline 110 & 2 -years & 29 \\ \hline\end{array}$

With there being an arbitrage opportunity, how do I know whether the first call should be 29, the second call should be 27.5, or the third call should be 24 (and the same situation for the 2 year calls)?
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How to explain actuarial exams to someone else...
Good Einstein quote - "One had to cram all this stuff into one's mind for the examinations, whether one liked it or not. This coercion had such a deterring effect on me that, after I had passed the final examination, I found the consideration of any scientific problems distasteful to me for an entire year."
#2
11-02-2007, 09:06 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,262

Hmm. I'm confused. Convexity seems to be fine for those. However there does seem to be a time-based arbitrage present.

For instance, we can sell 100 6-month calls with strike 100 and buy 100 2-year calls with strike 110. This results in an up-front profit of
$100 (30) - 100 (29) = 100$

After six months the worst case scenario is that the calls we sold are exercised, in which case we have to pay
$100 (S_{0.5} - 100)$

On the other hand we still have the 100 2-year calls with strike 110 that we purchased, which according to put-call parity are each worth
$Call(S, 110, 1.5 \,\text{years}) = Put(S, 110, 1.5 \,\text{years}) + S_{0.5} - 110 e^{-0.06(1.5)} \geq S_{0.5} - 100.5324$

So if we sold our calls to cut our losses then all told we would make at least
$100 (S_{0.5} - 100.5324) - 100 (S_{0.5} - 100) = -53.24$

This would make the present value of all our cash flows at least
$100 - 53.24 e^{-0.06(0.5)} = 48.33$

There are other arbitrages possible, but I think this one is probably the most profitable.
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#3
11-03-2007, 12:28 PM
 colby2152 Note Contributor SOA Join Date: Feb 2006 Location: Virginia Studying for FAP College: PSU '07 Favorite beer: Oskar Blues Old Chub Scotch Ale Posts: 4,171

Okay, so there are other arbitrage opportunities, but that is the best one.

I thought this was an example of convexity - the differing strike prices and premiums.
__________________
How to explain actuarial exams to someone else...
Good Einstein quote - "One had to cram all this stuff into one's mind for the examinations, whether one liked it or not. This coercion had such a deterring effect on me that, after I had passed the final examination, I found the consideration of any scientific problems distasteful to me for an entire year."
#4
11-03-2007, 02:48 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,262

Well, there are other arbitrage opportunities, but they're all time-based. There are no convexity or slope-based arbitrages in the problem.
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.

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