homework and exercises - How do you find the uncertainty of a 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.

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 $w_i = 1/\sigma_i^2$