What happens if there is no difference between the mean and the weighted mean?

Question

I was wondering. What happens if there is no difference between the arithmetic mean and the weighted mean? Why would that happen?

Answer 1

Let's look at a simple case, where you only have 2 values $x_1$ and $x_2$ with weights $w_1$ and $w_2$. So the unweighted mean is $\frac{1}{2}(x_1 + x_2)$, and the weighted mean is $\frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2)$.

So why would these be equal? Well, there are a lot of values involved, so it could just be a lucky coincidence - in fact, if you fix any three of the values, there is (usually) a value for the fourth that will make everything equal.

However, there are a few special cases that you can see:

1. The weights are equal. If $w_1 = w_2$ then the weighted mean is $\frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = \frac{1}{w_1 + w_1}(w_1 x_1 + w_1 x_2) = \frac{1}{2w_1} \times w_1 (x_1 + x_2) = \frac{1}{2}(x_1 + x_2)$, which is the unweighted mean.

2. The sample values are equal. If $x_1 = x_2$ then the weighted mean is $\frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = \frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_1) = \frac{w_1 + w_2}{w_1 + w_2} x_1 = x_1$, and it's pretty easy to show that the unweighted mean has the same value.

You can show that these two cases are true for any number of values, but I'll leave the working to you.