I think this works perfectly if your list has 2^n elements. Otherwise, you have to resort to multiplying by imprecise fractions.
\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)
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…
I think this works perfectly if your list has 2^n elements. Otherwise, you have to resort to multiplying by imprecise fractions.
\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)
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…