# 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.

• 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
• This is a rather specific question, sort of on the verge of being closeable as "too localized," but I edited it a little to try and keep it general enough to stay open. The thing is, because the answer key is wrong, there's not much to say in an answer except explaining the general procedure for combining uncertainties, and I'm pretty sure that's already covered by other questions here and on many other sites. – David Z Sep 29 '11 at 4:56

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 $w_i = 1/\sigma_i^2$

• luksen explained how to find a weighted average, but I did not see a direct answer to the question. If the weighted average is calculated by the precision of each measurement do we still find an error? Or because we used error to find the weighted average, it is included in the result? – Tsangares Jan 25 '17 at 1:20
• So what is the answer to the question? – ProfRob Apr 27 '17 at 18:37

It would be highly unreasonable to obtain the uncertainty of the average of the measurements to be greater than the uncertainty of any one measurement (√5/2 > 1). After all, what's the point in taking averages if it just makes your readings more uncertain?

I believe the ETS people used the argument that the harmonic sum of the individual variances should give the reciprocal of the average's variance i.e. 1/v = 1/v1 + 1/v2, as detailed here: http://en.wikipedia.org/wiki/Weighted_arithmetic_mean#Dealing_with_variance

Since other questions as well as Google point to this question, I would like to extend the already existing answer from luksen by a motivation from the stochastic for using the error propagation equation.

Let's assume we have $n$ random measurements $X_i$, typically denoted by $E\{X_i\} \pm \sigma_i$ (or $x_i \pm \Delta x_i$), whereas $E\{\cdot\}$ and $\sigma_i$ denote the expected value and the standard deviation, respectively. According to the question, we are interested in the weighted average of these measurements, calculated by $Y = \frac{\sum_i w_i X_i}{\sum_i w_i}$ ($=c$). Thanks to the linearity of the expected value, it is rather easy to get $$E\{Y\} = E\left\{\frac{\sum_i w_i X_i}{\sum_i w_i} \right\} = \frac{\sum_i w_i E\{ X_i\}}{\sum_i w_i}.$$ For the variance $\sigma^2$, we need a few more lines but no hacky tricks. \begin{align} \sigma^2 &= E\left\{(Y-E\{Y\})^2\right\} = E\left\{\left(\frac{\sum_i w_i X_i}{\sum_i w_i} - E\left\{\frac{\sum_i w_i X_i}{\sum_i w_i} \right\} \right)^2\right\} \\ &= \frac{1}{(\sum_i w_i)^2} E\left\{ \left(\sum_i w_i X_i\right)^2 - 2 \left(\sum_i w_i X_i\right) \left(\sum_j w_j E\{X_j\}\right) + \left( \sum_i w_i E\{X_i\} \right)^2 \right\} \\ &= \frac{1}{(\sum_i w_i)^2} E\left\{ \sum_{i,j} w_i X_i w_j X_j - 2 \sum_{i,j} w_i X_i w_j E\{X_j\} + \sum_{i,j} w_i E\{X_i\} w_j E\{X_j\} \right\} \\ &= \frac{1}{(\sum_i w_i)^2} \sum_{i,j} w_i w_j \left( E\{X_i X_j\} - 2 \cdot E\{X_i\} E\{X_j\} + E\{X_i\} E\{X_j\} \right) \\ &= \frac{1}{(\sum_i w_i)^2} \sum_{i,j} w_i w_j \left( E\{X_i X_j\} - E\{X_i\} E\{X_j\} \right) \\ &= \frac{1}{(\sum_i w_i)^2} \sum_{i,j} w_i w_j \cdot Cov(X_i, X_j) \end{align} The covariance holds $Cov(X_i, X_i) = \sigma_i^2$ and, as long as the original measurements $X_i$ and $X_j$ are independent, $Cov(X_i, X_j) = 0$ otherwise. This leads to the answer of the original question, the variance (squared standard deviation) of the average $Y$ of $n$ measurements $X_i$ with random uncertainty $\sigma_i$: $$\sigma^2 = \frac{1}{(\sum_i w_i)^2} \sum_i w_i^2 \sigma_i^2,$$ which yields $$\Delta c = \sigma = \frac{1}{\sum_i w_i} \sqrt{\sum_i (w_i \sigma_i)^2}.$$ That's exactly what the error propagation equation yields after inserting the derivations of $c$. Why is it exact while the error propagation equation is an approximation? The error propagation equation approximates by a first-order Tayler expansion. Since the average is a linear function, it is exact and not only approximated, here.

Additional information: For the unweighted average ($w_i = 1~ \forall i$), we get $$E\{Y\} = \frac{1}{n} \sum_i E\{ X_i\} \quad\quad\quad \sigma^2 = \frac{1}{n^2} \sum_i \sigma_i^2$$ If all original samples $X_i$ have the same variance $\tilde\sigma^2$, this leads also to the well know variance of the average: $$\sigma^2 = \frac{\tilde\sigma^2}{n}.$$