Distribution-Free Payoff Bound for an Unequal-Strike Call Butterfly
Summary
The document asks for the maximum price of a European call butterfly with strikes at 103, 106, and 108, assuming no fees and a zero interest rate, without specifying a distribution for the underlying. The accepted reply gives a maximum theoretical value of three dollars. For a standard long butterfly with one call at each outer strike and two short calls at the middle strike, the terminal payoff is capped by the smaller strike interval: the payoff peaks at the middle strike and cannot exceed three dollars. This payoff cap provides a distribution-independent upper bound on the price under the stated assumptions.
Another reply discusses option price bounds and a straddle/strangle construction, but it does not establish the quoted bound for this specific call butterfly and introduces Black–Scholes examples. The document offers no derivation of the accepted answer or discussion of discounting, settlement, or contract details. The bound depends on the standard butterfly position and stated assumptions; the prompt’s phrase about buying the wing is not clarified.
Key ideas
- A standard long call butterfly combines calls at two outer strikes with twice the short position at the middle strike.
- Its terminal payoff cannot exceed the narrower distance between adjacent strikes.
- For the stated strikes, the accepted answer gives a maximum theoretical value of three dollars.
- The result depends on the standard payoff structure and the assumptions about fees and interest rates.
Tags
Full text
# Butterfly price bound independent on underlying distribution # 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 ## Answer by Rodrigo (score 4, accepted) https://quant.stackexchange.com/a/74470 Maximum theoretical value is 3$ ## Answer by oronimbus (score 2) https://quant.stackexchange.com/a/74466 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.
Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.