Counting Statistics

Ho Kyung KimPusan National University

Radiation Measurement SystemsKnoll chap. 3

• Unavoidable statistical fluctuation in radiation measurements due to the randomness in radioactive decay

– Source of uncertainty (or imprecision or error)• Note: accuracy vs. inaccuracy (bias)

precision vs. imprecision (uncertainty)– Additional electronic noise

• The value of counting statistics (model)– We can say that some abnormality exists in the counting system if the measured

fluctuation is not consistent w/ predictions of statistical models– We can estimate an accuracy even in a single measurement by predicting inherent

statistical uncertainty

2

CHARACTERIZATION OF DATA

• Let’s consider N indep. measurements of the same physical quantity– x1, x2, x3 ,…, xi, …, xN

• Experimental mean:

• Example– N = 20– Σ = 176–

NN

xx

N

ii

eΣ

==∑=1

x1 = 8 x11 = 14 x2 = 5 x12 = 8 x3 = 12 x13 = 8 x4 = 10 x14 = 3 x5 = 13 x15 = 9 x6 = 7 x16 = 12 x7 = 9 x17 = 6 x8 = 10 x18 = 10 x9 = 6 x19 = 8 x10 = 11 x20 = 7

8.8=ex

3

• Frequency distribution function:

– Note that

–

• Example

NxC

NxxF )(

)( tsmeasuremen ofnumber value theof occurences ofnumber )( =

=≡

1)(0

=∑∞

=x

xF

∑∑∑ ∞

=

∞

=

= ==+++==00

211 )()(1)(1

xxN

N

ii

e xxFxxCN

xxxNN

xx

8.805.01400.0405.03

)14(14)4(4)3(3

)(0

=⋅++⋅+⋅=⋅++⋅+⋅=

=∑∞

=

FFF

xxFxx

e

4

• How to quantify the amount of internal fluctuation in the data set?

5

– Residual (exp. mean):

– Deviation (true mean):

• Sample variance

– But, we do not know the true mean!

– Also calculated from the FDF

eii xxd −=

01

=∑=

N

iid

xxii −=ε

∑=

−==N

ii xx

Ns

1

222 )(1ε

∑=

−−

=N

iei xx

Ns

1

22 )(1

1

∑∞

=

−=0

22 )()(x

xFxxs

222 xxs −=

Refer to Appendix B for the derivation

6

STATISTICAL MODELS

• Let’s define a measurement as counting the number of successes resulting from a given number of trials

– Each trial is a binary (success/fail) process– The probability of success is a constant, p

• Statistical models– The Binomial distribution: applicable to all constant-p processes– The Poisson distribution: if p is small & constant– The Gaussian or Normal distribution: if the avg. # of successes is large (≥ 25 or 30)

Trial Definition of success Prob. of success = p

Tossing a coin Heads 1/2

Rolling a die A six 1/6

Observation a given radioactive nucleus for a time t

The nucleus decays during the observation

1 – e-λt

7

The Binomial distribution

– n = # of trials– x = # of successes– p = success probability

• Example– Success: 3, 4, 5, or 6 for the six

possible results in rolling a die– p = 4/6 or 0.667– n = 10 rolls

xnxxnx ppxxn

nppxn

xP −− −−

=−

= )1(

!)!(!)1()(

50100

10694.131

32

!0)!010(!10)0( −

−

×=

−==xP

8

• Properties of the binomial distribution

– 0 ≤ 𝑃𝑃(𝑥𝑥) ≤ 1

– Normalization: ∑𝑥𝑥=0𝑛𝑛 𝑃𝑃(𝑥𝑥) = 1

– Mean (or average) value: �̅�𝑥 = ∑𝑥𝑥=0𝑛𝑛 𝑥𝑥𝑃𝑃(𝑥𝑥) = 𝑛𝑛𝑛𝑛

– Predicted variance: 𝜎𝜎2 = ∑𝑥𝑥=0𝑛𝑛 𝑥𝑥 − �̅�𝑥 2𝑃𝑃(𝑥𝑥) = �̅�𝑥(1 − 𝑛𝑛)

• For the previous example:

– �̅�𝑥 = 𝑛𝑛𝑛𝑛 = 10 23 = 6.67

– 𝜎𝜎2 = �̅�𝑥 1 − 𝑛𝑛 = 2.22 or 𝜎𝜎 = 1.49

Derive! (take home)

Derive! (take home)

9

The Poisson distribution

• A constant and small prob. (p << 1) in most nuclear counting experiment in which # of nuclei in the sample is large and the observation time is short compared with T1/2 of the radioactive species, as well as in a particle beam experiment

– 𝑃𝑃 𝑥𝑥 = (𝑝𝑝𝑛𝑛)𝑥𝑥𝑒𝑒−𝑝𝑝𝑝𝑝

𝑥𝑥!= (�̅�𝑥)𝑥𝑥𝑒𝑒−�𝑥𝑥

𝑥𝑥!

• Properties of the Poisson distribution

– Normalization: ∑𝑥𝑥=0𝑛𝑛 𝑃𝑃(𝑥𝑥) = 1

– Mean (or average) value: �̅�𝑥 = ∑𝑥𝑥=0𝑛𝑛 𝑥𝑥𝑃𝑃(𝑥𝑥) = 𝑛𝑛𝑛𝑛

– Predicted variance: 𝜎𝜎2 = ∑𝑥𝑥=0𝑛𝑛 𝑥𝑥 − �̅�𝑥 2𝑃𝑃(𝑥𝑥) = 𝑛𝑛𝑛𝑛 = �̅�𝑥

10

• Example: Asking 1,000 people, “Is your birthday today?”

– n = 1000, p = 1/365 = 0.00274 << 1

– Average number of people who say “Yes”

• �̅�𝑥 = 𝑛𝑛𝑛𝑛 = 2.74,𝜎𝜎 = �̅�𝑥 = 1.66

– What is the prob. that four people say “Yes”?

• 𝑃𝑃 𝑥𝑥 = 4 = (2.74)4𝑒𝑒−2.74

4!= 0.152

– Note that the distribution is asymmetric

11

The Gaussian or Normal distribution

• If p << 1 & n >> 1 or the mean is large (≥ 25 or 30),

– 𝑃𝑃 𝑥𝑥 = 12𝜋𝜋�̅�𝑥

𝑒𝑒𝑥𝑥𝑛𝑛 − (𝑥𝑥−�̅�𝑥)2

2�̅�𝑥

• Properties of the Gaussian distribution

– Normalization: ∑𝑥𝑥=0𝑛𝑛 𝑃𝑃(𝑥𝑥) = 1

– Mean (or average) value: �̅�𝑥 = 𝑛𝑛𝑛𝑛

– Predicted variance: 𝜎𝜎2 = 𝑛𝑛𝑛𝑛 = �̅�𝑥

12

• For the same previous example but w/ n = 10,000 people

– �̅�𝑥 = 27.4 > 25 =>reduced to the Gaussian distribution

– 𝜎𝜎 = �̅�𝑥 = 5.23

– e.g., 𝑃𝑃 𝑥𝑥 = 30 = 12𝜋𝜋(27.4)

𝑒𝑒𝑥𝑥𝑛𝑛 − (30−27.4)2

2(27.4)= 0.0673

– Note that the distribution is symmetric

• Continuous Gaussian distribution– Prob. of observing a deviation in 𝑑𝑑𝜖𝜖 about 𝜖𝜖(= 𝑥𝑥 − �̅�𝑥 )

– 𝐺𝐺 𝜖𝜖 d𝜖𝜖 = 22𝜋𝜋�̅�𝑥

𝑒𝑒𝑥𝑥𝑛𝑛 − 𝜖𝜖2

2�̅�𝑥d𝜖𝜖

13

14

• Universal Gaussian distribution

– Letting 𝑡𝑡 ≡ 𝜖𝜖𝜎𝜎 then 𝐺𝐺 𝑡𝑡 = 𝐺𝐺 𝜖𝜖 d𝜖𝜖

d𝑡𝑡 = 𝐺𝐺(𝜖𝜖)𝜎𝜎

– 𝐺𝐺 𝑡𝑡 d𝑡𝑡 = 2𝜋𝜋𝑒𝑒− ⁄𝑡𝑡2 2d𝑡𝑡 where 0 ≤ 𝑡𝑡 ≤ ∞

– The prob. that a random sample from a Gaussian distribution will show a normalized deviation 𝑡𝑡(= 𝜖𝜖

𝜎𝜎)

that is less than the assumed value 𝑡𝑡0

• 𝑓𝑓 𝑡𝑡0 = ∫0𝑡𝑡0 𝐺𝐺 𝑡𝑡 d𝑡𝑡

Prob. of occurrence of given deviations predicted by the Gaussian distribution

𝜖𝜖0 𝑡𝑡0 𝑓𝑓(𝑡𝑡0)

00.674σσ1.64σ1.96σ2.58σ3.00σ

00.6741.001.641.962.583.00

00.5000.6830.9000.9500.9900.997

15

APPLICATIONS OF STATISTICAL MODELS

• Why to apply the counting statistics in nuclear measurements?

– To determine whether a set of multiple measurements of the same physical quantity shows an amount of internal fluctuation that is consistent w/ statistical predictions

• e.g., A common quality control procedure in many counting laboratories– Periodically (perhaps once a month) records a series of 20-50 successive

counts from the detector syst.– Observe whether abnormal amounts of fluctuation can be detected that

could indicate malfunctioning of some portion of the counting syst.– See the abnormal fluctuation of data (sample variance) in comparison with

the statistical (predicted) variance» Abnormally small: systematically wrong» Abnormally large: terribly noisy

– To make prediction about the uncertainty one should associate w/ that single measurement to account for the unavoidable effects of statistical fluctuations

16

Checkout of the counting syst. to see whether observed fluctuations are consistent w/ expected statistical fluctuation

• Inspection of a set of data for consistency w/ a statistical model

“Qualitative” graphical test

“Quantitative” χ2 test to determineif the observed difference (btwn 𝑠𝑠2 & 𝜎𝜎2) is really significant

17

• Graphical test

– Plot the data distribution function w/ the predicted prob. function

• In this example, assuming that �̅�𝑥 = �̅�𝑥𝑒𝑒, the statistical model is chosen to be the Poisson distribution because �̅�𝑥 = 8.8 < 25

– Compare two distributions & see deviations

18

• Comparison of the sample variance w/ the predicted variance from the statistical model

– 𝑠𝑠2 = 7.36 vs. 𝜎𝜎2 = 8.80– Difficult to say whether the observed difference is really significant because of a

limited sample size (N = 20)

• Chi-squared test– Find the prob. 𝒑𝒑 that a random sample from a true Poisson distri. would have a

larger value of 𝜒𝜒2 than a specific value by using tables or graphs• 𝑛𝑛 ≤ 0.02: abnormally large fluctuations in the data (the syst. is terribly noisy)• 𝑛𝑛 ≥ 0.98: abnormally small fluctuations in the data (the syst. is in failure)• 𝑛𝑛 = 0.50: perfect fit to the Poisson distri. for large samples

– 𝜒𝜒2 = 1�̅�𝑥𝑒𝑒∑𝑖𝑖=1𝑁𝑁 (𝑥𝑥𝑖𝑖 − �̅�𝑥𝑒𝑒)2 = (𝑁𝑁−1)𝑠𝑠2

�̅�𝑥𝑒𝑒= 𝜈𝜈𝑠𝑠2

�̅�𝑥𝑒𝑒

19

• Example– Degree of freedom 𝜈𝜈 = 𝑁𝑁 − 1 = 19

– 𝜒𝜒2 = 𝜈𝜈𝑠𝑠2

�̅�𝑥𝑒𝑒= 15.89

– 𝜒𝜒2

𝜈𝜈= 0.84

20

21

Estimation of the precision of a single measurement

• The expected sample variance 𝑠𝑠2

– ≅ 𝜎𝜎2 of the statistical model from which we think the measurement 𝑥𝑥 is drawn

– = �̅�𝑥 provided the model is either Poisson or Gaussian

– ≅ 𝑥𝑥 because 𝑥𝑥 is our only measurement on which to base an estimate of �̅�𝑥

22

• 𝑠𝑠2 ≅ 𝜎𝜎 = 𝑥𝑥– Our best estimate of the deviation from the true mean that should typify our single

measurement 𝑥𝑥– 𝑥𝑥 ± 𝑥𝑥 will contain the true mean �̅�𝑥 with 68% probability

• Example

Examples of error intervals for a single measurement of 𝑥𝑥 = 100

IntervalProb. that the true mean

�̅�𝑥 is included

𝑥𝑥 ± 0.67𝜎𝜎𝑥𝑥 ± 𝜎𝜎

𝑥𝑥 ± 1.64𝜎𝜎𝑥𝑥 ± 2.58𝜎𝜎

93.3 – 106.790.0 – 110.083.6 – 116.474.2 – 125.8

50%68%90%99%

23

• Graphical display of error bars associated w/ experimental data

• In radioactive decay, or counting experiments– If the T1/2 is long, then p << 1 ==> Poisson– If the counted number x > 20 ==> Gaussian

==>so we can apply the error formula, 𝜎𝜎 = 𝑥𝑥

• Caution !!!– This definition of error 𝜎𝜎 = 𝑥𝑥 is applied only to the counts (of radiation by a

detector for a given time)– It is not applicable to any other derived quantities

• e.g.) counting rates, sums or differences of counts, averages of independent counts, & any derived quantities

24

ERROR PROPAGATION

• Let’s measure # of quanta emitted by a radioisotope source w/ some assumptions:

– Counts follow the Gaussian distribution => Then, we need only one parameter to fully specify the distribution (what is it?)

– Using a background-free detector system– Absolute detection efficiency of the syst. = 50%

• From the repeated measurements we can obtain a new distribution– The mean value of # of quanta from the source = 2 × the avg. # of counts, but what

will be its shape and σ?

– The derived distri. will be of Gaussian shape provided the distri. of the original data is also Gaussian

– If the errors are individually small & symmetric about zero, a general result can be obtained for the expected error to be associated w/ any quantity that is calculated as a function of any number of independent variables

– The variables should be truly independent in order to avoid the effects of correlation

25

Error propagation formula

• For a derived quantity 𝑢𝑢 = 𝑢𝑢 𝑥𝑥,𝑦𝑦, 𝑧𝑧,⋯

– 𝜎𝜎𝑢𝑢2 = 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

2𝜎𝜎𝑥𝑥2 + 𝜕𝜕𝑢𝑢

𝜕𝜕𝑦𝑦

2𝜎𝜎𝑦𝑦2 + 𝜕𝜕𝑢𝑢

𝜕𝜕𝑧𝑧

2𝜎𝜎𝑧𝑧2 + ⋯

• 𝑥𝑥,𝑦𝑦, 𝑧𝑧,⋯ = directly measured variables (e.g., counts)• 𝜎𝜎𝑥𝑥 ,𝜎𝜎𝑦𝑦,𝜎𝜎𝑧𝑧 ⋯ = the standard deviations of variables

26

Sums or differences of counts

• 𝑢𝑢 = 𝑥𝑥 ± 𝑦𝑦

– 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

= 1, 𝜕𝜕𝑢𝑢𝜕𝜕𝑦𝑦

= ±1

– 𝜎𝜎𝑢𝑢2 = 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

2𝜎𝜎𝑥𝑥2 + 𝜕𝜕𝑢𝑢

𝜕𝜕𝑦𝑦

2𝜎𝜎𝑦𝑦2 = 𝜎𝜎𝑥𝑥2 + 𝜎𝜎𝑦𝑦2

• 𝜎𝜎𝑢𝑢 = 𝜎𝜎𝑥𝑥2 + 𝜎𝜎𝑦𝑦2 called “combining in quadrature” or taking their “quadrature

sum”

– If each variable follows the Gaussian distri.,

• 𝜎𝜎𝑢𝑢 = 𝑥𝑥 + 𝑦𝑦

27

• Example: “net counts = total counts – background counts”– Total counts = x = 1071– Background counts = y = 521– Net counts = u = x – y = 550

– 𝜎𝜎𝑢𝑢 = 𝑥𝑥 + 𝑦𝑦 = 1071 + 521 = 39.9

– Therefore, the net counts = 550 ± 39.9 counts

28

Multiplication or division by a constant

• 𝑢𝑢 = 𝐴𝐴𝑥𝑥

– 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

= 𝐴𝐴

– 𝜎𝜎𝑢𝑢2 = 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

2𝜎𝜎𝑥𝑥2 = 𝐴𝐴2𝜎𝜎𝑥𝑥2 = 𝐴𝐴2𝑥𝑥

– 𝜎𝜎𝑢𝑢 = 𝐴𝐴𝜎𝜎𝑥𝑥 = 𝐴𝐴 𝑥𝑥

– 𝜎𝜎𝑢𝑢𝑢𝑢

= 𝜎𝜎𝑥𝑥𝑥𝑥

= 1𝑥𝑥

– No change in fractional error

• 𝑣𝑣 = 𝑥𝑥𝐵𝐵

– 𝜕𝜕𝑣𝑣𝜕𝜕𝑥𝑥

= 1𝐵𝐵

– 𝜎𝜎𝑣𝑣2 = 𝜕𝜕𝑣𝑣𝜕𝜕𝑥𝑥

2𝜎𝜎𝑥𝑥2 = 1

𝐵𝐵2𝜎𝜎𝑥𝑥2 = 𝑥𝑥

𝐵𝐵2

– 𝜎𝜎𝑣𝑣 = 𝜎𝜎𝑥𝑥𝐵𝐵

= 𝑥𝑥𝐵𝐵

– 𝜎𝜎𝑣𝑣𝑣𝑣

= 𝜎𝜎𝑥𝑥𝑥𝑥

= 1𝑥𝑥

– No change in fractional error

29

• Example: “counting rate = r = x/t”– x = 1120 counts– t = 5 s– r = 1120/5 = 224 s-1

– 𝜎𝜎𝑟𝑟 = 𝜎𝜎𝑥𝑥𝑡𝑡

= 11205

= 6.7 s−1

– Therefore, the counting rate = 224 ± 6.7 counts/s

30

Multiplication or division of counts

• 𝑢𝑢 = 𝑥𝑥𝑦𝑦

– 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

= 𝑦𝑦 𝜕𝜕𝑢𝑢𝜕𝜕𝑦𝑦

= 𝑥𝑥

– 𝜎𝜎𝑢𝑢2 = 𝑦𝑦2𝜎𝜎𝑥𝑥2 + 𝑥𝑥2𝜎𝜎𝑦𝑦2

– 𝜎𝜎𝑢𝑢𝑢𝑢

2= 𝜎𝜎𝑥𝑥

𝑥𝑥

2+ 𝜎𝜎𝑦𝑦

𝑦𝑦

2

• 𝑢𝑢 = 𝑥𝑥𝑦𝑦

– 𝜕𝜕𝑢𝑢𝜕𝜕𝑥𝑥

= 1𝑦𝑦

𝜕𝜕𝑢𝑢𝜕𝜕𝑦𝑦

= − 𝑥𝑥𝑦𝑦2

– 𝜎𝜎𝑢𝑢2 = 1𝑦𝑦

2𝜎𝜎𝑥𝑥2 + − 𝑥𝑥

𝑦𝑦22𝜎𝜎𝑦𝑦2

– 𝜎𝜎𝑢𝑢𝑢𝑢

2= 𝜎𝜎𝑥𝑥

𝑥𝑥

2+ 𝜎𝜎𝑦𝑦

𝑦𝑦

2

31

• Example: calculate the ratio of two source activities from independent counts taken for equal counting times (background is neglected)

– Counts from source 1 = N1 = 16,265– Counts from source 2 = N2 = 8,192– Activity ratio = R = N1/N2 = 1.985

– 𝜎𝜎𝑅𝑅𝑅𝑅

2=

𝜎𝜎𝑁𝑁1𝑁𝑁1

2

+𝜎𝜎𝑁𝑁2𝑁𝑁2

2

= 1.835 × 10−4

– 𝜎𝜎𝑅𝑅𝑅𝑅

= 0.0135

– 𝜎𝜎𝑅𝑅 = 𝑅𝑅 × 0.0135 = 0.027

– Therefore, R = 1.985 ± 0.027

32

Quiz (take home)

• Given– Counts for unknown activity source, x = 687– Counts for known activity source, y = 436– Count for background for equal time, z = 315

– Activity of known source, Ay = 625 kBq– Relative error of activity of known source = 10%

• Find the activity of the unknown source, Ax

– Hint: 𝐴𝐴𝑥𝑥 = 𝐴𝐴𝑦𝑦 × 𝑥𝑥−𝑧𝑧𝑦𝑦−𝑧𝑧

33

Mean value of multiple independent counts

• Example: 10 indep. counts per 1 min each

– Does �̅�𝑥 or the average value of 𝑥𝑥𝑖𝑖 correctly designate the error of the average count?

Count, 𝑥𝑥𝑖𝑖 Error, 𝑥𝑥𝑖𝑖

3003302129952965304530072990298230623037

54.855.054.754.555.254.854.754.655.355.1

Σ, Σ 30107 173.5

�̅�𝑥, �̅�𝑥 3010.7 54.87

⁄�̅�𝑥 𝑁𝑁 17.35

34

• Standard deviation of the averaged value

– Standard deviation (error, uncertainty) of the average is not just the square root of the value of average

– Average of counts is a derived quantity

– It means that “if you do the similar 10 measurements, its new average value will be within that error by the probability of 68.3 %”

– Similarly, for single count, a new count will be within ...

Count, 𝑥𝑥𝑖𝑖 Error, 𝑥𝑥𝑖𝑖

3003302129952965304530072990298230623037

54.855.054.754.555.254.854.754.655.355.1

Σ 30107 173.5

�̅�𝑥,𝜎𝜎𝑥𝑥𝑖𝑖 3010.7 17.35

35

• Suppose we have recorded N repeated counts from the same source for equal counting times

– Σ = 𝑥𝑥1 + 𝑥𝑥2 + ⋯+ 𝑥𝑥𝑁𝑁– 𝜎𝜎Σ2 = 𝜎𝜎𝑥𝑥1

2 + 𝜎𝜎𝑥𝑥22 + ⋯+ 𝜎𝜎𝑥𝑥𝑁𝑁

2 = 𝑥𝑥1 + 𝑥𝑥2 + ⋯+ 𝑥𝑥𝑁𝑁 = Σ

– 𝜎𝜎Σ = Σ

– The mean value of N indep. measurements, �̅�𝑥 = Σ𝑁𝑁

– 𝜎𝜎�̅�𝑥 = 𝜎𝜎Σ𝑁𝑁

= Σ𝑁𝑁

= 𝑁𝑁�̅�𝑥𝑁𝑁

= �̅�𝑥𝑁𝑁

• It is improved by a factor of 1𝑁𝑁

compared to a single measurement

• Because any typ. count will not differ greatly from the mean, 𝑥𝑥𝑖𝑖 ≅ �̅�𝑥, the mean value based on N indep. counts will have an expected error that is smaller by a factor 𝑁𝑁 compared with any single measurement

• Therefore, we wish to improve the statistical precision of a given measurement by a factor of 2, we must invest four times the initial counting time

36

Combination of indep. measurements w/ unequal errors

• A simple average no longer is the optimal way for N indp. measurements of the same quantity with different precision at each measurement

• Give more weight to measurements w/ small error & less weight to measurements w/ large error

– 𝑥𝑥 = ∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖𝑥𝑥𝑖𝑖∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖

, where 𝑎𝑎𝑖𝑖 = a weighting factor

– Letting α ≡ ∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖 and seek a criterion by which 𝑎𝑎𝑖𝑖 should be chosen in order to minimize the expected error in 𝑥𝑥

– 𝜎𝜎 𝑥𝑥2 = ∑𝑖𝑖=1𝑁𝑁 𝜕𝜕 𝑥𝑥

𝜕𝜕𝑥𝑥𝑖𝑖

2𝜎𝜎𝑥𝑥𝑖𝑖2 = ∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖

𝛼𝛼

2𝜎𝜎𝑥𝑥𝑖𝑖2 = 1

𝛼𝛼2∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖2𝜎𝜎𝑥𝑥𝑖𝑖

2 ≡ 𝛽𝛽𝛼𝛼2

– 0 =𝜕𝜕𝜎𝜎 𝑥𝑥

2

𝜕𝜕𝑎𝑎𝑗𝑗=

𝛼𝛼2 𝜕𝜕𝛽𝛽𝜕𝜕𝑎𝑎𝑗𝑗−2𝛼𝛼𝛽𝛽 𝜕𝜕𝛼𝛼

𝜕𝜕𝑎𝑎𝑗𝑗

𝛼𝛼4⇒ 𝑎𝑎𝑗𝑗 = 𝛽𝛽

𝛼𝛼� 1𝜎𝜎𝑥𝑥𝑗𝑗2

– If α ≡ ∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖 = 1 (normalization), 𝑎𝑎𝑗𝑗 = 𝛽𝛽𝜎𝜎𝑥𝑥𝑗𝑗2

– Then, β = ∑𝑖𝑖=1𝑁𝑁 𝑎𝑎𝑖𝑖2𝜎𝜎𝑥𝑥𝑖𝑖2 = ∑𝑖𝑖=1𝑁𝑁 𝛽𝛽

𝜎𝜎𝑥𝑥𝑖𝑖2

2

𝜎𝜎𝑥𝑥𝑖𝑖2 or β = ∑𝑖𝑖=1𝑁𝑁 1

𝜎𝜎𝑥𝑥𝑖𝑖2

−1

37

• Therefore,

– 𝑎𝑎𝑗𝑗 = 1𝜎𝜎𝑥𝑥𝑗𝑗2 ∑𝑖𝑖=1𝑁𝑁 1

𝜎𝜎𝑥𝑥𝑖𝑖2

−1

• Each data point should be weighted inversely as the square of its own error

– 𝜎𝜎 𝑥𝑥2 = 𝛽𝛽

𝛼𝛼2= 𝛽𝛽 for the normalized weighting coefficient

– Hence, the min. error in 𝑥𝑥 due to the optimal weighting: 1𝜎𝜎 𝑥𝑥2 = ∑𝑖𝑖=1𝑁𝑁 1

𝜎𝜎𝑥𝑥𝑖𝑖2

38

OPTIMIZATION OF COUNTING EXPERIMENTS

• Design of counting experiments to minimize the associated statistical uncertainty based on the principle of error propagation

• Consider the net counting rate from a long-lived radioactive source in the presence of a steady-state background

– S = counting rate due to the source alone w/o background– B = counting rate due to bgnd (= N2/TB)– S+B = N1/TS+B

– N1 = tot. counts obtained from counting the source + bgnd for a time TS+B

– N2 = tot. counts obtained from counting the bgnd alone for a time TB

• Then, we can have

– 𝑆𝑆 = 𝑁𝑁1𝑇𝑇𝑆𝑆+𝐵𝐵

− 𝑁𝑁2𝑇𝑇𝐵𝐵

– 𝜎𝜎𝑆𝑆 =𝜎𝜎𝑁𝑁1𝑇𝑇𝑆𝑆+𝐵𝐵

2

+𝜎𝜎𝑁𝑁2𝑇𝑇𝐵𝐵

2

= 𝑁𝑁1𝑇𝑇𝑆𝑆+𝐵𝐵2 + 𝑁𝑁2

𝑇𝑇𝐵𝐵2 = 𝑆𝑆+𝐵𝐵

𝑇𝑇𝑆𝑆+𝐵𝐵+ 𝐵𝐵

𝑇𝑇𝐵𝐵

39

Optimal allocation of time

• Assuming that a fixed time T = TS+B +TB, what is the optimal fraction of T allocated to TS+B or TB?

– Minimizing the uncertainty

• 2𝜎𝜎𝑆𝑆d𝜎𝜎𝑆𝑆 = − 𝑆𝑆+𝐵𝐵𝑇𝑇𝑆𝑆+𝐵𝐵2 d𝑇𝑇𝑆𝑆+𝐵𝐵 −

𝐵𝐵𝑇𝑇𝐵𝐵2 d𝑇𝑇𝐵𝐵

• d𝜎𝜎𝑆𝑆 = 0 (for optimization) & d𝑇𝑇𝑆𝑆+𝐵𝐵 + d𝑇𝑇𝐵𝐵 = 0 (∵ 𝑇𝑇 = 𝑐𝑐𝑐𝑐𝑛𝑛𝑠𝑠𝑡𝑡)

– Then, we can have

• �𝑇𝑇𝑆𝑆+𝐵𝐵𝑇𝑇𝐵𝐵 𝑜𝑜𝑝𝑝𝑡𝑡

= 𝑆𝑆+𝐵𝐵𝐵𝐵

40

Quiz

• If the ratio of the count rate of source and background is about 3 to 1, what is the optimal ratio of measurement time for minimization of the error in measuring source activity?

41

• For the accurate measurement of source rate, the measurement time should be long!

• Figure of merit:

– 1𝑇𝑇

= 𝜖𝜖2 𝑆𝑆2

( 𝑆𝑆+𝐵𝐵+ 𝐵𝐵)2

• Two extreme cases:

– Case 1) If 𝑆𝑆 ≫ 𝐵𝐵:

• 1𝑇𝑇≅ 𝜖𝜖2𝑆𝑆 => choosing exp. parameters to maximize 𝑆𝑆 for max. FOM

– Case 1) If 𝐵𝐵 ≫ 𝑆𝑆 (low-level radioactivity measurements):

• 1𝑇𝑇≅ 𝜖𝜖2 𝑆𝑆2

4𝐵𝐵=> maximizing 𝑆𝑆

2

𝐵𝐵(detector design)

Derive! (Take home)

42

LIMITS OF DETECTABILITY

• When we want to measure the small activity of a source, the background will hinder. So we want to set the trigger level (a specified count for a given time = specified count rate) high enough to minimize the background count.[ Hardware Setting Value ]

• When we invent or use a counting system, we should be able to tell the minimum detectable level or trustable count (or count rate), which is strongly related to the background (or blank source) count of the system.[ Value in Spec. Report ]

43

• Counting-syst. performance is based on a binary (yes or no) decision whether a given detector response is representative of bgnd, or alternatively whether that output indicates presence of a source above bgnd

• Definitions:– 𝑃𝑃𝐷𝐷 = detection prob. = the prob. of detecting a weak signal– 𝑃𝑃𝐹𝐹𝐹𝐹 = false alarm prob. = the prob. that a syst. will erroneously indicate source

detection due to statistical fluctuations in the bgnd or expected syst. fluctuations– 𝑃𝑃𝐹𝐹𝑁𝑁 = false negatives = no alarm when a source is present– 𝑃𝑃𝐹𝐹𝑃𝑃 = false positives = alarm when a source is absent– 𝑁𝑁𝑇𝑇 = # of recorded counts w/ a sample (S + B)– 𝑁𝑁𝐵𝐵 = # of recorded counts w/ a blank sample (B only)– 𝐿𝐿𝐶𝐶 = critical level (or threshold or alarm level) = representative of bgnd if # of counts

is < 𝐿𝐿𝐶𝐶 ; presence of a source if # of counts is > 𝐿𝐿𝐶𝐶

44

ROC curves

• Receiver operating characteristic curves: the relationship btwn 𝑃𝑃𝐷𝐷 & 𝑃𝑃𝐹𝐹𝐹𝐹

– Plot 𝑃𝑃𝐷𝐷 as a function of 𝑃𝑃𝐹𝐹𝐹𝐹 by varying 𝐿𝐿𝐶𝐶 from several std’s below �𝑁𝑁𝐵𝐵 to several std’sabove �𝑁𝑁𝑇𝑇

• 𝑃𝑃𝐷𝐷 = ∫𝐿𝐿𝐶𝐶∞𝑃𝑃(𝑁𝑁𝑇𝑇)d𝑁𝑁𝑇𝑇

• 𝑃𝑃𝐹𝐹𝐹𝐹 = ∫𝐿𝐿𝐶𝐶∞𝑃𝑃(𝑁𝑁𝐵𝐵)d𝑁𝑁𝐵𝐵

• 𝑃𝑃𝐹𝐹𝑁𝑁 = ∫0𝐿𝐿𝐶𝐶 𝑃𝑃(𝑁𝑁𝑇𝑇)d𝑁𝑁𝑇𝑇 = 1 − 𝑃𝑃𝐷𝐷

Prob. distributions

𝑃𝑃𝐹𝐹𝐹𝐹 𝑃𝑃𝐷𝐷

Best 𝐿𝐿𝐶𝐶

45

Minimum detectable amount (MDA)

• Curie definition of MDA– Specific choices (e.g., 𝐿𝐿𝐶𝐶 , 𝑁𝑁𝐷𝐷) to ensure that 𝑃𝑃𝐷𝐷 = 95% & 𝑃𝑃𝐹𝐹𝐹𝐹 = 5%

• 𝑁𝑁𝐷𝐷 = the min. mean number of counts needed from the source to ensure a false-negative rate no larger than 5% when the syst. is operated w/ 𝐿𝐿𝐶𝐶 that, in turn, ensures a false-positive rate of no greater than 5%

46

Setting of critical level

• Under condition of equal measurement time of source and bgnd• Under condition that “no real activity is present” (𝑁𝑁𝑆𝑆 = 0 & 𝜎𝜎𝑁𝑁𝑇𝑇 = 𝜎𝜎𝑁𝑁𝐵𝐵)

– 𝑁𝑁𝑆𝑆 = 𝑁𝑁𝑇𝑇 − 𝑁𝑁𝐵𝐵 = 0– 𝜎𝜎𝑁𝑁𝑆𝑆

2 = 𝜎𝜎𝑁𝑁𝑇𝑇2 + 𝜎𝜎𝑁𝑁𝐵𝐵

2 = 2𝜎𝜎𝑁𝑁𝐵𝐵2

– If the only significant fluctuations entering the measurement are from counting statistics,

• 𝜎𝜎𝑁𝑁𝑆𝑆2 = 2𝜎𝜎𝑁𝑁𝐵𝐵

2 = 2𝑁𝑁𝐵𝐵

– Set 𝐿𝐿𝐶𝐶 to ensure that 𝑃𝑃𝐹𝐹𝑃𝑃 will be no larger than 5%

• 𝐿𝐿𝐶𝐶 = 1.64𝜎𝜎𝑁𝑁𝑆𝑆 = 1.64 2𝜎𝜎𝑁𝑁𝐵𝐵 = 2.33𝜎𝜎𝑁𝑁𝐵𝐵 = 2.33 𝑁𝑁𝐵𝐵

47

Setting of MDA

• Let’s define the MDA level (𝑁𝑁𝐷𝐷) to be the source strength necessary to produce a mean value of 𝑁𝑁𝑆𝑆 that is high enough to reduce the false-negative rate to an acceptable level (𝑃𝑃𝐹𝐹𝑁𝑁 = 0.5 from the Curie definition)

• Equivalently, set 𝑁𝑁𝐷𝐷 to ensure that 95% of the area under the 𝑁𝑁𝑆𝑆 distri. lies above 𝐿𝐿𝐶𝐶 ;

– 𝑁𝑁𝐷𝐷 = 𝐿𝐿𝐶𝐶 + 1.64𝜎𝜎𝑁𝑁𝐷𝐷

– 𝜎𝜎𝑁𝑁𝐷𝐷2 = 𝜎𝜎𝑁𝑁𝑇𝑇

2 + 𝜎𝜎𝑁𝑁𝐵𝐵2 = 𝑁𝑁𝑇𝑇 + 𝑁𝑁𝐵𝐵 = 𝑁𝑁𝑇𝑇 − 𝑁𝑁𝐵𝐵 + 2𝑁𝑁𝐵𝐵 = 𝑁𝑁𝐷𝐷 + 2𝑁𝑁𝐵𝐵

– 𝑁𝑁𝐷𝐷 = 𝐿𝐿𝐶𝐶 + 1.64𝜎𝜎𝑁𝑁𝐷𝐷 = 1.64 2𝑁𝑁𝐵𝐵 + 1.64 𝑁𝑁𝐷𝐷 + 2𝑁𝑁𝐵𝐵 ⇒ 𝑁𝑁𝐷𝐷 = 4.65 𝑁𝑁𝐵𝐵 + 2.71• “Curie equation”: the min. mean number of counts needed from the source to

ensure a false-negative rate no larger than 5% when the syst. is operated w/ 𝐿𝐿𝐶𝐶(“trigger point”) that, in turn, ensures a false-positive rate of no greater than 5%

48

• 𝑁𝑁𝐷𝐷 in counts to min. detectable activity 𝛼𝛼

– 𝛼𝛼 = 𝑁𝑁𝐷𝐷𝑓𝑓𝜖𝜖𝑇𝑇

• 𝑓𝑓 = radiation yield per disintegration• 𝜖𝜖 = absolute detection efficiency• 𝑇𝑇 = counting time per sample

49

DISTRIBUTION OF TIME INTERVALS

• Poisson random process– A memoryless (or independent) process with a constant probability of event

occurrence per unit time– 𝑟𝑟d𝑡𝑡 = the diff’l prob. of occurrence of an event w/i a diff’l time interval d𝑡𝑡

• 𝑟𝑟 = the average rate of occurrence• 𝑟𝑟d𝑡𝑡 = 𝑟𝑟𝑇𝑇 = the avg. # of events occurring for a finite time interval 𝑇𝑇

• Radioactive decay of any amount of radioactive material (a series of events) can be classified statistically as a Poisson random process, because the probability of event occurrence is constant

• Radiation counting (especially for a short time) can be considered also as a Poisson random process, for the similar reason

50

Intervals btwn successive events

• Time interval, t, between any two events in counting is a random variable

• The probability of observing an interval between two pulses with a length between “t” and “t+dt”

51

• Derive a distri. function to describe the time intervals btwn adjacent random events assuming that an event has occurred at time t =0

• 𝑃𝑃 0 = (𝑟𝑟𝑡𝑡)0𝑒𝑒−𝑟𝑟𝑡𝑡

0!= 𝑒𝑒−𝑟𝑟𝑡𝑡

– 𝐼𝐼1 𝑡𝑡 d𝑡𝑡 = 𝑟𝑟𝑒𝑒−𝑟𝑟𝑡𝑡d𝑡𝑡• 𝐼𝐼1 𝑡𝑡 = the distri. function for intervals btwn adjacent random events• Most probable interval = 0

• Avg. interval length, ̅𝑡𝑡 = ∫0∞ 𝑡𝑡𝐼𝐼1 𝑡𝑡 d𝑡𝑡

∫0∞ 𝐼𝐼1 𝑡𝑡 d𝑡𝑡

= ∫0∞ 𝑡𝑡𝑟𝑟𝑒𝑒−𝑟𝑟𝑡𝑡d𝑡𝑡 = 1

𝑟𝑟

prob. of next event taking place in d𝑡𝑡after delay of 𝑡𝑡

𝐼𝐼1 𝑡𝑡 d𝑡𝑡

prob. of no events during

time from 0 to 𝑡𝑡𝑃𝑃(0)

prob. of an event

during d𝑡𝑡𝑟𝑟d𝑡𝑡

= ×

52

Time-to-next-event

• Randomly selected intervals

– Selecting a random point in time is not equivalent to randomly selecting an interval from the distribution function of 𝐼𝐼1 𝑡𝑡 d𝑡𝑡

– Selecting a random point in time produces another distribution of intervals• Randomly selected time will more likely fall within a long interval than a short

one• Therefore, the distribution has more of long intervals

– Selected interval is defined due to selecting a random point in time

• 𝐼𝐼𝑆𝑆 𝑡𝑡 = 𝑡𝑡̅𝑡𝑡𝐼𝐼1 𝑡𝑡 d𝑡𝑡 = 𝑟𝑟𝑡𝑡𝐼𝐼1 𝑡𝑡 d𝑡𝑡 = 𝑟𝑟2𝑡𝑡𝑒𝑒−𝑟𝑟𝑡𝑡d𝑡𝑡

• Most probable interval = 0

• Avg. selected interval, ̅𝑡𝑡 = 2𝑟𝑟

53

Intervals btwn scaled events

• A “scaler” is a useful device to reduce the rate at which data are recorded from a detector syst. by producing an output pulse only when N input pulses have been accumulated

• Distri. function for “N-scaled intervals”:

– 𝐼𝐼𝑁𝑁 𝑡𝑡 d𝑡𝑡 = 𝑃𝑃 𝑁𝑁 − 1 × 𝑟𝑟d𝑡𝑡 = (𝑟𝑟𝑡𝑡)𝑁𝑁−1𝑒𝑒−𝑟𝑟𝑡𝑡

(𝑁𝑁−1)!𝑟𝑟d𝑡𝑡

– Most probable interval, |𝑟𝑟 𝑚𝑚𝑜𝑜𝑠𝑠𝑡𝑡 𝑝𝑝𝑟𝑟𝑜𝑜𝑝𝑝𝑎𝑎𝑝𝑝𝑝𝑝𝑒𝑒 = 𝑁𝑁−1𝑟𝑟

from d𝐼𝐼𝑁𝑁 𝑡𝑡d𝑡𝑡

= 0

– Avg. N-scaled interval, ̅𝑡𝑡 = 𝑁𝑁𝑟𝑟

54