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I have a set of independent measurements with associated measurement errors. However, each of the errors is substantially different. For example:

10 ± 0.8

12 ± 1.2

7 ± 0.5

8 ± 1.5

I want to calculate the mean of these samples along with the error of this mean. Calculating the mean is straightforward, but I'm having trouble figuring out how to properly propagate the measurement errors. Any clues?

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3 Answers 3

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Usually, measurement errors are reported as some multiple of the standard deviation: often the SD itself, or twice the SD, or occasionally some other multiple. To cover all these cases, let's just call that multiple $\kappa.$

In this fashion, taking the measurement errors to be independent, we may conceive of the numbers $x_i$ in your dataset as being of the form

$$x_i = y_i + \varepsilon_i$$

where the $y_i,$ $(i=1,2,\ldots, n)$ are the true underlying values modeled as a sample from a population or process with a distribution function $F,$ the $\sigma_i$ (when multiplied by $\kappa$) are the "plus or minus" values as given in the question, the $\varepsilon_i$ are independent random variables (and independent of the $y_i$) with variances $\sigma_i^2,$ and the $x_i$ are the measurements. Let's further assume the measurement method is unbiased: this means the expectations of the $\varepsilon_i$ are all zero.

Another assumption is needed: that the $\sigma_i$ are accurately known. (This often is not the case: the $\sigma_i$ often are just (fairly crude) estimates based on repeated measurements of each value. But sometimes the $\sigma_i$ are determined by a measurement system that has been calibrated and tested so often that its statistical properties are very well known and the $\sigma_i$ vary due to modifications of that system. For instance, the measurement error in a laboratory analysis of an aqueous concentration may depend on how much the sample was diluted during a preparation phase.)

With all these assumptions in place--the data are a sample from a distribution $F$ with mean $\mu$ and variance $\sigma^2;$ and their measurements are independent and unbiased with known standard deviations--we may estimate the mean with a formula like

$$\bar x = \sum_{i=1}^n \omega_i\,x_i = \sum_{i=1}^n \omega_i\left(y_i + \varepsilon_i\right)$$

where the $\omega_i$ are suitable positive numbers. To make them suitable, we need the expectation of $\bar x$ to equal the expectation of $F,$ which implies the $\omega_i$ must sum to unity; and among all such possible $\omega_i,$ we may select the ones that minimize the variance of $\bar x,$ given by

$$\operatorname{Var}(\bar x) = \operatorname{Var}\left(\sum_{i=1}^n \omega_i\left(y_i + \varepsilon_i\right)\right) = \sum_{i=1}^n \omega_i^2 \left(\sigma^2 + \sigma_i^2\right).$$

This is minimized when the $\omega_i$ are inversely proportional to the factors $\sigma^2 + \sigma_i^2.$ Choosing this solution gives

$$\bar x = \frac{\sum_{i=1}^n \frac{x_i}{\sigma^2 + \sigma_i^2}}{\sum_{i=1}^n \frac{1}{\sigma^2 + \sigma_i^2}};\quad \operatorname{Var}(\bar x) = \frac{\sum_{i=1}^n \frac{1}{\sigma^2 + \sigma_i^2}}{\left(\sum_{i=1}^n \frac{1}{\left(\sigma^2 + \sigma_i^2\right)}\right)^2}.$$


That's a correct answer, insofar as it goes: but we usually don't know the value of $\sigma^2.$ It needs to be estimated. That raises the question of how best to estimate it.

One reasonable (although not necessarily optimal) procedure is to estimate $\sigma$ and the weights iteratively. That is, given some estimate $\hat\sigma_j^2$ of $\sigma^2$ obtained after $j$ iterations, compute the weights

$$\omega_{i;j+1} = \frac{1}{\sum_{i=1}^n \frac{1}{\hat\sigma_j^2 + \sigma_i^2}}\left(\frac{1}{\hat\sigma_j^2 + \sigma_i^2}\right)$$

and with them update the estimated (weighted) mean

$$\bar x_{j+1} = \sum_{i=1}^n \omega_{i;j+1}\,x_i$$

and then update the estimate $\hat\sigma^2$ to a new value $\hat\sigma^2_{j+1}.$ Extensive simulations suggest a good update would be a bias-adjusted weighted variance from which the contributions of the $\sigma_i^2$ have been removed:

$$\hat\sigma^2_{j+1} = \frac{n}{n-1}\sum_{i=1}^n \omega_{i;j+1}(x_i - \bar x_{j+1})^2 - \frac{1}{n} \sum_{i=1}^n \sigma_i^2.$$

(This formula is inspired by the weights that are applied when there is no measurement error: in that case, they are all equal to $1/(n-1),$ which is $n/(n-1)$ times the weights $\omega_i = 1/n$ that would be used. Using values of $0$ for all the $\sigma_i$ will thereby cause the algorithm to converge immediately to the usual unbiased variance estimate.)

Whenever this value is negative, set it to zero. Iterate until the $\hat\sigma^2_j$ converge, as tested by comparing $\hat\sigma^2_{j+1} / \hat\sigma^2_j$ to $1.$

The iteration may begin by setting $\hat\sigma^2_0$ to be the naive (unweighted) variance estimate of the $x_i$ (computed by setting all weights to $1/n$).

This algorithm appears always to converge and to produce nearly unbiased estimates of $\sigma^2$ provided $\sigma^2$ is not close to the root mean square of the $\sigma_i^2.$ Specifically,

  • When the variation among the measured values $x_i$ can be explained by the measurement errors, $\sigma^2$ is likely small and often is estimated as $0.$

  • When the measurement errors are small compared to the variation among the measured values, this procedure estimates $\sigma^2$ accurately on average.

  • When the spread of the true values is comparable to the measurement errors, this procedure tends to underestimate $\sigma^2.$

Such behavior looks unavoidable in small datasets due to the difficulty of estimating $\sigma^2$ precisely even with zero measurement error.

For the data in the question, interpreting the $\pm$ values as standard deviations of the measurements (that is, $\kappa=1$), this procedure converges after three iterations (to a precision of one in a million) to the estimates $\bar x_3 = 9.167$ and $\hat\sigma_3^2 = 3.801.$ We may take $\hat \sigma = \sqrt{\hat\sigma_3^2} = 1.95.$ Because this exceeds most of the measurement standard deviations, it is probably an accurate estimate.

At the end the weights are

$$\omega = (\omega_{i;3}) = (0.2719, 0.2304, 0.2981, 0.1996).$$

As we expect, lower weights are associated with the measurements having greater error--but the variation in the weights is not great, given that the measurement SDs vary by a factor of three.

As a further test, I performed a parametric bootstrap, assuming the $y_i$ are sampled from a Normal distribution. This procedure drew 5,000 values of the $y_i$ from a Normal distribution having the same (weighted) mean and variance reported above, applied random measurement errors to those $y_i$ with the given standard deviations, and ran the proposed procedure to estimate the mean and variance of this Normal distribution. Here is a graphical summary of the results:

Figure

The vertical colored bars show the true underlying values in the bootstrap. They are positioned squarely in the middle of the bootstrap distributions of their estimates, indicating a lack of bias. The bar plot of iteration counts at the right indicates that few iterations were needed in the great majority of cases and often 2 - 5 iterations sufficed.

The spread of estimates of $\bar x$ indicates the error of the mean. The middle 95% of these values lie in the interval $[7.0, 11.3],$ for instance. We may communicate this by saying we are 95% confident the true mean (of the $y_i$) lies between $7.0$ and $11.3.$


To make the details perfectly clear, here is the R code implementing the variance estimation algorithm (as var.wt) and performing the parametric bootstrap.

#
# An iterative procedure to estimate mean and variance from a set of 
# measurements `x` made with known measurement errors `sigma.i`.
#
var.wt <- function(x, sigma.i, thresh=1e-6, iter.max=200) {
  #
  # To avoid giving undue weight to any subset, all SDs should be nonzero.
  #
  if (missing(sigma.i)) sigma.i <- rep(0, length(x)) # No measurement error
  tol <- sum(abs(x)) * 1e-16
  sigma.i[sigma.i <= 0] <- min(c(tol, sigma.i[sigma.i > 0])) / 2
  #
  # Initialize.
  #
  v <- var(x)
  n <- length(x)
  S <- mean(sigma.i^2) # Avoids recomputing this constant within the loop
  for (i in 1:iter.max) {
    w <- 1/(v + sigma.i^2); w <- w / sum(w)
    m <- sum(x * w)
    v.0 <- max(0, sum((x-m)^2 * w) * n / (n-1) - S) 
    if (v <= v.0 * (1+thresh) && v.0 <= v * (1+thresh)) break
    v <- v.0
  }
  if (i >= iter.max) warning("Maximum iteration count exceeded.")
  list(Variance=v, Mean=m, Iterations=i)
}
#
# The data in the question.
#
x <- c(10, 12, 7, 8)
sigma.i <- c(0.8, 1.2, 0.5, 1.5)
(s <- var.wt(x, sigma.i))
#
# A parametric bootstrap.
#
n <- length(x)
sigma <- sqrt(s$Variance)
mu <- s$Mean
set.seed(17)
X <- replicate(5e3, {
  y <- rnorm(n, mu, sigma)
  epsilon <- rnorm(n, 0, sigma.i)
  x <- y + epsilon
  var.wt(x, sigma.i)
})
X <- matrix(unlist(X), 3)
#
# Display the results.
#
par(mfrow=c(1,3))
hist(sqrt(X[1,]), main="Weighted SD", col="#f0f0f0", xlab="")
abline(v = sigma, lwd=2, col="#d01010")
hist(X[2,], main="Weighted Mean", col="#f0f0f0", xlab="")
abline(v = mu, lwd=2, col="#1010d0")
plot(table(X[3,]), log="x", main="Iterations", xlab="Count", ylab="Frequency")
par(mfrow=c(1,1))

(quantile(X[2,], c (0.025, 0.975))) # A central 95% interval of the weighted means
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  • $\begingroup$ Can you provide some reference to all this? $\endgroup$
    – Walter
    Nov 17, 2022 at 6:30
  • $\begingroup$ This is all very nice. But how does it generalize for the multi-variate case, when the data are $k$-tuples and the $\sigma_i^2$ are symmetric $k\times k$ matrices? $\endgroup$
    – Walter
    Nov 17, 2022 at 12:10
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Ideally, you need the original measures. For example, for your 7 +- 0.5, you should have something like. [6.5, 7, 7.5, 7 , 6.5, 7.5 etc]

If you do not have them then try to "Unroll" them yourself in a similar way, for example take 1.000 random samples with mean 7 and variance 0.5.

Then you will have one bigger sample, which can be analyzed, further. Find its distribution, get the mean and the variance, or do a One Sample T-test which is more statistically correct for the mean.

From this you can easily derive the range of uncertainty of the calculated sampled mean for a given level of uncertainty (usually you take the 0.95)

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  • $\begingroup$ This proposal is erroneous because the testing will be applied to a much larger dataset than is actually available. In particular, p-values in the t-test will be far too small and standard errors of the mean will be far too narrow. $\endgroup$
    – whuber
    Mar 15, 2020 at 17:29
  • $\begingroup$ The OP never gave the same size, so there is nothing wrong with this answer which suggests a simulation using 1000 realizations - purely Monte Carlo uncertainty analysis. In fact, you could use 10,000 realizations. Standard deviation doesn't change with sample size, but power depends on effect size, sample size, and significance. I don't recommend (other comment) turning the OP into a power and sample size question. $\endgroup$
    – user32398
    Mar 15, 2020 at 18:01
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    $\begingroup$ @whuber Ideally you need the original measurements. I guess that those errors come from some tests which also have a distribution. $\endgroup$ Mar 15, 2020 at 18:07
  • $\begingroup$ I agree that if the measurement SDs were obtained from repeated measurements, then it would be ideal to have the original measurements. The appropriate analysis in this case is a components of variance analysis, carried out using ANOVA machinery. However, note that if we only had (a) the number of measurements in each case and (b) an indication of exactly how the $\pm$ values were computed then that would suffice to do all the necessary calculations. $\endgroup$
    – whuber
    Mar 15, 2020 at 20:38
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Using straightforward quadrature sum, the standard deviation (error) for the average, $\bar{x}$, of your 10, 12, 7, and 8 would be

\begin{equation} \sigma_{\bar{x}} = \sqrt{0.8^2 + 1.2^2 + 0.5^2 + 1.5^2}, \end{equation}

so the result is $\bar{x} \pm \sigma_{\bar{x}}$.

Calculating this out, the result is (7.1,11.4).

Another approach for propagation, called weighted means, would suggest that the weights are the inverse squares of the corresponding uncertainties, $w_j=1/\sigma_j^2$, and the best estimate, $x_{best}$, is

\begin{equation} x_{best} = \frac{\sum_{j=1}^p w_jx_j}{\sum_{j=1}^p w_j}, \end{equation}

and the overall uncertainty is

\begin{equation} \sigma_{x,best} = \left(\sum_{j=1}^p w_j \right) ^{-1/2}. \end{equation}

Upon substitution with results for $p=4$ measurements, we get

\begin{align} x_{best} &= \frac{1.56*10 + 0.694*12 + 4*7 + 0.444*8}{1.56 + 0.694 + 4 + 0.444} \\ &= 8.28,\\ \end{align}

and the uncertainty is

\begin{align} \sigma_{x,best} &= \frac{1}{\sqrt{1.56 + 0.694 + 4 + 0.444}}\\ &=\frac{1}{\sqrt{6.7}}\\ &=0.386, \end{align}

yielding a result of 8.3(7.9,8.7).

For Monte Carlo simulation, since no distributional information is available, I would employ triangle ("lack-of-knowledge") distributions (5000 realizations, zero correlation). The resulting mean $y$ is normally distributed, and the sigma (0.217) is smaller leading to (9,9.5), while normal distributions resulted in (8.7,9.8), where sigma was 0.531. Below are the histograms for the triangle distribution and the normally-distributed mean distribution (Shapiro-Wilk $P$=0.2382).

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    $\begingroup$ This is incorrect because (1) it ignores the variation between the data and (2) it does not account for the weights of $1/4$ applied to the data in computing the mean. As a result--purely because these two gross errors nearly cancel out in this case--the result of the first calculation is close to a reasonable one for these data. $\endgroup$
    – whuber
    Mar 15, 2020 at 14:36
  • $\begingroup$ The OP states that the measurements were independent, so saying that the covariance between the variables needs be looked at is "changing the parameters" - i.e., making the question into something it's not. Both equations above for quadrature sum assume independence, and the OP states that the measurements were independent. $\endgroup$
    – user32398
    Mar 15, 2020 at 17:46
  • $\begingroup$ Your interpretation is untenable. Your formula gives the variance of the sum of the measurement errors. That is a small part of answering the questions that have explicitly been formulated about estimating the mean of the values and determining its probable error. $\endgroup$
    – whuber
    Mar 15, 2020 at 18:02

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