# what is the exchange rate Expectation? I am confused about it

Suppose now the rate of EUR/USD is 1. and in half-year, maybe become 1.25EUR/USD, or 0.8 EUR/USD, the probability of each case is 0.5. what's the Expectation of EUR/USD in half-year? And, what's the Expectation of USD/EUR in half-year?

E(EUR/USD) = 0.5*1.25 + 0.5*0.8 = 1.025 EUR/USD。


answer 2: since 1.25EUR/USD is equivalent to 0.8 USD/EUR , and 0.8EUR/USD is equivalent to 1.25 USD/EUR, so

E(USD/EUR) = 0.5*1.25 + 0.5*0.8 = 1.025 USD/EUR。


Both USD and EUR ate expected to be appreciated. It's confusing me.

• In general, forgetting FX, do we expect E(1/X) to equal 1/E(X)? For example, if X can be 1 or 2 with the same probability 1/2, then E(1/X) = (1+1/2)/2 = 3/4 ; But E(X) = (1 + 2)/2 = 3/2 and 1/E(X) = 2/3. – Dimitri Vulis Nov 18 '20 at 5:20
• Yes concavity leads to some counterintuitive results for FX rates, see also here: en.wikipedia.org/wiki/…. Your result vanishes when using log-exchange rates though. – fesman Nov 18 '20 at 8:11
• If random variable X is strictly positive non-constant, and E(X) is non-zero, the function f(x)=1/x is strictly convex, by Jensen's inequality 1/E(X) is strictly less than E(1/X). – Dimitri Vulis Nov 18 '20 at 11:52
• Thanks I meant convexity. Why many find these cases puzzling is that here it looks like both Americans and Europeans could gain by swapping currencies (assuming risk neutrality). At least on the surface it seems like they are creating value from nothing. – fesman Nov 18 '20 at 13:00
• Hi: This sounds similar to a game I was told once. You pick an envelope and contains a dollar. The choice you have is to A) keep the dollar and stop or B) pick one of 2 envelopes. One mvelope contains 2 dollars and the other contains 50 cents. It turns out that you should always switch because you're expected value is greater than 1.0. But if you change the problem so that log(1) is in the envelope and you get log(1/2) or log(2), then its a fair game. – mark leeds Nov 19 '20 at 2:11