New answers tagged binomial-tree
0
"Cannot figure out where this reasoning fails. The question is: am I misunderstanding the real meaning of arbitrage-free pricing or the only reason for which this it seems to be strange to me is that I miss some economical / "real markets world related" point of view?"
In opposite to the first comment, I think you do miss something a bit subtle here.
First ...
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Yes it is equivalent of the dividend rate. The b in the function is cost of carry, so here it would be:
$b=r_{USD}-r_{AUD}$
And r in the function is $r_{USD}$.
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What you say is perfectly true and there is no contradiction. Arbitrage means risk free profit , so your ‘statistical arbitrage’ is not arbitrage at all. It just says that if you take risk, your expected returns can be higher than the risk free rate. How much higher depends in the risk aversion of market participants.
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