# Is a weighted average unambiguously superior to an unweighted average?

I want to describe a statistic on my data. For example, I have data on firms (companies), and I have a measure of $x$ for each of them. Now, firms differ in size, which could be understood in terms of number of employees, or value of sales, or other metric.

Say I want to compute $\bar x$. A simple arithmetic average will treat small and large firms identically. However, a weighted average (using a measure of relative size as weight) will give more weight to the large firm and less weight to the small firm. In a sense, a weighted average divides large firms into smaller units, making them all of equal size, and assigning an identical $x$ to them.

Now, because the unweighted mean is contained in the weighted one (in the sense that, if firms have identical size, the two are the same), it might seem logical to assume that the weighted average is unambiguously superior to the unweighted one. But is this the case?

I think a counterexample would suffice here. Is there any sense in which we are better off knowing the unweighted average of $x$ rather than a weighted one?

PS: surely, one might say "compute both". But say you are conducting a study were if you use both definitions, you are essentially duplicating the number of tables, graphs, and their analysis. In this case, it might be better just to focus on one definition of $\bar x$.

## 1 Answer

It depends on what exactly you are estimating. The sample average and the sample weighted average are estimators of some population quantity, and which one is "superior" depends on what quantity that is.

For example, if you are estimating the population mean of workers' salary and $x_i$ is the average salary of workers in a firm $i$, weighting the observations by number of workers $n_i$ in a given firm $i$ will give a more precise (lower variance) estimate than weighting them equally. In the latter case, workers from large companies would be underrepresented while those from small companies overrepresented. If in addition workers at large companies are paid more than those in small companies, equal weighting will produce a negative bias (the estimate will tend to be lower than the true value).

But if you are estimating the population mean of CEOs' salary and $x_i$ is the salary of the CEO of firm $i$, then unweighted average will be more accurate (have lower variance). By using weights proportional to firm size or the like you would inflating the influence of large firms and probably overestimating the population mean (if large companies' CEOs get higher salaries).

To formalize, if the sample elements are drawn with equal probabilities from the population (are equally representative), then unweighted average will have the lowest variance among unbiased estimators of the population mean and the lowest mean squared error. If the sample elements are drawn with unequal probabilities (are not equally representative), then weighting by the inverse probability (rather than weighting equally) will be optimal.

• Thanks! In which sense a worker from a smaller firm has different probability of being drawn than one in a larger firm? Assume firms' probability of selection is independent of size. Also, what if my data is a census of firms? (and thus probability of selection is 1 for all). – luchonacho Mar 24 '18 at 13:32
• If $x_i$ is the average salary for the worker in firm $i$, then $x_i$ represents many workers for a large firm but few workers for a small firm. Then it should be weighted proportionally to how many workers there are in the firm. If it is weighted equally instead, the salary of a worker from a 100-worker firm contributes 5 times less than the salary from a 20-worker firm, so the workers from a large firm are underrepresented. If the data is a census, this still applies. (Simple example: 2 firms with 2 and 1 workers earning {3, 3} and 2. $x_1=3$, $x_2=2$, unw.avg. 2.5, w.avg. 2.67, truth 2.67.) – Richard Hardy Mar 24 '18 at 13:37