Consider a market with a risk-free asset such that $A(0) = 100, A(1) = 110, A(2) = 121$ dollars and a risky asset, the price of which can follow three possible scenarios

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Is there an arbitrage strategy if short selling of a stock is allowed, but transaction costs of 5% of the transaction volume apply whenever stock is traded?

How can I solve this?

I know the the No-Arbitrage Principle would be violated if there was a self-financing predictable strategy with initial value $V(0) = 0$ and final value $0 \neq V(2) \geq 0 $ such that $V(1)<0$ with positive probability


At time 0 you do not know if the asset price will rise to 120 or fall to 90 so you cannot be assured of a profit, in which case there is no arbitrage strategy available at time 0.

At time 1 if the price is 120 you, again, cannot be assured of a profit since the price may fall to 96 or risk to 144.

At time 1 if the price is 90 you can be assured that the price will rise to 96 but this is useless since the 5% commission and weaker than risk free accumulation would preclude you from executing this strategy.

  • $\begingroup$ thank you. when i attempted this, I got: The risk free rate is 10%. The guaranteed return on the stock from periods 1 to 2 in scenario 3 would be $\frac{1}{95} = 1.05%$ The net difference would be 8.95%. A transaction cost of 5% would make the difference 3.95%. The difference is still positive, so an arbitrage opportunity is still possible. $\endgroup$ – user477465 Nov 5 '18 at 19:33
  • $\begingroup$ but this is incorrect right? $\endgroup$ – user477465 Nov 5 '18 at 19:33
  • $\begingroup$ I don't know the definition of an arbitrage opportunity in your context. Your return in scenario 3 is less than the risk free rate, which under my own practical definition is not an arbitrage opportunity since you earn less than otherwise if you did nothing. If I say I will guarantee to give you usd10 in a week would you consider it an arbitrage opportunity to forgo my offer and instead guarantee yourself usd2? But perhaps my practical understanding is not precisely the same as the mathematical one, albeit seems strange to me. $\endgroup$ – Attack68 Nov 5 '18 at 19:38
  • $\begingroup$ oh, I think I misunderstood. thank you again. $\endgroup$ – user477465 Nov 5 '18 at 19:49

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