Error propagation
Weighted mean of three observations of differing precision
Three crews, three standard errors, one distance. Weights as the inverse of the variance, the weighted mean, the standard deviation of unit weight, and what it means when that comes out well above one.
Given
- The same distance was measured by three parties using different equipment and different procedures. Each reported a value and its own standard error.
- Party 1: 1842.36 ft, σ₁ = ±0.03 ft
- Party 2: 1842.29 ft, σ₂ = ±0.05 ft
- Party 3: 1842.41 ft, σ₃ = ±0.02 ft
- The three determinations are independent. All were reduced to the same horizontal plane and the same units, US survey feet.
- No party's result is to be discarded; the task is to combine them.
Required
- The weight of each observation and the weighted mean.
- The standard deviation of unit weight, σ₀.
- The standard deviation of the weighted mean, and a judgement on whether the reported standard errors are credible.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Weights are the inverse of the variance, not of the standard deviation
A weight expresses how much an observation should count, and it is inversely proportional to the variance: w = 1/σ². The squaring is the part that surprises people. Party 3, with half the standard error of Party 2, counts not twice as much but four times as much — and in fact 6.25 times, since (0.05/0.02)² = 6.25.
The reason is that variances, not standard deviations, are what add when independent quantities are combined. Any weighting that was linear in σ would fail to minimise the variance of the result, which is the entire purpose of weighting.
The absolute size of the weights does not matter — only their ratios do — so it is common to scale them to convenient numbers. Here they are left as computed.
| Party | Value v (ft) | σ (ft) | Weight w = 1/σ² | w · v |
|---|---|---|---|---|
| 1 | 1842.36 | 0.03 | 1111.111 | 2,047,066.667 |
| 2 | 1842.29 | 0.05 | 400.000 | 736,916.000 |
| 3 | 1842.41 | 0.02 | 2500.000 | 4,606,025.000 |
| Σ | 4011.111 | 7,390,007.667 |
The weighted mean
The weighted mean is the sum of the weighted values divided by the sum of the weights. It comes to 1842.384 ft.
Sanity-check its position before going further. The unweighted mean of the three values is 1842.353 ft, and the weighted mean sits 0.031 ft above it, pulled toward Party 3's 1842.41 — which is what should happen when the most precise observation is the highest one. A weighted mean that fell outside the range of the observations, or that moved away from the best observation, would mean a weight was inverted.
weighted mean = Σ(w·v) / Σw = 7,390,007.667 / 4011.111 = 1842.38418 ft
unweighted mean for comparison = (1842.36 + 1842.29 + 1842.41) / 3 = 1842.35333 ft
the weighted mean lies 0.031 ft above the unweighted, toward the best observation ✓Residuals and the standard deviation of unit weight
Take residuals about the weighted mean, then form the weighted sum of their squares. Dividing by the degrees of freedom — three observations of one unknown, so two — and taking the square root gives σ₀, the standard deviation of an observation of unit weight.
σ₀ is the diagnostic. It answers the question 'were the reported standard errors telling the truth?'. If the three parties' claims about their own precision were correct, σ₀ would come out near 1.0, because the weights were built from those very claims.
It comes out at 1.71. The three determinations disagree with one another about 1.7 times more than their own claimed standard errors say they should. Something in the three procedures is not accounted for — most likely a systematic effect that differs between the parties, such as a scale or an atmospheric reduction.
| Party | Value (ft) | Residual v (ft) | Weight w | w · v² (ft²) |
|---|---|---|---|---|
| 1 | 1842.36 | −0.02418 | 1111.111 | 0.649788 |
| 2 | 1842.29 | −0.09418 | 400.000 | 3.548162 |
| 3 | 1842.41 | +0.02582 | 2500.000 | 1.666316 |
| Σ | 4011.111 | 5.864266 |
The standard deviation of the weighted mean
There are two ways to state the precision of the weighted mean and they give different answers, which is the point of this step.
The a priori route uses only the claimed weights: σ = 1/√Σw = 0.016 ft. It is what the three parties' own claims imply, and it takes no account of whether the three results actually agree.
The a posteriori route scales that by σ₀, giving σ₀/√Σw = 0.027 ft. It is what the observed scatter says. Because σ₀ is 1.71, the a posteriori figure is 1.71 times the a priori one, and it is the honest number to publish: the three parties disagree by more than they claimed to, so their combination is less well determined than their claims imply.
Quote 1842.384 ± 0.027 ft. Quoting ±0.016 ft would be asserting a precision that the observations themselves contradict.
σ₀ = √( Σw·v² / (n − 1) ) = √( 5.864266 / 2 ) = √2.932133 = 1.712
a priori σ of the mean = 1 / √Σw = 1 / √4011.111 = 1 / 63.333 = 0.0158 ft
a posteriori σ of the mean = σ₀ / √Σw = 1.712 / 63.333 = 0.0270 ft
ratio = σ₀ = 1.71 — the observations scatter 71 % more than claimedWhat weighting actually bought
Compare the weighted result with what a simple average would have given. The unweighted mean is 1842.353 ft with a standard error of the mean of 0.035 ft, computed from the scatter of the three values in the ordinary way.
The weighted mean is 1842.384 ± 0.027 ft. Weighting moved the value by 0.031 ft — nearly one standard error — and improved the precision by 23 per cent. Both of those are worth having, and neither is dramatic: with only three observations, weighting is a refinement, not a rescue.
The larger message is in σ₀. A σ₀ of 1.71 on three observations is not statistically damning — with two degrees of freedom the estimate is very rough — but it is a flag. If a fourth determination were available it should be sought before this value is used for anything the parcel depends on.
| Statistic | Weighted | Unweighted |
|---|---|---|
| Mean (ft) | 1842.384 | 1842.353 |
| σ of the mean (ft) | 0.027 | 0.035 |
| Basis | w = 1/σ², σ₀ = 1.71 applied | equal weight, σ = 0.060 over √3 |
Answer
- Weights: w₁ = 1111.11, w₂ = 400.00, w₃ = 2500.00, summing to 4011.11. Party 3 counts 6.25 times Party 2, because weight goes as the inverse square of the standard error.
- Weighted mean = 7,390,007.667 / 4011.111 = 1842.384 ft.
- Standard deviation of unit weight σ₀ = √(5.864266 / 2) = 1.712.
- Standard deviation of the weighted mean = σ₀ / √Σw = 1.712 / 63.333 = ±0.027 ft.
- Report 1842.384 ± 0.027 ft. The a priori value 1/√Σw = ±0.016 ft understates the uncertainty by the factor σ₀ = 1.71 and should not be published.
- σ₀ of 1.71 rather than about 1.0 says the three parties disagree by considerably more than their own reported standard errors allow, which points to an unmodelled systematic difference between the procedures.
Check
Recompute the weighted mean on shifted data. Subtract 1842.00 ft from every value, giving 0.36, 0.29 and 0.41. Then Σ(w·v) = 1111.111 × 0.36 + 400 × 0.29 + 2500 × 0.41 = 400.000 + 116.000 + 1025.000 = 1541.000, and 1541.000 / 4011.111 = 0.384182. Adding 1842.00 back gives 1842.384182, matching the direct computation exactly on four-figure arithmetic instead of ten-figure.
Check that the weighted residuals close: Σ(w·v) taken about the weighted mean must be zero. 1111.111(−0.02418) + 400(−0.09418) + 2500(+0.02582) = −26.867 − 37.672 + 64.550 = +0.011, which is zero to the rounding of the residuals at five decimals. That is the defining property of a weighted mean and it fails immediately if a weight is misapplied.
Check the weight ratios directly against the standard errors: w₃/w₂ = 2500/400 = 6.25, and (σ₂/σ₃)² = (0.05/0.02)² = 2.5² = 6.25. Consistent. If the weights had been taken as 1/σ the ratio would have been 2.5 and the mean would have landed at a different value.
Bracket check: the weighted mean must lie between the smallest and largest observation, 1842.29 and 1842.41. It does, at 1842.384, and closer to the 0.02 ft observation than to the 0.05 ft one.
More error propagation
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- Error propagated into a computed area from errors in two sidesA rectangular parcel measured on two sides, and the uncertainty those two measurements put into the acreage — with the fractional form of the same result as a check and the worst case as a bound.