Building a Butterfly Spread from Call Options by Matching Payoff Slopes
Summary
The document presents a constructive way to represent a piecewise linear European payoff as a portfolio of calls. Begin to the left of all strike points, where call payoffs are zero, and move through the underlying price levels where the target payoff’s slope changes. At each strike, add or subtract calls so the portfolio’s slope matches the target payoff after that point.
For a butterfly with lower, middle, and upper strikes, the procedure yields a long call at each outer strike and twice as many short calls at the middle strike. The answer also uses a long put to illustrate that a short zero strike call followed by a long call at the put strike reproduces its payoff, connecting the construction to put-call parity. This method applies to continuous piecewise linear European payoffs; the examples omit transaction costs and do not address early exercise or market frictions.
Key ideas
- A call contributes zero payoff below its strike and a constant positive payoff slope above it.
- Match a target payoff by tracking its slope from left to right and changing the call position at each kink.
- A standard butterfly uses long calls at the outer strikes and a larger short call position at the middle strike.
- The call portfolio can represent a put using a zero strike call and a call at the put strike.
- The construction is presented for continuous European payoffs and ignores transaction costs.
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Full text
# Construction of Butterfly Spread as sum of Call Options
# Construction of Butterfly Spread as sum of Call Options
I have rigorously stated my problem here.
The task at hand is to express a butterfly spread [no transaction fees] as a sum of long and short call options.
I have found the solution on Wikipedia: $$\big(K-|V-a|\big)^+ = \big(V-(a+K)\big)^+ + \big(V-(a-K)\big)^+ - 2\big(V-a\big)^+,$$ for the underlying $V$, $K>0$ and $a\in \mathbb{R}$. However, in the book [Föllmer, Schied] it's an exercise to come up with the long and short call options [i.e. the righthand side] without any hint. For me, it turned out to be insolvable in a constructive way.
Would anyone even be able to come up this by themselves? What am I missing?
Crossposting in the hopes of finding people who have thought about this before.
Thanks for reading.
## Answer by will (score 4, accepted)
https://quant.stackexchange.com/a/50415
If you're reconstructing a payoff as a linear sum of call options, then the procedure is quite simple -> since the payoff of a call is zero up to the strike, and then linear, you start on the left (i.e. most negative value) and move to the right.
example 1: long butterfly spread, strikes of 80, 100, 120.
- start at zero. gradient of payoff is zero: hold no zero strike calls.
- Move to next change of gradient: 80. Gradient is now 1, so we must be long 1 call at strike 80.
- Move to next change of gradient: 100. Gradient is now -1, but the gradient of our current portfolio of calls is 1 since we are already holding a call at a lower strike. Solution: sell 2 calls, portfolio gradient is now -1.
- Move to next change of gradient: 120. Gradient is now zero, but the gradient of our current portfolio of calls is -1 since we are already holding (net) negative one call at a lower strikes. Solution: buy 1 call, portfolio gradient is now 0.
result: long one 80 strike call, short two 100 strike calls, long one 120 strike call.
You can apply the same logic to any (continuous) european payoff.
silly example: long put, strike 100.
- start at zero. gradient of payoff is -1. short 1 zero strike call.
- move to next change of gradient: 100. payoff is now flat, so we must buy a 100 strike call to flatten out the gradient of our call option portfolio.
result: short 1 zero strike call, long a 100 strike call. note that a zero strike call is just the same as holding the stock - we're saying that a put is equivalent to a call plus a short position in the stock -> i.e. put call parity.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.