umdiff

5 ms·

1. 1.▲
   by umdiff 7y ago

   I think this works perfectly if your list has 2^n elements. Otherwise, you have to resort to multiplying by imprecise fractions.
2. 2.▲
   by umdiff 7y ago

   \left( \sum_{i=1}^m x_i/m + \sum_{i=m+1}^{2m} x_i/m \right) / 2 = \sum_{i=1}^{2m} x_i /(2m)
3. 3.▲
   by umdiff 8y ago

   I'd add that sampling is still very useful in contexts when the partition function is known or efficient to calculate. What's generally interesting is the general character of the posterior -- its modes, the shape of variance arou