Counting Statistics and Error Prediction in Radiation Measurements

Foundations of Counting Statistics and Error Prediction

  • Radiation events are inherently stochastic in nature, meaning they occur randomly over time. Because of this, it is necessary to analyze radiation measurement data using statistical methods.

  • The study of counting statistics and error prediction is based on the book Flynn Naw, tablet three.

  • Error prediction is derived from standard relations used for error propagation.

  • In any research or equipment operation measurement, reporting a single mean value is insufficient. Data should be reported as the mean plus or minus a scatter band (e.g., xˉ±σx \bar{} \pm \sigma).

Characterization of Experimental Data

  • When measuring decays over a specific time interval (e.g., in seconds), data is collected as a series of counts.

  • An example dataset of measured decays includes the following values: 8,10,3,13,9,7,12,9,6,10,10,11,78, 10, 3, 13, 9, 7, 12, 9, 6, 10, 10, 11, 7.

  • Three primary properties are used to characterize such a dataset:

    1. Experimental Mean (xex_e): The average value of the measurements.

    2. Distribution Function (f(x)f(x)): Defined as the number of occurrences of a particular value xx divided by the total number of measurements (nn).

    3. Sample Variance (s2s^2): A measure of the internal scatter within the recorded data.

    4. Standard Deviation (σ\sigma): The square root of the sample variance, defining the scatter band for the reported measurement.

Mathematical Framework for Data Properties

  • Experimental Mean (xex_e):

    • Calculated as the sum of all measurement values divided by the total number of measurements:     xe=∑i=1nxinx_e = \frac{\sum_{i=1}^n x_i}{n}

  • Distribution Function (f(x)f(x)):

    • Also known as the frequency distribution function or data distribution function.

    • It represents normalized occurrences. For a dataset of nn measurements, the sum of all distribution function values must equal one:     ∑x=0∞f(x)=1\sum_{x=0}^{\infty} f(x) = 1

    • The experimental mean can be calculated as the first moment of the distribution function:     xe=∑x=0∞xf(x)x_e = \sum_{x=0}^{\infty} x f(x)

  • Residuals (did_i):

    • The difference between an individual measurement and the experimental mean:     di=xi−xed_i = x_i - x_e

    • The sum of all residuals in a dataset is always zero:     ∑i=1ndi=0\sum_{i=1}^n d_i = 0

  • True Mean (xˉ\bar{x}):

    • The true mean represents the actual value if one were to measure every possible instance (e.g., inspecting all 10,000 fuel rods produced by a factory).

    • Because inspecting every item is expensive and time-consuming (total quality control), companies use sampling (e.g., inspecting 20 rods), which gives the experimental mean (xex_e).

  • True Deviation (ϵi\epsilon_i):

    • The difference between a measurement and the true mean:     ϵi=xi−xˉ\epsilon_i = x_i - \bar{x}

  • Sample Variance (s2s^2):

    • Based on the true mean, variance is defined as:     s2=1n∑i=1n(xi−xˉ)2s^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2

    • Since the true mean is often unknown, the sample variance is calculated using the experimental mean:     s2=1n−1∑i=1n(xi−xe)2s^2 = \frac{1}{n - 1} \sum_{i=1}^n (x_i - x_e)^2

    • The use of n−1n-1 instead of nn accounts for the fact that the experimental mean is an estimate. As nn becomes very large, the difference between the two definitions becomes negligible.

    • Sample variance measures internal scatter and does not depend significantly on the number of measurements nn; the scatter band of a system remains relatively constant regardless of whether 20 or 40 measurements are taken.

Statistical Models for Trials and Success

  • Statistical models are used to determine if measured sample variance matches predicted variance. Significant discrepancies indicate systematic errors, procedural issues, or system artifacts.

  • Models are defined by three parameters:

    1. nn: The number of trials.

    2. pp: The probability of success in a single trial.

    3. xx: The number of successful events.

Binomial Distribution

  • Generally complex to solve for radiation measurements because nn is typically very large.

  • It assumes probability pp is constant and does not change with time or the number of trials.

  • Distribution Function:     P(x)=n!x!(n−x)!px(1−p)n−xP(x) = \frac{n!}{x! (n - x)!} p^x (1 - p)^{n - x}

  • Mean (xˉ\bar{x}):     xˉ=n×p\bar{x} = n \times p

  • Predicted Variance (σ2\sigma^2):     σ2=n×p(1−p)=xˉ(1−p)\sigma^2 = n \times p (1 - p) = \bar{x} (1 - p)

  • Example: Rolling a Die:

    • Rolling a die 10 times (n=10n = 10).

    • Success defined as rolling a 3, 4, 5, or 6 (p=46=0.667p = \frac{4}{6} = 0.667; non-success 1−p=26=0.3331-p = \frac{2}{6} = 0.333).

    • Mean success: xbar=10×0.667=6.67x_{bar} = 10 \times 0.667 = 6.67.

    • Variance: σ2=10×0.667×0.333=2.22\sigma^2 = 10 \times 0.667 \times 0.333 = 2.22.

    • Standard Deviation: σ=2.22=1.49\sigma = \sqrt{2.22} = 1.49.

Poisson Distribution

  • Simplified model widely used for radiation measurements where success probability pp is very small.

  • Often used when the mean value (xˉ\bar{x}) is less than 20.

  • Distribution Function:     P(x)=(np)xe−npx!=xˉxe−xˉx!P(x) = \frac{(np)^x e^{-np}}{x!} = \frac{\bar{x}^x e^{-\bar{x}}}{x!}

  • Mean (xˉ\bar{x}):     xˉ=n×p\bar{x} = n \times p

  • Predicted Variance (σ2\sigma^2):     σ2=xˉ\sigma^2 = \bar{x}

  • Standard Deviation:     σ=xˉ\sigma = \sqrt{\bar{x}}

  • Example: Birthday Survey:

    • Survey 1,000 people (n=1,000n = 1,000) to see if their birthday is today (p=1365p = \frac{1}{365}).

    • xˉ=1,000×1365=2.74\bar{x} = 1,000 \times \frac{1}{365} = 2.74.

    • Calculated probabilities P(x)P(x):

      • P(0)=0.064P(0) = 0.064

      • P(1)=0.177P(1) = 0.177

      • P(2)=0.242P(2) = 0.242

      • P(3)=0.221P(3) = 0.221

      • P(4)=0.152P(4) = 0.152

      • P(5)=0.083P(5) = 0.083

      • P(6)=0.038P(6) = 0.038

    • The Poisson distribution for small means is asymmetric.

Gaussian (Normal) Distribution

  • A continuous function used when p≪1p \ll 1 and the mean xˉ>20\bar{x} > 20.

  • Distribution Function:     P(x)=12πxˉe−(x−xˉ)22xˉP(x) = \frac{1}{\sqrt{2 \pi \bar{x}}} e^{-\frac{(x - \bar{x})^2}{2 \bar{x}}}

  • Normalized Deviation (tt):

    • Defined as the deviation divided by the standard deviation:     t=x−xˉσt = \frac{x - \bar{x}}{\sigma}

    • The Gaussian distribution in terms of tt is:     G(t)=2πe−t22G(t) = \sqrt{\frac{2}{\pi}} e^{-\frac{t^2}{2}}

  • Confidence Levels and Standard Deviations:

    • 0.674σ0.674 \sigma: 50% probability.

    • 1.00σ1.00 \sigma: 68.3% probability.

    • 1.64σ1.64 \sigma: 90% probability (often used in experiment design).

    • 1.96σ1.96 \sigma: 95% probability.

    • 3.00σ3.00 \sigma: 99.7% probability.

Practical Examples and Comparative Analysis

Example: Probability of 8 Heads in 12 Coin Tosses

  • n=12n = 12, x=8x = 8, p=0.5p = 0.5.

  • Using Binomial Distribution:     P(8)=12!8!4!(0.5)8(0.5)4=0.121P(8) = \frac{12!}{8! 4!} (0.5)^8 (0.5)^4 = 0.121

Example: Inspecting Corn for Insects

  • A farmer pulls 200 ears of corn (n=200n = 200). Probability of insects p=0.01p = 0.01.

  • Binomial Analysis:

    • Probability of exactly 3 ears having insects:     P(3)=200!197!3!(0.01)3(0.99)197=0.181355P(3) = \frac{200!}{197! 3!} (0.01)^3 (0.99)^{197} = 0.181355

    • Mean: xˉ=2\bar{x} = 2. Standard deviation: σ=200×0.01×0.99=1.407\sigma = \sqrt{200 \times 0.01 \times 0.99} = 1.407.

  • Poisson Analysis:

    • P(3)=23e−23!=0.18045P(3) = \frac{2^3 e^{-2}}{3!} = 0.18045

    • Mean: xˉ=2\bar{x} = 2. Standard deviation: σ=2=1.414\sigma = \sqrt{2} = 1.414.

Example: Radioactive Decay Over Time

  • Given decay constant λ=1.3188×10−7 min−1\lambda = 1.3188 \times 10^{-7} \, \text{min}^{-1}.

  • Success defined as 120 decays (x=120x = 120) in 1 minute (Δt=1 min\Delta t = 1 \, \text{min}) from a sample of 10910^9 atoms (n=109n = 10^9).

  • Average success xˉ=n×p=109×(1.3188×10−7)=131.88\bar{x} = n \times p = 10^9 \times (1.3188 \times 10^{-7}) = 131.88.

  • Poisson probability: P(120)=0.02096P(120) = 0.02096 (approx. 2%).

  • Gaussian probability: P(120)=0.0203P(120) = 0.0203.

Chi-Squared Value and Single Measurements

  • Chi-Squared (χ2\chi^2):

    • A method to compare predicted variance and experimental variance:     χ2=(n−1)s2xˉ\chi^2 = \frac{(n - 1)s^2}{\bar{x}}

    • Alternatively expressed as:     χ2=∑i=1n(xi−xˉ)2xˉ\chi^2 = \sum_{i=1}^n \frac{(x_i - \bar{x})^2}{\bar{x}}

  • Single Experimental Values:

    • If only one experimental value xx is available, the true mean and experimental mean are unknown.

    • In this common situation, the standard deviation is approximated as x\sqrt{x}.

    • The result is reported as: x±xx \pm \sqrt{x}.