Hey guys! I've got a question regarding option arbitrage: Suppose the continuously compounded risk-free rate is 5% for all maturities. The current level of the index is 1000. Dividends are assumed to be re-invested 1) A European call option on this index with strike price equal to 1000 and time to maturity of 1 year is priced at c = $80: 2) A European put option on the same index with strike price equal to 1822 and time to maturity of 12 years is also priced at p = $80 There is apparently an arbitrage opportunity but I'm struggling to understand it. I've calculated the implied volatility of both the call and the put options. After, I calculated the price of put with strike 1000 and price of call with strike 1822 with the implied volatility. But then I don't know how to continue... Any help is very much appreciated!
You seem to propose a non existent hypothetical case. IE, your assumptions are false. Locate actual products, you think fit your idea, then the rotten tomatoes can appear.
What happens to the put at the end of one year? Rinse&Repeat. (Which is to say, it sounds like you're trying to get it all in your head at the same time. *Don't.* The call dies in one year -- what happens(ed) to it? But the put has 11 years to expiration! What happens to the put? If you take the next course in this class, you will do all of this in one equation and a whole lot of parentheses.
Is anyone planning to actually answer your question? As pointed out above the put is underpriced, which is easily discernible through visual inspection. Note that the 1yr call is priced at ~20% vol or 0.04 var (s/b atmf at k~=1050 instead of k=1000, but just ignore that for now). Since 1822 is the approximate 12 year atmf of S=1000 at r=0.05, the 80 price implies var (and thus vol) is priced at zero for years 2 through 12. That's your arb. Edit: the actual price level is not germane to the analysis. The arb works as long as the price is the same for both options and the price is greater than zero for the 1yr call.
I was going to, but I am pretty certain this is a problem from OPs homework assignment, despite his assurances to the contrary.
It does though. It helps the OP understand what to look at first in a problem like this. First thing I noticed was that 1000*exp(0.05*12) ~= 1800 which is pretty close to 1822. Second I recalled that a 1 year atmf call is ~= S*0.4*v, and, since vol is linear in root time, a 12 year atmf call at the same vol should be about 3.5 times that figure. The rest of the analysis flows from those two observations. The answer also implicitly informs the OP that, for understanding options, knowing a few logs (e.g. ln(1800/1000) equals ln(9) minus ln(5) equals 0.60 equals 0.05*12) is useful, as is the ability to mentally calc square roots (which becomes straight division by two in log terms).