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Suppose you go through every country's currencies and look at the change WRT all other currencies that day. Assuming there is a change in value that day, does it stand to reason that there will always be a certain currency that goes up WRT all others and one that goes down WRT to all others everyday?

Derek Seabrooke
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This is true and the reason is that currency exchange rates are transitive. That means if you know the exchange rate between A and B, and the rate between B and C, the you can compute the exchange rate between A and C (we are assuming no conversion fees as is the case for published conversion rates).

This transitivity also implies that if for example A increased relative to B and B increased relative to C than A must have increased relative to C as well. Once you know that it is straightforward that there every day there must be one currency that increased relative to all others (assuming all exchange rates changed).

quarague
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    Except that transitivity is enforced by (threat of) arbitrage, not structurally... – Ian Sep 26 '22 at 13:09
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    You are assuming a no-arbitrage assumption that was not stated (and doesn't have to hold, even in the real world). I agree that the OP probably meant to assume it, but the issue was raised in the comments and the OP declined to add it to the problem. – lulu Sep 26 '22 at 13:09
  • Great answer! Would this not imply unlimited earning potential as one could simply look for this currency ever day and move all of your money into it? – Derek Seabrooke Sep 26 '22 at 13:09
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    @DerekSeabrooke If you knew the future, then yes. If you don't know the future, making that trade after the change doesn't help you, because the stronger currency now also costs more to buy. – Ian Sep 26 '22 at 13:10
  • How would arbitrage change the answer? – Derek Seabrooke Sep 26 '22 at 13:11
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    If arbitrage is possible then things don't have to behave consistently, so the prices can just do whatever, they don't have to be related to one another at all. In effect you have ${n \choose 2}$ prices varying independently (one for each pair). By contrast in the no-arbitrage system you effectively have $n$ independent "true values" and the ${n \choose 2}$ prices are determined by just taking ratios of the "true values". (I'm not an expert on this domain, so I'm just making up terminology willy-nilly.) – Ian Sep 26 '22 at 17:56