The worst-case scenario

Consider a population that consists of M age fractions and define relevant fractions to be those fractions that are greater than f. For a given M (assuming M$\leq$1/f), the worst-case scenario is that M-1 of the population fractions are of size f, and one fraction is of size 1-f(M-1). The probability p that at least one fraction $\geq$f of the population was missed is given by:

$$p = \sum_{n=1}^{M-A}(-1)^{n-1} \left( \binom{M-A}{n} (1-nf)^k + B \binom{M-1}{n-1} ((M-n)f)^k \right)$$ (2)

with

$$A = 0 Mf = 1$$

$$A = 1 Mf \neq 1$$

$$B = 0 Mf \geq 1$$

$$B = 1 Mf < 1$$

This is a combinatoric expression where $\binom{x}{y}$ is the binomial coefficient. Each term in the summation adds a correction to the previous terms. Equation 2 is derived in Appendix A. For a given number of relevant fractions m (m$\leq$1/f), a best-case scenario can also be calculated (Appendix A):

$$p = \sum_{n=1}^{m}(-1)^{n-1} \binom{m}{n}\left(1-\frac{n}{m}\right)^k$$ (3)

Exploration of equations 2 and 3 over M and m, and for different values of f and k, is shown in Figure 1. The maximum number of (relevant) fractions for which Equations 2 and 3 are valid is 1/f. At larger values of M (or m), p is kept constant. The shaded region on Figure 1a marks the area where this is the case. One way to reduce the probability that fractions $\geq$f are missed when only k grains are dated is to reduce the number of bins in the sample histogram. For example, if k=60, f=0.05, and p=20%, then M$_{opt}$=6 (Figure 1). A detrital age-histogram that is constructed in this way conveys as much information about the population as can be inferred from the sample and is statistically "allowed" by p and f. However, it is less well suited for showing the sample distribution. Therefore, such a histogram should be used in conjunction with markers for the sample data, or better still, a probability density plot [6]. Such a combined plot carries an optimal amount of information: the histogram represents the population with the resolution that the data and the parameters p and f allow, while at the same time, the probability density plot represents the data itself and the uncertainties that are associated with it (Figure 2). M$_{opt}$ usually is a rather small number, much smaller than commonly used guidelines for the number of histogram bins such as Sturges' rule [7,8]. Using M$_{opt}$ will tend to oversmooth the histogram, so although it theoretically is a viable way to reduce the chance of missing significant fractions of the population, there are better methods for dealing with datasets that contain fewer than the optimal number of measurements. These methods are discussed in the following paragraph and the Conclusions section.

Rather than reducing m, a much better way to reduce p is to increase k or f. We now define p$_{max}$ as the maximum value of p, reached when M=m=[1/f], where square brackets mark truncation to the nearest integer. The equation for p$_{max}$ is a special case of (2):

$$p_{max} = \sum_{n=1}^{[1/f]}(-1)^{n-1} \binom{[1/f]}{n} (1-nf)^k$$ (4)

Figure 3 shows the evolution of p$_{max}$ as a function of f and k. Note the discrete "knee" in the p$_{max}$ vs. f curve wherever M = 1/f. Figure 3 can be used for a quick assessment of the number of grains that are needed for a provenance study, and of the risk of information loss that is caused by smaller samples. For example, if 60 grains are dated, then $p_{max}$=64%. Therefore, in the worst-case scenario (which, at m=20, is a perfectly uniform population) there is 64% chance that at least one fraction $\geq$0.05 of the population is missed. This is a dramatically different result from the 5% probability suggested by Equation 1. Furthermore, the actual fraction f$_{act}$ that we can be sure not to have missed with 95% certainty is not 0.05, but 0.085, as can be read from Figure 3. Finally, and perhaps most importantly, Figure 3 also shows that in order to be 95% confident that no fraction $\geq$0.05 was missed, at least k=117 grains must be dated. Table 1 can be used to choose k, the number of grains required to lower p and f to some desired limits. If fewer than this optimal number of grains have been dated, Table 2 can be used to estimate the actual levels of p and f that have been achieved with that k. The same table also lists the value of $M_{opt}$ in the unlikely event that the user prefers to reduce the resolution of the age histogram, rather than to increase the desired p and/or f. Table 1 should be used before embarking on a provenance study to determine how many grains are needed. Alternatively, Table 2 can be used for the interpretation of provenance data with less than the optimal number of grains. For example, if only 30 grains have been dated, Table 2 says that f$_{act}$=0.15 is the smallest fraction not missed at a 95% confidence level. Likewise, there is 20% chance of missing at least one fraction representing $\geq$0.12 of the total population, and the probability of missing at least one fraction $\geq$0.1 when 30 grains were dated is 37%. Finally, to reduce the chance of missing at least one fraction $\geq$0.2 of the population to less than 10%, and still only use 30 grains, the age-histogram cannot have more than M$_{opt}$=5 bins. As an alternative to Figure 3, and to Tables 1 and 2, an online web-form [9] is available for the calculation of k, p$_{max}$, f$_{act}$ and M$_{opt}$.