How Transaction Costs Widen Option No-Arbitrage Bounds

# Why do transaction costs increase the range of the no-arbitrage bounds for an option's price?


# Why do transaction costs increase the range of the no-arbitrage bounds for an option's price?












I am reading this book by Mark S. Joshi. Can you help me make sense of one of the exercise questions? Here is the question (from page 40 of the book):

> Exercise 2.5 Suppose no-arbitrage bounds for an option price show that the price lies between $L_1$ and $L_2$ in a world without transaction costs. What can we say about the bounds if we take transaction costs into account?

Here are the solutions (from page 474):

> Exercise 2.5 Increasing transaction costs can only decrease the number of arbitrage portfoliois, so the bounds will be at least as wide.

I don't understand why the bounds will be at least as wide. I understand that transaction costs make it harder to find an arbitrage. Shouldn't decreasing the number of arbitrage portfolios act to decrease the no-arbitrage price range?

## Answer by Pontus Hultkrantz (score 8, accepted)

https://quant.stackexchange.com/a/60001

Assume a store is fairly pricing a bottle of water at \$1. Now another store is pricing the same bottle of water for \$1.2. Assuming it is possible, you can buy the water at the first store, end sell it to the second store for a \$0.2 risk-less profit, as a consequence forcing him to lower the price to \$1. At this point you no longer arbitrage. In this example $$ L_1=L_2=$1.0. $$

Now assume instead that the second store is a long journey away, costing you \$0.1 in petrol (transaction cost). Hence, you will only buy from the cheap store and sell to the expensive store as long as you make money from it. Initially you make 1.2-(1+0.1) = \$0.1 a bottle, but the more you arbitrage the more the second store will lower their price. In fact, they will lower it until you stop, which occurs when the price is \$1.10 (you make zero profit). In this example $$ L_1=$0.90, \;L_2=$1.10. $$

So including costs will widen the bounds for where there is no arbitrage, simply because no one can bother making pennies when they are eaten up by costs.

## Answer by Valometrics.com (score 1)

https://quant.stackexchange.com/a/59998

If the arbitrage portfolios P1 and P2 that gives L1 and L2 prices are not valid no more, L1 and L2 will be higher and as the transaction costs increases with the invested amount. the difference between L1 and L2 will also increase.

## Answer by Con Fluentsy (score 1)

https://quant.stackexchange.com/a/59999

The book extract does leave his proposition confusing, from Taleb's book on Dynamic Hedging he quotes a model based on research where, they use transaction costs as a percentage of trade, and it affects the volatility estimate you plug into the pricing formula, and hence affecting the profitability (or alternately arbitrage outcomes) for the option.Leland's breakeven volatility model including transaction costs. It does not directly address the issue, of bounded transaction costs but how to trade successfully given trading costs, but I did point out this the bounds will be different due to an options gamma whether you are an option seller or buyer. Short sellers are more affected by transaction costs as they are asking a price so have to bear the transaction cost more because of negative gamma, narrowing their arbitrage opportunities, whereas bidders are offering a price, so transaction costs do not significantly alter ther bounds, they have positive gamma,which keeps bounds in parity even as transaction costs may rise.