CFA Reading 18 problem 16 cross-rate

Given USD:CHF, USD:AUD and CHF:AUD fx rates, your supposed to find out argitrage opportunity , i.e. calculate cross rate. How do you decide which cross to calculate? I started calculating USD:CFH, found out CHF overvalued. The correct cross to calculate was CHF:AUD, results in CHF being overvalued anyway. So how do you know which cross to start calculating? Might well calculate all, but on exam day that might take too much of the precious time. Thanks!

Rocci- To check if triangle arbitrage is possible, first you need to determine what cross rate you get for CHF:AUD. USD:CHF= 1.5971 USD:AUD=1.8215 CHF:AUD=1.145 1) CHF/USD x USD/AUD= CHF/AUD 1.5971 1/1.8215= .8768 AUD/CHF so inverse is 1.1405 CHF/AUD Since 1.405 is lower that 1.145 CHF is overvalued and there is arbitrage. So you want to sell CHF and buy AUD. 2) You have $1,000,000 USD, you sell USD and buy CHF at 1.5971 and you get 1,597,100CHF. 3) Sell CHF to get AUD using 1.145 as CHF:AUD. 1,597,100CHF x 1.145 AUD/CHF = 1,828,679.50 AUD 4) Then sell AUD and buy back USD at 1.8215. 1,828,6179.50 AUD/ 1.8215 AUD/USD= 1,003,941.53 USD. You net this with the $1,000,000 and you have 3,941.53 as profit. Hope this helps.

Thanks adee 1031, I get the math, so that’s no problem. What I was wondering is how you decide, in the first place, you need to start calculating the cross for CHF:AUD (eg. why not USD:CHF like I started)?

this is from an earlier post on triangular arbitrage EUR/USD = x JPY/EUR = y USD/JPY = z arrange the three rates in 2 different sequences such that the rates cancel each other out and the LHS = 1, for example A. (EUR/USD) * (JPY/EUR) * (USD/JPY) = 1 (sell USD to buy EUR, use EUR to buy JPY, use JPY to buy USD back) B. (JPY/USD) * (EUR/JPY) * (USD/EUR) = 1 (sell USD to buy JPY, use JPY to buy EUR, use to EUR to buy USD back) In each path A & B, USD is the base you start with and then end up with. Now, for the path with RHS of the equation : xyz > 1; Arbitrage opportunity exists and the profit to be made starting with $1 = xyz - 1 (path A) profit with path B = [1/(xyz)] - 1 (note you might have to invert the quotes in path B. This method works with mid quotes as well as with bids and asks. You can start with either all bids or asks but will have to switch to the other each time you need to flip the given quote). Hope this helps. USD/CHF:: 0.9350 - 0.9352 JPY/USD :: 91 - 92 JPY/CHF :: 83 - 84 CHF is the base currency in this example. Suppose you start with CHF to buy USD, then use USD to buy JPY and finally use JPY to buy CHF back. This wil result in (USD / CHF) * (JPY / USD) * (CHF / JPY) start with all bids using bid for CHF/JPY = 1 / (ask for JPY/CHF) 0.9350 x 91 x 1/84 = 1.012917 > 1 (Arbitrage opportunity exists) You can make (1.012917 - 1) * 1000000 = CHF 12917 starting with CHF 1m Alternatively, if you start with CHF to buy JPY, then use JPY to buy USD and finally use USD to buy CHF back. This wil result in (JPY / CHF) * (USD / JPY) * (CHF / USD) = 83 * 1/92 * 1/.9352 = 0.9646855 (<1 No arbitrage profit)

Thanks, got it now. I did the triangle in two ways: one way the result (for initial position of 1USD) was > 1(sell USD for CHF, sell CHF for AUD and sell AUD for USD)- arbitrage profit exists and the other way the result < 1 (sell USD for AUD, sell AUD for CHF and sell CHF for USD)- arbitrage “loss”. Thanks!

thanks for this, I was having a similar issue.

One note - if you have 3 market rates given by a question, it’s impossible to say which currency is ‘cheap’ or ‘expensive,’ as each currency’s value is relative to to the other two. It’s not useful to calculate all the implied cross rates as their value is also relative. You can only say whether an arbitrage profit exists or doesn’t exist. Interestingly, the same % arbitrage profit can be earned by starting with any of the three currencies and using the rates currently offered on the market, as long as you go the same direction around the triangle with each. As a result, the practice questions will typically note the currency you should start with (and thus end with) to solve a triangular arbitrage problem.