Non-Poisson Counting Uncertainty, or “What’s this J Factor All About?”

Keith D. McCroan US EPA National Air and Radiation Environmental Laboratory Radiobioassay and Radiochemical Measurements Conference October 29, 2009 Non-Poisson Counting Uncertainty, or“What’s this J Factor All About?”

Counting uncertainty • Most rad-chemists learn early to estimate “counting uncertainty” by square root of the count C. • They are likely to learn that this works because C has a “Poisson” distribution. • They may not learn why that statement is true, but they become comfortable with it.

“The standard deviation of C equals its square root. Got it.”

The Poisson distribution • What’s special about a Poisson distribution? • What is really unique is the fact that its mean equals its variance: μ = σ2 • This is why we can estimate the standard deviation σ by the square root of the observed value – very convenient. • What other well-known distributions have this property? None that I can name.

The Poisson distribution in Nature • How does Nature produce a Poisson distribution? • The Poisson distribution is just an approximation – like a normal distribution. • It can be a very good approximation of another distribution called a binomial distribution.

Binomial distribution • You get a binomial distribution when you perform a series of N independent trials of an experiment, each having two possible outcomes (success and failure). • The probability of success p is the same for each trial (e.g., flipping a coin, p = 0.5). • If X is number of successes, it has the “binomial distribution with parameters N and p.” X ~ Bin(N, p)

Poisson approximation • The mean of X is Np and the variance is Np(1 − p). • When p is tiny, the mean and variance are almost equal, because (1 − p) ≈ 1. • Example: N is number of atoms of a radionuclide in a source, p is probability of decay and counting of a particular atom during the counting period (assuming half-life isn’t short), and C is number of counts.

Poisson counting • In this case the mean of C is Np and the variance is also approximately Np. • We can consider C to be Poisson: C ~ Poi(μ) where μ = Np

Poisson – Summary • In a nutshell, the Poisson distribution describes occurrences of relatively rare (very rare) events (e.g., decay and counting of an unstable atom) • Where significant numbers are observed only because the event has so many chances to occur (e.g., very large number of these atoms in the source)

Violating the assumptions • Imagine measuring 222Rn and progeny by scintillation counting – Lucas cell or LSC. • Assumptions for the binomial/Poisson distribution are violated. How? • First, the count time may not be short enough compared to the half-life of 222Rn. • The binomial probability p may not be small. • If you were counting just the radon, you might need the binomial distribution and not the Poisson approximation.

More importantly... • We actually count radon + progeny. • We may start with N atoms of 222Rn in the source, but we don’t get a simple “success” or “failure” to record for each one. • Each atom might produce one or more counts as it decays. • C isn’t just the number of “successes.”

Lucas 1964 • In 1964 Henry Lucas published an analysis of the counting statistics for 222Rn and progeny in a Lucas cell. • Apparently many rad-chemists either never heard of it or didn’t fully appreciate its significance. • You still see counting uncertainty for these measurements being calculated as .

Radon decay • Slightly simplified decay chain: • A radon atom emits three α-particles and two β-particles on its way to becoming 210Pb (not stable but relatively long-lived). • In a Lucas cell we count just the alphas – 3 of them in this chain.

Thought experiment • Let’s pretend that for every 222Rn atom that decays during the counting period, we get exactly 3 counts (for the 3 α-particles that will be emitted). • What happens to the counting statistics?

Non-Poisson counting • C is always a multiple of 3 (e.g., 0, 3, 6, 9, 12, ...). • That’s not Poisson – A Poisson variable can assume any nonnegative value. • More important question to us: What is the relationship between the mean and the variance of C?

Index of dispersion, J • The ratio of the variance V(C) to the mean E(C) is called the index of dispersion. • Often denoted by D, but Lucas used J. • That’s why this factor is sometimes called a “J factor” • For a Poisson distribution, J = 1. • What happens to J when you get 3 counts per decaying atom?

Mean and variance • Say D is the number of radon atoms that decay during the counting period and C is the number of counts produced. • Assume D is Poisson, so V(D) = E(D). • By assumption, C = 3 × D.So, E(C) = 3 × E(D) V(C) = 9 × V(D) J = V(C) / E(C) = 3 × V(D) / E(D) = 3

Index of dispersion • So, the index of dispersion for C is 3, not 1 which we’re accustomed to seeing. • This thought experiment isn’t realistic. • You don’t really get exactly 3 counts for each atom of analyte that decays. • It’s much trickier to calculate J correctly.

Technique • Fortunately you really only have to consider a typical atom of the analyte (e.g., 222Rn) at the start of the analysis. • What is the index of dispersion J for the number of counts C that will be produced by this hypothetical atom as it decays? • Easiest approach involves a statistical technique called conditioning.

Conditioning • Consider all the possible histories for the atom – i.e., all the different ways the atom can decay. • It is convenient to define the histories in terms of the states the atom is in at the beginning and end of the counting period. • Calculate the probability of each history • typically using Bateman equations

Conditioning - Continued • For each history, calculate the conditional expected values of C and C2 given that history (i.e., assuming it occurs). • Next calculate the overall expected values E(C) and E(C2) as probability-weighted averages of the conditional values. • Calculate V(C) = E(C2) − E(C)2 . • Finally, J = V(C) / E(C). • Details left to the reader.

Radium-226 • Sometimes you measure radon to quantify the parent 226Ra. • Let J be the index of dispersion for the number of counts produced by a typical atom of the analyte 226Ra – not radon. • Technique for finding J (conditioning) is the same, but the details are different. • Value of J is always > 1 in this case.

Thorium-234 • If you beta-count a sample containing 234Th, you’re counting both 234Th and the short-lived decay product 234mPa. • With ~50 % beta detection efficiency, you have non-Poisson statistics here too. • The counts often come in pairs. • The value of J doesn’t tend to be as large as when counting radon in a Lucas cell or LSC (less than 1.5).

Gross alpha/beta? • If you don’t know what you’re counting, how can you estimate J? • You really can’t. • Probably most methods implicitly assume J = 1. • But who really knows?

Simplification • Assume every radiation of the decaying atom has detection efficiency ε or 0. Then where • m1 is expected number of detectable radiations from • an atom of analyte during the counting interval • m2 is expected square of this number

Bounds for J • m1 ≤ m2 ≤ Nm1, where N is the maximum number of counts per atom. So, 1 − ε × m1 ≤ J ≤ 1 + ε × (N − m1 − 1) • In many situations m1 is very small. Then 1 ≤ J ≤ 1 + ε × (N − 1) • E.g., for 226Ra measured by 222Rn in a Lucas cell, N = 3. So, 1 ≤ J ≤ 1 + 2ε

Remember • Suspect non-Poisson counting if: • One atom can produce more than one count (N>1) as it decays through a series of short-lived states • Detection efficiency (ε) is high • Together these effects tend to give you on average more than one count per decaying atom. • In many cases, 1 ≤ J ≤ 1 + ε × (N − 1).

Questions?

Reference • Lucas, H.F., Jr., and D.A. Woodward. 1964. Journal of Applied Physics 35:452.

Testing for J > 1 • You can test J > 1 with a χ2 test, but you may need a lot of measurements.

Non-Poisson Counting Uncertainty, or “What’s this J Factor All About?”