# arb between 2 options with K1, K2

It’s not easy to get intuitive feel about arb inequality involving European put/calls of 2 strikes K1 < K2. No stoch or Black/Scholes required. Just use hockey stick diagram i.e. range-of-possibilities payoff diagram.

Essential rule #1 to internalize — if a (super-replicating) portfolio A has terminal value dominating B, then at any time before maturity, A dominates B. Proof? arbitrage. If at any time A is cheaper, we buy A and sell B and keep the profit. At maturity our long A will adequately cover our short B.

—-Q1: Exactly four assets are available: The bond Z with Z0 = \$0.9; and three calls. The underlying is not available for you to trade. The calls have identical expiry T, strike K = 20; 22.5; 25, and time-0 price C0(K), where C0(20) = 6.40; C0(22.5) = 4.00; C0(25) = 1.00. Any arbitrage

Consider the portfolio B {long C(22.5) short C(25)} and A {25 Z – 22.5 Z} = {2.5 Z}

At maturity, A is worth \$2.5 since the bond has maturity value \$1 by definition.
At maturity, B < \$2.5, obvious by hockey stick. This is the part to get intuitive with.

By Rule #1, any time before maturity, A should dominate B. In reality, A0 = 2.5 Z0 but B0 = \$3 -> arb.

This trick question has some distractive information!

—-Q2: P141 [[xinfeng]] has a (simple) question — put {K=80} is worth \$8 and put {K=90} is worth \$9. No dividend. Covered on Slide 30 by Roger Lee. Need to remember this rule — 80/90*P(90) should dominates P(80) at time 0 or any time before expiry, otherwise arb.

Consider portfolio A {80/90 units of the K=90 calls} vs B {1 unit of the K=80 call}

If stock finishes above \$80 then B = 0 so A >= B i.e. dominates
If stock finishes below \$80, then B = 80-S. Hockey stick shows A = 80 – S*80/90 so A dominates.

Therefore at expiration A dominates and by Rule #1 A should be worth more at time 0. In reality, A0=B0.