mt1_fall_2008_q3_sol - 2. (a) Denote the USD-denominated...

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2. (a) Denote the USD-denominated value of the basket as V and the number of EUR in the basket, per unit of USD, as n . Then, V = 1 + n S usd/eur , and what it means for the basket to be 40% USD and 60% EUR is 0 . 4 = 1 V 0 . 6 = n S usd/eur V Solving these equations for n gives n = 1. This just says that, since EUR is 3/2 times more valuable than USD, then a basket of 1 USD and 1 EUR is worth 2.5 dollars and 40% of this value comes from the USD 1.0 and 60% from EUR 1.0, which is worth USD 1.5. (b) The KWD-denominated value of the basket is K = 1 . 0 × S kwd/usd + n S kwd/eur . (1) No-arbitrage means that the following must hold: S kwd/eur = S usd/eur × S kwd/usd , (2) so equation (1) can be written K = S kwd/usd ( 1 + n S usd/eur ) . (3) Given n = 1, S usd/eur = 1 . 5, and K = 10, this means that S kwd/usd = 4 . 0 If either of equation (2) or equation (3) do not hold, then there is an arbi- trage opportunity. In the former case the arbitrage is the usual triangular version. In the latter case the arbitrage would involve buying or selling
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mt1_fall_2008_q3_sol - 2. (a) Denote the USD-denominated...

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