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Little question on best-of of 2 calls
- Thread starter lukaluke
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Presumably you sell the best of two calls for more than either of the calls. So when you hedge, if you can hedge for the cost of either of the calls, then you've made a tidy profit. If the underlyings are X and Y, then you will have long positions in X and Y. If X and Y have correlation -1 always, then when you have to increase your position in X, you also have to decrease the position in Y. You can use some of the money from selling Y to buy X. This mitigates the normal cost of hedging with X. So it looks like you can hedge for less than the cost of hedging one option.
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It's not clear at all, nor does it look correct. For one thing, the first two sentences appear to violate no free lunch.
1) Presumably your question concerns what is a more attractive for someone selling the option assuming they want to hedge the risk (assuming something like Black-Scholes, the only profit is from what you charge in excess of the fair option value).
Then the seller wants to minimize Gamma in order to minimize the hedging costs from discrete rebalancing of the hedge. C.S. observed that when the underlyings move in opposite directions, the hedging cost should be reduced. This should be the case since the errors should cancel each other, but it requires a formal argument if you want to prove it.
2) Another interpretation of your question is why the value of the option is maximized when the correlation is -1. I believe this is true, and should be reasonable to show mathematically.
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Hello friends.
I think the clearest way to understand the correlation exposure of a best-of is to look at it through the lens of a worst-of. The worst-0f is replicated by vanilla call 1 + vanilla call 2 - best-of 1 and 2 calls. Go through a couple at-expiry scenarios to convince yourself that that is in fact the case. Neither of the vanillas on their own have any correlation exposure, so the worst-of and best-of have exactly opposite correlation exposure, which I think is pretty intuitive anyway. The worst-of is long correlation. This is true because if the correlation were very negative, that would mean one call would go in the money just as the other call would go out of the money, which in turn would mean the worst-of would pay off zero. You need to the correlation to be as high as possible so that both calls can go as far in the money as possible at the same time. Because buying the worst-of makes you long correlation, if you were to instead buy a best-of you would then be short correlation (and want correlation to be -1). If you were to sell a best-of, you would actually want the correlation to be as high as possible, i.e. as close to +1 as possible, so that you would have the opportunity for both calls to go out of the money and for the best-of to therefore be worth zero.
I think the clearest way to understand the correlation exposure of a best-of is to look at it through the lens of a worst-of. The worst-0f is replicated by vanilla call 1 + vanilla call 2 - best-of 1 and 2 calls. Go through a couple at-expiry scenarios to convince yourself that that is in fact the case. Neither of the vanillas on their own have any correlation exposure, so the worst-of and best-of have exactly opposite correlation exposure, which I think is pretty intuitive anyway. The worst-of is long correlation. This is true because if the correlation were very negative, that would mean one call would go in the money just as the other call would go out of the money, which in turn would mean the worst-of would pay off zero. You need to the correlation to be as high as possible so that both calls can go as far in the money as possible at the same time. Because buying the worst-of makes you long correlation, if you were to instead buy a best-of you would then be short correlation (and want correlation to be -1). If you were to sell a best-of, you would actually want the correlation to be as high as possible, i.e. as close to +1 as possible, so that you would have the opportunity for both calls to go out of the money and for the best-of to therefore be worth zero.