Weighted average using numpy.average

Question

I have an array:

In [37]: bias_2e13 # our array
Out[37]: 
[1.7277990734072355,
 1.9718263893212737,
 2.469657573252167,
 2.869022991373125,
 3.314720313010104,
 4.232269039271717]

The error on each value in the array is:

In [38]: bias_error_2e13 # the error on each value
Out[38]: 
array([ 0.13271387,  0.06842465,  0.06937965,  0.23886647,  0.30458249,
        0.57906816])

Now I divide the error on each value by 2:

In [39]: error_half # error divided by 2
Out[39]: 
array([ 0.06635694,  0.03421232,  0.03468982,  0.11943323,  0.15229124,
        0.28953408])

Now I calculate the average of the array using numpy.average, but using the errors as weights.

First I am using the full error on the values, then I am using half the error, i.e. the error divided by 2.

In [40]: test = np.average(bias_2e13,weights=bias_error_2e13)

In [41]: test_2 = np.average(bias_2e13,weights=error_half)

How do both the averages give me the same result when one array has errors which are half of that of the other?

In [42]: test
Out[42]: 3.3604746813456936

In [43]: test_2
Out[43]: 3.3604746813456936

Answer 1

Because all of the errors have the same relative weight. Supplying a weight parameter does not change the actual values you are averaging, it just indicates the weight with which each value value contributes to the average. In other words, after multiplying each value passed by its corresponding weight, np.average divides by the sum of the weights provided.

>>> import numpy as np
>>> np.average([1, 2, 3], weights=[0.2, 0.2, 0.2])
2.0
>>> np.average([1, 2, 3])
2.0

Effectively, the average formula for an n-dimensional array-like container is

                               

where each weight is assumed to be equal to 1 when not provided to numpy.average.