Tag: weighted mean

Often we want to compare the variability of a variable in different contexts – say, the variability of unemployment in different countries over time, or the variability of height in two populations, etc. The most often used measures of variability are the variance and the standard deviation (which is just the square root of the variance). However, for some types of data, these measures are not entirely appropriate. For example, when data is generated by a Poisson process (e.g. when you have counts of rare events) the mean equals the variance by definition. Clearly, comparing the variability of two Poisson distributions using the variance or the standard deviation would not work if the means of these populations differ. A common and easy fix is to use the coefficient of variation instead, which is simply the standard deviation divided by the mean. So far, so good. Things get tricky however when we want to calculate the weighted coefficient of variation. The weighted mean is just the mean but some data points contribute more than others. For example the mean of 0.4 and 0.8 is 0.6. If we assign the weights 0.9 to the first observation [0.4] and 0.1 to the second [0.8], the weighted mean is (0.9*0.4+0.1*0.8)/1, which equals to 0.44. You would guess that we can compute the weighted variance by analogy,  and you would be wrong. For example, the sample variance of {0.4,0.8} is given by [Wikipedia]: or in our example ((0.4-0.6)^2+(0.8-0.6)^2) / (2-1) which equals to 0.02. But, the weighted sample variance cannot be computed by…