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PU-106 / L09 · lecture

Combining Measurements: The Weighted Mean

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable calculating the variance of a linear combination of independent random variables and understand why the sample variance uses \(N-1\) in the denominator (Bessel's correction).

Why this matters

When multiple experiments measure the same physical constant, simply averaging their results ignores the fact that some measurements are inherently more precise than others.

In experimental physics, we rarely have just one measurement. We often have a set of independent estimates for a single true value, each with its own uncertainty. If one measurement has a standard deviation of 0.1 and another has a standard deviation of 10, the first is clearly more reliable. A simple arithmetic mean treats them as equal, which is statistically inefficient. The weighted mean provides the optimal way to combine these data points, yielding the most precise estimate possible from the available information. This technique is foundational for combining results from different laboratories, different instruments, or different time periods in high-precision physics.

The Intuition Behind Weighting

Imagine you are trying to determine the exact length of a table. You have two friends helping you. Friend A uses a high-quality laser rangefinder and reports a length with a very small uncertainty. Friend B uses a worn-out tape measure and reports a length with a large uncertainty. If you simply add their two numbers and divide by two, you are giving Friend B's noisy result the same influence as Friend A's precise result. This is illogical. We want our final answer to be "pulled" more strongly by the precise measurement and less by the noisy one. The "weight" assigned to each measurement should reflect its reliability. In statistics, reliability is inversely related to variance: the smaller the variance, the more reliable the measurement. Therefore, we assign higher weights to measurements with smaller variances and lower weights to those with larger variances. The goal is to find the specific set of weights that minimizes the total uncertainty of the final combined result.

Mathematical Formulation

Let \(x_1, x_2, \dots, x_N\) be \(N\) independent, unbiased measurements of a true quantity \(\mu\). Each measurement \(x_i\) has a known variance \(\sigma_i^2\). We seek a linear combination of these measurements:

\[ \hat{x} = \sum_{i=1}^N w_i x_i \]

For \(\hat{x}\) to be an unbiased estimator of \(\mu\), the weights must sum to one:

\[ \sum_{i=1}^N w_i = 1 \]

The variance of this linear combination, assuming independence, is:

\[ \mathrm{Var}(\hat{x}) = \sum_{i=1}^N w_i^2 \sigma_i^2 \]

Our objective is to minimize this variance subject to the constraint that the weights sum to unity. This is a classic constrained optimization problem. The solution, derived rigorously in the cited derivation card, shows that the optimal weights are proportional to the inverse of the variances:

\[ w_i = \frac{1/\sigma_i^2}{\sum_{j=1}^N 1/\sigma_j^2} \]

Substituting these weights back into the linear combination gives the weighted mean:

\[ \hat{x} = \frac{\sum_{i=1}^N \frac{x_i}{\sigma_i^2}}{\sum_{i=1}^N \frac{1}{\sigma_i^2}} \]

The variance of this optimal estimator is given by the reciprocal of the sum of the inverse variances:

\[ \frac{1}{\sigma_{\hat{x}}^2} = \sum_{i=1}^N \frac{1}{\sigma_i^2} \]

This result is powerful because it shows that the precision of the combined measurement is the sum of the precisions of the individual measurements. Precision, defined as \(1/\sigma^2\), is additive for independent measurements.

Interpreting the Result

The formula for the weighted mean looks complex, but its structure is intuitive. The numerator is the sum of each measurement divided by its variance. The denominator is the sum of the inverse variances. If all variances are equal, say \(\sigma_i^2 = \sigma^2\), then the term \(1/\sigma^2\) cancels out from the numerator and denominator, and the weighted mean reduces to the simple arithmetic mean \(\frac{1}{N}\sum x_i\). This is a crucial consistency check.

Consider the limit where one measurement is extremely precise (\(\sigma_k^2 \to 0\)). The term \(1/\sigma_k^2\) becomes very large, dominating both the numerator and the denominator. The weighted mean \(\hat{x}\) approaches \(x_k\), and the combined variance \(\sigma_{\hat{x}}^2\) approaches \(\sigma_k^2\). This makes sense: if one measurement is infinitely precise, it determines the value, and the other noisy measurements become irrelevant. Conversely, if one measurement is extremely noisy (\(\sigma_k^2 \to \infty\)), its weight \(w_k\) approaches zero, and it contributes nothing to the final result.

The combined variance \(\sigma_{\hat{x}}^2\) is always smaller than the smallest individual variance \(\min(\sigma_i^2)\). This is because we are combining information. Even if one measurement is the best, the others still provide some information about \(\mu\), thereby reducing the uncertainty further. The more independent measurements we have, the smaller the combined variance becomes, approaching zero as \(N \to \infty\) (assuming the variances are bounded).

Practical Considerations

In practice, the variances \(\sigma_i^2\) are often not known exactly but are estimated from the data. If the variances are estimated from the same data used to calculate the mean, the resulting uncertainty in the weighted mean must account for the uncertainty in the weights themselves. However, for the standard treatment in this unit, we assume the variances are known a priori or estimated from independent sources (such as calibration data or repeated trials).

It is also important to check for consistency. If the measurements are truly independent and unbiased, the weighted mean should be close to the individual measurements, within their respective uncertainties. If one measurement deviates significantly from the weighted mean, it may indicate a systematic error or an outlier. In such cases, one might consider using a robust estimator or investigating the source of the discrepancy rather than blindly applying the weighted mean formula.

Another practical aspect is the reporting of the result. The final value should be reported as \(\hat{x} \pm \sigma_{\hat{x}}\), where \(\sigma_{\hat{x}}\) is the square root of the combined variance. This provides a complete description of the best estimate and its uncertainty.

The derivation

◆ DERIVATION EMBED · D-114 · not re-derived here

The Inverse-Variance Weighted Mean

Given N independent, unbiased measurements xi of a single quantity μ, each with variance σi2, the linear combination x̂ = Σi wi xi that is unbiased and has the smallest possible variance is the one whose weights satisfy wi ∝ 1/σi2. The resulting minimum variance is 1/σ2 = Σi 1/σi2.

wi = (1/σi2) / Σj(1/σj2)   ⇒   x̂ = Σi(xi/σi2) / Σi(1/σi2),    1/σ2 = Σi 1/σi2
Open full derivation →
Tier 3 ยท watch out

A common mistake is to assume that the weighted mean is always more precise than the best single measurement. While the combined variance is smaller than the smallest individual variance, this assumes that all measurements are unbiased and independent. If there are correlated errors or systematic biases, the weighted mean can be misleading. Always verify the independence and unbiasedness assumptions before applying this method. C

Worked example

Problem. Three independent measurements of the speed of light in a vacuum are obtained: \(x_1 = 2.9979 \times 10^8\ \mathrm{m/s}\) with \(\sigma_1 = 1000\ \mathrm{m/s}\), \(x_2 = 2.9981 \times 10^8\ \mathrm{m/s}\) with \(\sigma_2 = 2000\ \mathrm{m/s}\), and \(x_3 = 2.9978 \times 10^8\ \mathrm{m/s}\) with \(\sigma_3 = 500\ \mathrm{m/s}\). Calculate the weighted mean and its standard deviation.

  1. Calculate the inverse variances (precisions): \(1/\sigma_1^2 = 1/(1000)^2 = 10^{-6}\ \mathrm{s^2/m^2}\) \(1/\sigma_2^2 = 1/(2000)^2 = 2.5 \times 10^{-7}\ \mathrm{s^2/m^2}\) \(1/\sigma_3^2 = 1/(500)^2 = 4 \times 10^{-6}\ \mathrm{s^2/m^2}\)
  2. Sum the precisions: \(\sum 1/\sigma_i^2 = 10^{-6} + 0.25 \times 10^{-6} + 4 \times 10^{-6} = 5.25 \times 10^{-6}\ \mathrm{s^2/m^2}\)
  3. Calculate the weighted mean numerator: \(\sum x_i/\sigma_i^2 = (2.9979 \times 10^8)(10^{-6}) + (2.9981 \times 10^8)(2.5 \times 10^{-7}) + (2.9978 \times 10^8)(4 \times 10^{-6})\) \(= 299.79 + 74.9525 + 1199.12 = 1573.8625\ \mathrm{m/s}\)
  4. Calculate the weighted mean: \(\hat{x} = 1573.8625 / (5.25 \times 10^{-6}) = 2.997833 \times 10^8\ \mathrm{m/s}\)
  5. Calculate the combined variance: \(\sigma_{\hat{x}}^2 = 1 / (5.25 \times 10^{-6}) = 190476.19\ \mathrm{m^2/s^2}\)
  6. Calculate the standard deviation: \(\sigma_{\hat{x}} = \sqrt{190476.19} \approx 436.4\ \mathrm{m/s}\)
\[ \hat{x} = (2.997833 \pm 0.000004) \times 10^8\ \mathrm{m/s} \]

Units check. The numerator has units of \(\mathrm{m/s} \cdot \mathrm{s^2/m^2} = \mathrm{s/m}\). The denominator has units of \(\mathrm{s^2/m^2}\). The ratio is \(\mathrm{s/m} / (\mathrm{s^2/m^2}) = \mathrm{m/s}\). The variance has units of \(\mathrm{m^2/s^2}\), and the standard deviation has units of \(\mathrm{m/s}\). The units are consistent.

CHECK YOURSELF

Three quick tests

  1. If all measurements have the same variance, what does the weighted mean reduce to?
    AnswerThe simple arithmetic mean.
  2. Why is the combined variance always smaller than the smallest individual variance?
    AnswerBecause combining independent measurements adds precision, and precision is additive.
  3. What happens to the weight of a measurement if its variance becomes very large?
    AnswerThe weight approaches zero, meaning the measurement contributes negligibly to the final result.
PROBLEMS

Problem set for PU-106 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-114 · Threads: chance