Triangular Arbitrage Confusion

Hi Everyone! First time poster, glad to join the ranks. :slight_smile:

Reading 13 EOC Question 15 is a triangular arbitrage problem with two dealers. One is told they have to convert pounds to pesos in order to pay an invoice. After all the calculations, you are left with the following:

Dealer A Bid/Ask = .0366/.0370

Dealer B Bid/Ask = .0366/.0372

The question asks if an arb op exists.

The solution to the problem states that there is no arb op because although one can purchase from Dealer A at a lower price, one must sell to both dealers at the same price.

My question is how is it not profitable to use fewer pounds to get the same amount of pesos regardless of the sales price?

Thank you!

This really isn’t triangular arbitrage, because there are only two currencies involved, not three.

Starting with GBP, there are four possibilities:


Here, →_i_ means exchanging through dealer i.

Start with GBP 1,000,000 (or whatever), try each of these conversions, and see if you end up with more GBP than you started with. If so, there’s an arbitrage opportunity. If not, there isn’t.

(Note: you can start with MXN if you prefer, but the results will be the same.)

I see, thank you S2000magician. I stopped at the comparison of the two sets of numbers. Had I continued down the path, I would have realized that one side of the spread never comes into play. Tricky tricky…

You’re welcome.

If the situation is :

Dealer A Bid/Ask = .0366/.0370

Dealer B Bid/Ask = .0370/.0372

we still can not make an arbitrage profit. Am i right?

No. You can proceed with the math given by s2000magician above and you will see that there is not really a way for you to make riskless profit. The best alternative you can get is a break-even position (buy from dealer A at .0370 and sell to dealer B, also at .0370).

A quick way to identify if there is any arbitrage opportunity for this kind of problem: literally draw a number line for each of the margin. If there is no overlap in the number line (i.e. if they contain completely different sets of numbers), then there is an arbitrage opportunity.

same idea as you,thanks for answering.