I really have tough time trying to figure this out.

An investor observes the following prices in the market: Euro-Stoxx-Future DEC 148.02-148.03; Euro-Stoxx-Future Call-Option DEC 148.00 1.13-1.15; Euro-Stoxx-Future Put-Option DEC 148.00 1.19-1.21. What is the maximum profit the investor can achieve with the right arbitrage strategy, when only trading one contract each?

I am assuming that all the options mentioned in the question are european and I think this has something to do with the put-call-parity: $$c+Ke^{-rT}=p+F_0e^{-rT}$$ relation. But from the task at hand we don't know the rate $r$ nor the Time to maturity $T$.

Another thing that came to mind was the reverse conversion; long call, short put and short future. Which would pay: $$-148-1.15+1.19+148.02=0.06$$ But I feel like walking on a thin ice here because I don't really/fully understand where am I making the profit in the reverse conversion trade.

So please, if anyone has time to give me any feedback with the problem, comment on the current solution and explain the reverse conversion trade (or provide some futher reading link etc.) I would very much appreciate it.



1 Answer 1


You're on the right track.

The idea is to buy (sell) a synthetic forward (short put + long call) and sell (buy) the future.

You can re-arrange the formula as c - p = (F0 - K) * e^(-rT)

For simplicity's sake, lets assume r = 0:

c - p = F0 - K

If everything were priced correctly, both sides of the equation would be equal. To figure out the max profit, you need to consider the bid-ask spreads. Then you sell the one that's rich and buy the one that's cheap for a risk-less profit.

Here are a few links:

https://www.investopedia.com/articles/optioninvestor/05/011905.asp https://www.investopedia.com/university/conversion-arbitrage/conversion-arbitrage3.asp


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