Posted by econ on Wednesday, May 28, 2008 at 7:49pm.
True of False?
E=exchange rate
Suppose E $/euros = 3/2 and E $/pounds = 2, and there is no trade friction, the no arbitrage condition implies
that E pounds/euros = 2

econ  economyst, Thursday, May 29, 2008 at 9:29am
False. Think it through.
You have $/euros = 3/2 which means $3 = 2E. Also, you have 2E=1L. Ergo, $3=1L.

econ  econ, Thursday, May 29, 2008 at 10:17am
so the arbitrage condition should be E pounds/euros=3/4. ( i got this by multiplying 1/2 and 3/2 )
Is this correct?
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