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Showing that the existence of risk neutral probabilities implies and is implied by no arbitrage

+1 vote
asked May 17, 2017 in BUS 4028F - Financial Economics by anonymous
"Part of the fundamental theorem of asset pricing says that a market model is arbitrage free iff risk neutral probabilities exist: prove this for the one period binomial model"

the way I went about this was to use the condition d<\(e^r\)<u for no arbitrage and that probabilities must lie in the interval [0,1], this worked except that I didn't get strict inequalities on the arbitrage condition
commented May 23, 2017 by Rowan (4,010 points)

Hi Anonymous.

A tutor has looked at your question, but was not sure of the necessity regarding strict bounds. I have asked Alex and will post an Answer once I have one for you.

1 Answer

+4 votes
answered May 26, 2017 by ABackwell (580 points)
selected May 30, 2017 by Richard van Gysen
Best answer

You seem to be happy with the bulk of the answer: if e^r is outside of the interval [u,d], there is an easy long-short arbitrage, and the risk-neutral probabilities don't exist (because they are outside of [0,1] and therefore aren't probabilities); and if e^r is strictly inside [u,d], there is no arbitrage and the probabilities exist. This all agrees with the fundamental theorem of asset pricing. So far so good.

There is a subtlety involved in the boundary cases of e^r=u or =d, which you would not have been able to spot until now. In these cases, we get q=1 or =0, which appear to be valid probabilities. However, we require the full definition of a risk-neutral probability measure, which I only give in slide set (4) (slide 20). Part of this definition is that the risk-neutral measure must be equivalent to the original, real-world measure. If q=1 or =0, the measure Q does not agree with P about what has probability and what does not, and is therefore not equivalent. In these cases, therefore, the risk-neutral measure does not exist, strictly speaking. Furthermore, the long-short arbitrages are available in these cases (make sure you can see why), so the fundamental theorem holds again.

(This is all assuming that p is strictly inside [0,1], and not 0 or 1 exactly).