# How do you find the uncertainty of an weighted average?

The following is taken from a practice GRE question:

Two experimental techniques determine the mass of an object to be $11\pm 1\, \mathrm{kg}$ and $10\pm 2\, \mathrm{kg}$. These two measurements can be combined to give a weighted average. What is the uncertainty of the weighted average?

What's the correct procedure to find the uncertainty of the average?

I know what the correct answer is (because of the answer key), but I do not know how to obtain this answer.

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Forget that it is an average. Do you know how to combine errors for sums (and products)? –  genneth Sep 29 '11 at 1:22
I added homework tag because it is of that nature even though it is not literally homework –  genneth Sep 29 '11 at 1:23
Yes, but doing this, I arrive at $\sqrt{5}/2$. According to the practice test, this is incorrect. –  Jonathan Gleason Sep 29 '11 at 2:12
The people who make up these tests at ETS are usually incompetent, and it is often impossible to read their minds. You should just say what answer they got, so people can try to work backwards to extract whatever flawed reasoning they used. There is no other way to solve ETS problems other than learning the psychology of the testers. –  Ron Maimon Sep 29 '11 at 3:51
The correct error is as you state it. It could be a typing error in the key. I think human typing error can be up to 5%. This was studied during the early computer days when we were feeding the data to computers with punched cards. First we punched in the data with a punching machine, a large desk with a keyboard etc., and then the punched cards were taken to the checking machine and retyped/punched so that errors showed up and were corrected. That brought the error level to an acceptable number. –  anna v Sep 29 '11 at 4:07

## 1 Answer

I agree with @Ron Maimon that these ETS questions are problematic. But this is (i think) the reasoning they go with. Unlike @Mike's assumption you should not take the normal average, but as stated in the question the weighted average. A weighted average assigns to each measurement $x_i$ a weight $w_i$ and the average is then

$$\frac{\sum_iw_ix_i}{\sum_i w_i}$$

Now the question is what weights should one take? A reasonable ansatz is to weigh the measurements with better precision more than the ones with lower precision. There are a million ways to do this, but out of those one could give the following weights:

$$w_i = \frac{1}{(\Delta x_i)^2},$$ which corresponds to the inverse of the variance.

So plugging this in, we'll have

$$c = \frac{1\cdot a+\frac{1}{4}\cdot b}{1+\frac{1}{4}}= \frac{4a+b}{5}$$

Thus,

$$\Delta c = \sqrt{\left(\frac{\partial c}{\partial a}\Delta a\right)^2+\left(\frac{\partial c}{\partial b}\Delta b\right)^2}$$

$$\Delta c = \sqrt{\left(\frac{4}{5}1\right)^2+\left(\frac{1}{5}2\right)^2}=\sqrt{\frac{16}{25}+\frac{4}{25}}=\sqrt{\frac{20}{25}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt5}$$

which is the answer given in the answer key.

### Why $w_i=1/\sigma_i^2$

The truth is, that this choice is not completely arbitrary. It is the value for the mean that maximizes the likelihood (the Maximum Likelihood estimator).

$$P(\{x_i\})=\prod f(x_i|\mu,\sigma_i)=\prod\frac{1}{\sqrt{2\pi\sigma_i}}\exp\left(-\frac{1}{2}\frac{\left(x_i-\mu\right)^2}{\sigma_i^2}\right)$$. This expression maximizes, when the exponent is maximal, i.e. the first derivative wrt $\mu$ should vanish:

$$\frac{\partial}{\partial\mu}\sum_i\left(-\frac{1}{2}\frac{\left(x_i-\mu\right)^2}{\sigma_i^2}\right) = \sum_i\frac{\left(x_i-\mu\right)}{\sigma_i^2} = 0$$

Thus, $$\mu = \frac{\sum_i x_i/\sigma_i^2}{\sum_i 1/\sigma_i^2} = \frac{\sum_iw_ix_i}{\sum_i w_i}$$ with $w_i = 1/\sigma_i^2$

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