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Statistics Decay Probability • Radioactive decay is a statistical process. – Assume N large for continuous function • Express problem in terms of probabilities for a single event. – Probability of decay p – Probability of survival q – Time dependent N N 0 e t p 1 e t q e t p q 1 Combinatorics • The probability that n specific occurrences happen is the product of the individual occurrences. – Other events don’t matter. – Separate probability for negative events • Arbitrary choice of events require permutations. • Exactly n specific events happen at p: P pn • No events happen except the specific events: N n Pq • Select n arbitrary events from a pool of N identical types. N N! n n!( N n)! Bernoulli Process • Treat events as a discrete trials. – N separate trials – Trials independent – Binary outcome of trial – Probability same for all trials. • This defines a Bernoulli process. Typical Problem • 10 atoms of 42K with a half-life of 12.4 h is observed for 3 h. What is the probability that exactly 3 atoms decay? Answer • Probability of 1 decay, p 1 e t 1 e(t ln 2) / T 0.154 • And 3 arbitrary atoms decay from the 10 and 7 do not: 10 3 7 P p q 0.136 3 Binomial Distribution • The general form of the Bernoulli process is the binomial distribution. – Terms same as binomial expansion N n N n Pn p q n • Probabilities are normalized. N P n 0 n ( p q) N 1 mathworld.wolfram.com Mean and Standard Deviation • The mean m of the binomial distribution: N N N n N n m nPn n p q n 0 n 0 n • The standard deviation s of the binomial distribution: • Consider an arbitrary x, and differentiate, and set x = 1. N N n n N n N ( px q) p x q n 0 n s 2 (n2 Pn 2mnPn m 2 Pn ) N Np( px q ) N 1 nx n 1 Pn n 0 N s n m 2 Pn 2 n 0 s 2 n2 Pn 2m nPn m 2 Pn s 2 [ N ( N 1) p 2 m ] 2m 2 m 2 s 2 N 2 p 2 Np 2 Np N 2 p 2 N Np nPn m n 0 s Np(1 p) Npq Disintegration Counts • In counting experiments there is a factor for efficiency e. – Probability that a measurement is recorded p e (1 e t ) q 1 p 1 e ee t Typical Problem • A sample has 10 atoms of 42K in an experiment with e = 0.32. What is the expected count rate over 3 h? Answer • Use the mean of the observable count, convert to rate. rc m / t Ne (1 e t ) / t • 10(0.32)(0.154)/3 h = 0.163 h-1. More Counts • Consider a source of 42K with an activity of 37 Bq, in a counter with e = 0.32 measured in 1 s intervals. • The mean disintegration rate is just the activity, rd = 37 Bq. – The count rate is • What is the mean count rate? • Decay constant is = ln2 / T = 0.056 h-1 = 1.55 x 10-5 s-1. – The probability of decay is • What is the standard deviation of the count rate? rc erd 11.8 s 1 p 1 e t 1.55 105 • Number of atoms is N = rd / = 2.4 x 106. s c Nep(1 ep) 3.4 s1 Poisson Distribution • Many processes have a a large pool of possible events, but a rare occurrence for any individual event. – Large N, small n, small p N N! Nn n n!( N n)! n! • This is the Poisson distribution. – Probability depends on only one parameter Np – Normalized when summed from n =0 to . q N n q N (1 p) N N ( N 1) 2 p 2! ( Np) 2 1 Np e Np 2! q N n 1 Np q N n N n N n ( Np) n Np Pn p q e n! n Poisson Properties • The mean and standard deviation are simply related. – Mean m = Np, standard deviation s2 = m, s m • Unlike the binomial distribution the Poisson function has values for n > N. Poisson Away From Zero • The Poisson distribution is based on the mean m = Np. – Assumed N >> 1, N >> n. • Now assume that n >> 1, m large and Pn >> 0 only over a narrow range. • This generates a normal distribution. • Let x = n – m. m m xe m m m m xe m Px ( m x)! m![( m x)! / m!] • Use Stirling’s formula m! 2mm m e m Px Px mx 2m[( m 1)...( m x)] 1 2m[(1 1 / m )...(1 x / m )] e x / 2m Px 1/ m x/m 2m[(e )...(e )] 2m 1 2 Normal Distribution • The full normal distribution separates mean m and standard deviation s parameters. f ( x) e x m 2 2 s / 2s 2 • Tables provide the integral of the distribution function. • Useful benchmarks: – P(|x - m| < 1 s = 0.683 – P(|x - m| < 2 s = 0.954 – P(|x - m| < 3 s = 0.997 Typical Problem • Repeated counts are made in 1min intervals with a long-lived source. The observed mean is 813 counts with s = 28.5 counts. What is the probability of observing 800 or fewer counts? Answer • This is about -0.45s. • Look up P((x-m)/s < -0.45) – P = 0.324 Cumulative Probability • Statistical processes can be described for large numbers. – Can we model one event? – No two events are equal • Consider an event with a 500 keV incident photon on soft tissue with attenuation m = 0.091 cm-1. • Probability distributions typically reflect incidence in an infinitessimal region. – Integrate over a range • The probability of an interaction in 2 cm is x Pc P1 ( x)dx x Pc ( x) memx dx 1 emx 0 • P = 1 – 0.834 = 0.166 0 • How does one simulate this? Random Numbers • To simulate a statistical process one needs a random selection from the possible choices. • Algorithms can generate pseudo-random numbers. – Uncorrelated over a large range of trials. – Randomness limited for large sets or fixed starts • Linear Congruential Generator • Start with a seed value, X0. – Select integers a, b. • For a given Xn, – Xn+1 = (aXn + b) mod m • The maximum number of random values is m. Random Distribution • A selected random number is usually generated in a large range of integers. – Uniform over the range • To select a number with a normal distribution: – Take two random numbers R1, R2 from 1 to N. ri Ri / N • Normalize values to select a narrow range. – Usually from 0 to 1 – Apply algorithm with m, s • Convert range to match a distribution. – (Box-Mueller algorithm) x m s cos2r1 2 ln r2 Monte Carlo Method • The Monte Carlo method simulates complicated systems. • Use random numbers with distribution functions to select a value. • Test that value to see if it meets certain conditions. • Simple Monte Carlo for . – Select a pair of random numbers from 0 to 1. – Sum the squares and count if it’s less than 1. – Multiply the fraction that succeed by 4. Interaction Simulation Typical Problem • A 100 keV neutron beam is incident on a mouse (3 cm thick). Calculate the energy deposited at different depths. Data Table • H m=0.777 cm-1 • O m=0.100 cm-1 • C m=0.0406 cm-1 • N m=0.00555 cm-1 • Total m=0.92315 cm-1 • Simulate one neutron. • Find the distance of penetration by inverting the probability. x 1 mtot ln 1 ri • Find the nucleus struck. – Normalize the mi to the total possible mtot. • Select the energy of recoil and angle. • Repeat for new distance. Fitting Tests • Collected data points will approximate the physical relationship with large statistics. • Limited statistics require fits of the data to a functional form. Least Squares • Assume that the data fits to a straight line. y ax b Q 1 a N 2 y x y a x 2j b x j j • Use a mean square error to determine closeness of fit. 1 Q N 1 N y N j 1 y j y( x j ) 2 ax j b 2 j • Minimize the mean square error. j Q 1 b N y a b j j ax j b x j 0 2y j x j b 0 a x j b a x j Nb N x j y j x j y j N x 2j x j 2 1 N y j a N x j Polynomial Fit • The procedure for a least squares fit applies to any polynomial. – n+1 parameters ak y ( x j ) ak x kj • Minimize error expression Q. n 1 N Q y j ak x kj N j 1 k 0 n k 0 2 n Q 2 N k m y j ak x j x j 0 am N j 1 k 0 • Requires simultaneous solutions to a set of n+1 equations. x j m j y j ak x kj x mj ak x kj m k j k j Exponential Fit • The least squares fit can be applied to other functions. • For a single exponential a fit can be made on the log. y aebx ln y ln a bx n y ( x ) ak e bk x • For the sum of exponentials consider constants a, b. – Select initial values – Taylor’s series to linearize – Find hk that minimize Q k 1 y hk k 1 bk n y( x j ; bk ) y( x j ; bk 0 ) bk bk 0 hk Chi Squared Test • Fitting is based on a limited statistical sample. • A chi-squared test measures the data deviation from the fit. – Normally distributed – Mean k for k degrees of freedom • Divide the sample into n classes with probabilities pi and frequencies mi. n mi Npi 2 i 1 Npi s • The test is k 1 s2 ( , ) 2 2 2 2 P( s ) k 1 ( ,0) 2 (a, x) t a 1et dt x