
Will the exponential mean of this market's probability be greater than ln(e-1)?
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270Ṁ5524resolved May 2
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After the market closes, I will calculate the time-weighted exponential mean of its probability (using the exact probabilities, not the rounded ones in the display). Here, "exponential mean" means the natural log of the time-weighted average of e^P, where P is the market's probability. In other words, it is the generalized f-mean with f = exp.
I will resolve it to YES if the final mean is greater than ln(e-1) ≈ 0.541, which is the exponential mean of all values between 0 and 1 and therefore theoretically the fairest cutoff point.
This question is managed and resolved by Manifold.
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