Option contingent upon exercise of another option

Factual Questions
1

OK, I have an option I’m pricing that’s at the money. There is also another option that’s at the money, but that option cannot be exercised unless the first option is exercised. It’s not a two way street though, the first option can be exercised wihtout exercising the second. Both options have the same time of expiry and must be exercised at the same time (that is, once you decide to exercise option 1, you must either exercise option 2 or forfeit the option).

This is not homework (I pity the person who gets this for homework :()

I can price each option seperately using a basic Black Scholes approach.

Assumptions are a vol of 0.12, a risk free rate of 0.002, strike price of 41, a market price of 41, and a time to expiration of 0.63 years for the first option. The second option is exactly the same except both the market and strike price are 42.

So, for the price of the first option (N(x) means standard cumulative normal distribution):

d(1) = (ln[41/41]+[0.63{0.002+([0.12^2]/2)}])/(0.12sqrt[0.63]) = 0.060859
d(2) = d(1) - 0.12sqrt(0.63) = -0.0344
price = 41N(d(1)) - 41(e^[-0.002063])*N(d(2)) = 1.582476

So, the price of the first option is about $1.58.
The second works out to be about $1.62 (you just sub 42 for 41 in the equation above).

So, if you just had each option, and they were independent, you’d say that, together, they’re worth about $3.20. But since option 2 is restricted, it can only be exercised if option 1 is exercised, then it can’t be worth its full independent value. So, the question is: how do you discount it? The underlying for both of the options are very highly correlated - probably around 0.9 to 0.95.

Any thoughts?

eta: I plan on doing some binomial modeling and simulation approaches on Monday, but I was wondering if anyone had an algebraic apporach.

2

What if you considered options 1 and 2 together as a distinct option (call it option B). Then you can just compare option 1 and option B. Of course, you can only choose 1 or B, (or neither) but not both.

3

Well, 1 and 2 have different underlying products. I guess I could assume that I could view them as 1 product with a current market price of 83 and a strike 83 and then go back and try to find the volatility of returns on the combined products - but that would take a bit of data mining. A little bit of math may allow the volatility of the combined products to solved be for using an assumption on correlation. Interesting - I’ll have to think on it.

Running the numbers, I see that using 83 as the strike and market in the formula gives 3.20 as the value, which makes perfect sense. If the products had perfect correlation, then they are in fact the same product and it doesn’t matter if you express the change in volume as a change in the number of units or the price of the unit. So, the question really becomes about correlation and volatility, which makes sense. The question is, how do we calculate the relationship between the correlation of the two underlying products and the volatility of the combined product.

4

Missed edit window - would still have to figure out what to do with option 1 and b…

5

Well, messed around with some calcs where:

volatility of combined product = .06 + (.06 * (1 - correlation)

This means that when they’re perfectly correlated, the volatility remains .12, and when they are perfectly negatively correlated, then the vol is 0, which is right. It also means that when the correlation is -1, that the option is worthless, which makes perfect sense because whenever one would make money, the other would lose money. I am ignoring the slight difference in price here. If they are uncorrelated (zero), then vol equals 0.06, which makes sense because the returns of one whould randomly cancel out the returns of the other about half of the time.

This allows the b portion of the option to valued. We still need to value option 1 given that (under this scenario) we can accept only 1 or b, bot both. So, the idea really should be to figure out the basic value of 1, which we’ve already done and the probability that you would exercise 1 instead of b.

The only time you would want to exercise 1 instead of b would be when option 1 was in the money and option 2 wasn’t (I do mean 2 here, and not b). The further the correlation is from 1, the more likely that you would exercise 1 and not b. For example, if they were perfectly negatively correlated, then you would want to exercise 1 whenever 2 wasn’t in the money.

I’ve used the formula (Price * (1 - cor))/2 to estimate the value of option 1 given that you can’t exercise it if you exercise option b.

Price being the full value of the option (the 1.58 calculated above), cor being the correlation between the underlying products of options 1 and 2.

For example, if the correlation is 1, then we get a value of zero (1.58 * (1-1))/2)- which makes sense, because if they move together then you always prefer to exercise both and not just 1.

If they have a correlation of 0, then you get 1.58*(1-0)/2 = 0.79, which is half the value of the original option. This seems right given that if there’s no correlation between the products then there should be a 50% chance that you’d prefer to exercise option 1 and not option 2.

If they have a correlation of -1, then you get 1.58 * (1- -1)/2 = 1.58, the full price of the option. This makes sense too, since when the underlying assets have perfectly negative correlation, then option b is worthless, and so the value of 1 + b should just equal the value 1.

Adding together the values of options 1 and b, we get something that looks like this, depending on the correlations:


-0.95	 $1.68 
-0.75	 $1.83 
-0.50	 $2.03 
-0.25	 $2.22 
0.00	 $2.42 
0.25	 $2.62 
0.50	 $2.81 
0.75	 $3.01 
1.00	 $3.20
6

There is no simple Black-Scholes-like formula for this. The problem arises because there are situations when the first option is out-of the-money, but the second option is in the money sufficiently that you should exercise the first in order to enable you to exercise the second. This occurs when S1 < K1 and S1 + S2 > K1 + K2. To compute this portion of the present value, we need to know E[Max(S1 + S2 - K1 - K2, 0)]. Knowing that required knowing the probability distribution of the sum of two lognormals which is an unsolved problem in statistics.

You can however get good approximations to the answer.