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in this paper : https://link.springer.com/content/pdf/10.1023/A:1022058209073.pdf at page 111, left part, we pass from the distribution p to price p. But I don't see why it's the case. I've searched, and it seems like we can see the LMSR as a n Arrow-Debreu security or a binary option. But I don't really see why it's the case. The transition in this paper is quite fast an unexplained. Thank you for any kind of help!

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    $\begingroup$ Yes I think they mean that the risk neutral price of an Arrow-Debreu security is the probability that the event occurs. This sets the expected return to zero. $\endgroup$ – fesman Dec 5 '20 at 19:27
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    $\begingroup$ Can you please add a quote of the paragraph? It is behind a paywall (at least for me). Of course, @fesman’s comment sounds like it’s the right answer... $\endgroup$ – Kermittfrog Dec 5 '20 at 19:53
  • $\begingroup$ guys @fesman I have uploaded a photo if someone want to see :) $\endgroup$ – wainwain Dec 6 '20 at 9:13
  • $\begingroup$ @Kermittfrog :) $\endgroup$ – wainwain Dec 6 '20 at 9:14
  • $\begingroup$ Thank you for your answers guys! $\endgroup$ – wainwain Dec 6 '20 at 9:22
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I didn't want to pay to download the paper, but the intuitive answer is that if the probability of event X is P, then a binary option on X (with a 0-vs-1 payoff) will have a fair price of FV = E(P).

The distribution around P will obviously differ in finite samples; but as sample size -> infinity, this will/should converge towards E(P). Anything else would represent a FV estimate that is biased with respect to P. In truth, FV does not care about the distribution around P. It cares only for the expected value of P = E(P).

This one is easy to overthink ;-) Formally tying everything back to Arrow-Debreu very often just serves to complicate the simple, giving the same answer. My answer is "wrong", if different segments of the distribution of P have different utility values. Then FV is an integral function of F(X,P,utility). Which differs from case to case with respect to one's subjective utility function in any case.

Assuming that the fair price of any distribution P is E(P) is a quite standard financial assumption, which is probably why your article didn't justify its assumptions in any detail.

hope this helps, DEM

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  • $\begingroup$ Thank you for complete answer! $\endgroup$ – wainwain Dec 6 '20 at 9:22

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