Predicting a range is clearly useful, but having the value fall into that range or hit the mean value of the range does not always mean the prediction is better.
Example:
Prediction 1: 90-100; Prediction 2: 50-150. Actual: 101
The first is clearly the better prediction, although the actual is outside the range and further from the mean than the second.
What you really want to do is 1) reward a predicted mean value (i.e. max+min/2) that is close to the actual; 2) reward a smaller range preferentially to a larger range.
Example:
Prediction 1: 90-100; Prediction 2: 50-150. Actual: 101
The first is clearly the better prediction, although the actual is outside the range and further from the mean than the second.
What you really want to do is 1) reward a predicted mean value (i.e. max+min/2) that is close to the actual; 2) reward a smaller range preferentially to a larger range.
Actually, it is not clear that the first prediction is better. It depends. Also, to the point 1), consider that the probability density function can be asymmetric within the range, i.e. the mean can be different from 0.5*(max+min).
No. D is the deviation from the interval edge and is added to the weighted interval width. I think you misread it.
Sorry, I misread your AND as an OR in the conditional and I didn't look too closely at the inequalities.In that case, this proposal would give the same score to two forecasts with the same I, regardless of where within the interval the actual outcome falls. For example: a prediction of 90-110 would be ranked equal to a prediction of 99-119 with an outcome of 100. I don't find that intuitive either.
Ok in that case people will have to guess distributions rather than intervals, then just compare the probability density at the price. If people want to do intervals, then they can use uniform distributions.
