# Butterfly price bound independent on underlying distribution

Assuming no fees and interest rate $$r=0$$%, what is the most you would be willing to pay for a \$103/\$106/\$108 European call fly, regardless of the underlying distribution? Buying the fly in this case mean buying the wing of the fly. Source: interview • Interesting question. Some thoughts although not sure if this is correct: the price of any vanilla option is bound between$[0,S]$where$S$is the stock price. The option itself can't be worth more than the stock. Let's assume the worst case scenario: the underlying literally doesn't move. Since we're interested in the maximum price of the strategy, we'd pay a maximum of$ATM_P+ATM_C=S+S=2S$. Jan 27 at 13:37 • Hey, thanks for the idea! One question: why are you only considering the sum of one call and one put? From the wiki page en.wikipedia.org/wiki/Butterfly_(options) a butterfly option is obtained by buying two calls and selling one call (twice) Jan 27 at 13:53 • I assume a BF is a combination of a straddle and strangle. In this example the straddle is ATM and strangle is OTM. I assume the strangle in the worst case is worth$0$and straddle$2S$. You can construct it from calls only or puts only, doesn't really matter. In the ATM case the highest you'd pay for a call (put) is the stock price$S$. I don't know if this is ultimately the answer the interviewers were looking for but it makes intuitively sense to me. Jan 27 at 14:00 ## 2 Answers Maximum theoretical value is 3$

• Agreed, maximum you can make here is $3, most you can lose is$0 so you cant pay more $3 for the option. Jan 27 at 21:52 • Could you explain further please? Jan 27 at 22:28 • Draw the payoff profile at maturity. You are long a 103 strike call, short 2x 106 strike calls and long a 108 strike call. Jan 28 at 14:16 • The whole point here is that the question states that it wants the max price irrespective of underlying distribution, so lets suppose we know for a fact that spot = 106 at maturity. In this case: K = 103 is worth (106-103) x 1 = 3 K = 106 os worth (106 -106) x 2 = 0 K =108 is worth max(0, 106-108) =0 So the structure is worth 3$, which is the maximum it could potentially be worth given you know the spot price at expiration. Of course it will always trade below that because what you can in fact estimate is the probability distribution. Jan 29 at 14:36
• Unless stated otherwise, it is 1-2-1. Jan 30 at 12:01

I'll put my comment as an answer. A butterfly is a combination of a straddle and a strangle. Let's assume the straddle is ATM and the strangle OTM. The price of an option $$V$$ is bounded $$0 \leq V \leq S$$. It can't exceed the value of the underlying.

The worst that can happen for an option holder is that the underlying doesn't move. Assume that the underlying doesn't move at all in which case the OTM options are rendered worthless, $$OTM_C=OTM_P=0$$. Also assume that the ATM option has maximum value $$S$$. That yields a total cost of $$ATM_C+ATM_P-OTM_C-OTM_P=2S$$.

I know this is making some distributional assumptions but you can verify the thought process using Black Scholes. I.e. put the vol of the OTM call and put equal to zero (or an extremely small value). This is because $$N(d_1)$$ and $$N(d_2)$$ both become zero since the $$d_1$$ and $$d_2$$ values become extremely negative (in case of the OTM call). Repeat the exercise and put the vol for the ATM to a very large value, resulting in an option price equal to that of the underlying. See below example from a pricer: If the OTM wings have any value larger than zero, the cost of the strategy will cheapen and be less than $$2S$$. Similarly, if the ATM options are worth less than $$S$$ the total cost will become cheaper.