Probability Distributions

Transcription

1 Chapter Probability Distributions In this chapter we will describe the most common probability distribution functions encountered in high energy physics.. Discrete Distributions.. Combinatorial Given the importance of combinatorial calculus for what follows, we summarize in this section a few of the main results. Typically we will be talking about sequences of objects: a distinction has to be made on whether we care or not about the order of the element in a sequence. If we care about the order we talk about permutations, otherwise we talk about combinations (permutations are ordered combinations). Example: take the set of letters {abc}. Thesequences{cab} and {bac} are considered equivalent combinations but distinct permutations. Permutations with repetitions: pick r objects from a set of n and put them back each time. The number of permutations (ordered sequences) is: n r (..) Example: A byte is a sequence of 8 bits (0/): the number of permutations with repetitions is 8 = 56. A lock with three digits (form 0 to 9) has 0 3 permutations. Permutations without repetitions: pick r objects from a set of n and don t put them back. At each pick you ll have less objects to choose from, so the number number of permutations is reduced with respect to the permutations with repetitions. The number of permutations (ordered sequences) is: n! (..) (n r)! 5

2 6 CHAPTER. PROBABILITY DISTRIBUTIONS Example: take all permutations without repetitions of the 5 cards in the deck: 5! (first pick you choose among 5 cards, second pick among 5 etc...). Take all permutations of the first 4 picks in a deck of 5 cards: = 5!/(5 4)! (first you pick from 5, then from 5, then from 50, then form 49). Combinations without repetitions: pick r objects from a set of n and don t put them back. At each pick you ll have less objects to choose from as for permutations, but this time all sequences that di er only by their order are considered to be the same. The number of combinations (non-ordered sequences) is the number of permutations corrected by the the factor that describes the number of ordered sequences (i.e. r!): n! n (n r)! r! = (..3) r These numbers are the so-called binomial coe (p + q) n = nx r=0 cients, which appear in the binomial theorem: n p r q n r (..4) r It follows that the number of combinations extracting r objects from n or (n-r) from n is the same! Example: lotto (six-numbers lottery game): 6 numbers are extracted (without putting them back) from a set of 90. The order of the extraction is irrelevant. The probability to win (when all tickets are sold) is / n r (which is o(0 7 )). Combinations with repetitions: pick r objects from a set of n and put them back. As in the case of permutations with repetition but this time without considering the order. (n + r )! n + r = (..5) (n r)!r! r Example: take r-scoops from n-icecream flavours. You can take them all the same or repeat them as you like. Factorial For large n, thestirling formula can be used to approximate n!: ln n! (n +/) ln n n + ln p (..6) n n p n! n (..7) e The first term in the second line, (n/e) n, is called the zero-th approximation, whereas the whole term in the above equation is the first approximation. The factorial n! can be extended for non-integer arguments x by the gamma function (x):

3 Bernoulli trials 7.. Bernoulli trials x! = Z 0 u x e u du = (x + ) (..8) (x + ) = x (x) (..9) A Bernoulli trial is an experiment with only two outcomes (success/failure or /0) and where success will occur with constant probability p and failure with constant probability q = p. Examples are again the coin toss, or the decay of K + into either µ + or all other channels. The random variable r {0, } is the outcome of the experiment and its p.d.f. (see Fig... is: f(r; p) =p r q ( r) (..0) The p.d.f. is simply the probability for a single experiment to give success/failure. The first two central moments of the distribution are: µ = p (..) V (r) =p( p) (..) Figure..: Bernoulli trials distribution for a fixed p = 0.3 and 000 trials...3 Binomial Given n Bernoulli trials with a probability of success p, the binomial distribution gives the probability to observe r successes and, consequently, n r failures independently of the order

4 8 CHAPTER. PROBABILITY DISTRIBUTIONS with which they appear. The random variable is again r but this time r {0,n}, i.e. the maximum is given when all trials give a success. The p.d.f. is: n P (r; n, p) = p r ( p) n r. (..3) r Equation..3 can be motivated in the following way: the probability that we get the result Sinthefirstr attempts (and not in the last n r attempts) is given by p r ( p) n r ;butthis n sequential arrangement is only one of a total of r possible arrangements. The distribution for di erent values of the parameters is plotted in figure... The important properties of the binomial distribution are: It is normalized to, i.e. P n r=0 P (r) =. The mean of r is <r>= P n r=0 r P (r) =np. The variance of r is V (r) =np( p). The binomial distribution (like several others we will encounter) has the reproductive property. IfX is binomially distributed as P (X; n, p) and Y is binomially distributed (with the same probability p) as P (Y ; m, p), then the sum is binomially distributed as P (X + Y ; n + m, p). Example: what is the probability to get out of 0 coin tosses 3 times a head? Solution: P (3; 0, 0.5) = ( 0.5) 0 3 = 0! 3!7! =0. Example: a detector with 4 layers has an e ciency per layer to detect a traversing particle of 88%. To reconstruct the complete track of the particle, we need at least three hits (i.e. three out of the four layers have to detect the particle). What is the probability to reconstruct the track? We need at least 3 hits (i.e. 3 or 4), so we have to sum the probability to have 3 hits to the probability to have 4 hits: P (r 3; n =4,p=0.88) = P (r = 3; n =4,p=0.88) + P (r = 4; n =4,p=0.88) = = What if we have 3 or 5 layers? For 3 layers = P (r = 3; n =3,p=0.88) = For 5 layers = P (r 3; n =5,p=0.88) = = How big should be the e ciency be for 3, 4 or 5 layers to have an overall probability for track reconstruction greater than 99%? Given r 3, n =3, 4, 5 and P 0.99 we need to get p. This can be done calculating P (r 3; n, p) for the various values of n solving for p. For n =3wefindp = 0.996; n =4! p = 0.940; n =5! p = Multinomial Distribution The precedent considerations can directly be generalized for the multidimensional problem. Assume we have n objects of k di erent types, and r i is the number of objects of type i. The

5 Poisson Distribution 9 Figure..: The binomial distribution for a fixed p =0.4 and di erent values for n. n! number of distinguishable arrangements is then given by r!r! r k!. If we choose now randomly r objects (putting them back every time), then the probability of getting an arrangement of r i objects of type i is given by p r pr pr k k. The overall probability is therefore simply the probability of our arrangement, multiplied with the number of possible distinguishable arrangements: r! P (r,..,r k ; N,p,p k )= p r r!r!r 3! r k! pr pr k k. (..4) This distribution is called the multinomial distribution and it is what describes the probability to have r i events in bin i of an histogram with n entries. The corresponding properties are:..5 Poisson Distribution <r i >= np i and V (r) =np i ( p i ). (..5) The Poisson p.d.f. applies to the situations where we detect events but do not know the total number of trials. An example is a radioactive source where we detect the decays but do not detect the non-decays.

6 0 CHAPTER. PROBABILITY DISTRIBUTIONS The distribution can be obtained as a limit of the binomial: let be the probability to observe a radioactive decay in a period T of time. Now divide the period T in n time intervals T = T/n small enough that the probability to observe two decays in an interval is negligible. The probability to observe a decay in T is then /n, while the probability to observe r decays in the period T is given by the binomial probability to observe r events in n trials each of which has a probability /n. n! r n r P r; n, = (..6) n (n r)! r! n n Under the assumption that n>>rthen: n! (n r)! = n(n )(n )...(n r + ) nr (..7) and n r n! e for n! (..8) n n and replacing this in the binomial expression in eq...6 we obtain the Poisson p.d.f: P (r; )=. (..9) r! The Poisson distribution can so be seen as the limit of the binomial distribution when the number n of trials becomes very large and the probability p for a single event becomes very small, while the product pn = remains a (finite) constant. It gives the probability of getting r events if the expected number (mean) is. Properties of the Poisson distribution: It is normalized to : P r=0 P (r) =e P r=0 The mean <r>is : <r>= P r=0 r e r r! =. The variance is V (r) =. ( r+ s) r e r r! = e e + = P (r + s; r, s) = ( r+ s)(r+s) e (r+s)! : the p.d.f. of the sum of two Poisson distributed random variables is also Poisson with equal to the sum of the s of the individual Poissons Example An historical example is the number of deadly horse accidents in the Prussian army. The fatal incidents were registered over twenty years in ten di erent cavalry corps. There was a total of fatal incidents, and therefore the expectation value per corps per year is given by = /00 = 0.6. The probability that no soldier is killed per year and corps is P (0; 0.6) = e /0! = To get the total events (of no incidents) in one year and per corps, we have to multiply with the number of observed cases (here 00), which yields = The total statistics of the Prussian cavalry is summarized in table.., in agreement with the Poisson expectation. The Poisson distribution is very often used in counting experiments: This example is mentioned for the first time in the book from L. von Bortkiewicz in the year 898: Das Gesetz der kleinen Zahlen.

7 .. CONTINUOUS DISTRIBUTIONS Fatal incidents per corps and year Reported incidents Poisson distribution Table..: The total statistics of deadly accidents of Prussian soldiers. Number of particles which are registered by a detector in the time interval t, iftheflux and the e ciency of the detector are independent of time and the dead time of the detector is su ciently small, such that. Number of interactions caused by an intense beam of particles which travel through a thin foil. Number of entries in a histogram, if the data are taken during a fixed time interval. Number of flat tires when traveling a certain distance, if the expectation value f lats/distance is constant. Some counter-examples, in which the Poisson distribution cannot be used: The decay of a small amount of radioactive material in a certain time interval, if this interval is significant compared to the lifetime. The number of interactions of a beam of only a few particles which pass through a thick foil. In both cases we have the event rate is not constant (in the first it decreases with time, in the second with distance) and therefore the Poisson distribution cannot be applied. The Poisson p.d.f. requires that the events be independent. Consider the case of a counter with a dead time of µsec. This means that if a second particle passes through the counter within µsec after one which was recorded, the counter is incapable of recording the second particle. Thus the detection of a particle is not independent of the detection of other particles. If the particle flux is low, the chance of a second particle within the dead time is so small that it can be neglected. However, if the flux is high it cannot be. No matter how high the flux, the counter cannot count more than 0 6 particles per second. In high fluxes, the number of particles detected in some time interval will not be Poisson distributed. Picture..3 shows the Poisson distribution for some values of.. Continuous Distributions.. Uniform Distribution The probability density function of the uniform distribution in the interval [a, b] is given by (see Fig...4):

8 CHAPTER. PROBABILITY DISTRIBUTIONS Figure..3: The Poisson distribution for di erent values of. f(x) = ( b a if a apple x apple b 0 else. (..0) The expectation value and the variance are given by <x> = Z b a x b a dx = (a + b) (..) Var(x) = (b a) (..) Example Consider a detector built as a single strip of silicon with a width of mm. If a charged particle hits it, the detector reads otherwise zero (binary readout). What is the spacial resolution of the detector? Estimating the resolution as the variance of the corresponding uniform distribution, we get 90µm.

9 Gaussian or Normal Distribution 3 Figure..4: The uniform distribution... Gaussian or Normal Distribution The Gaussian or normal distribution is probably the most important and useful distribution we know. It is widely used in practice 3. The probability density function is f(x; µ, )= p e (x µ). (..3) The Gaussian distribution is described by two parameters: the mean value µ and the variance,where denotes the standard deviation. By substituting z =(x µ)/ we obtain the so-called normal or standardized Gaussian distribution: N(0, ) = p e z /. (..4) It has an expectation value of zero and standard deviation. Properties of the normal distribution are: It is normalized to : R + P (x; µ, )dx = µ is the expectation value of the distribution: R + xp (x; µ, )dx = µ, as well as (being a symmetric distribution) its mode and median. is the standard deviation and the variance is : R + (x µ) P (x; µ, )dx =. If X and Y are two independent r.v.s distributed as f(x; µ x, x) and f(y; µ y, y) thenz=x + Y is distributed as f(z; µ z, z) withµ z = µ x + µ y and z = x + y. C.F. Gauss did not discovery it all alone. Independently, Laplace and de Moivre knew about this distribution. 3 A legend says that Gauss did describe the size of bread loaves in the city of Königsberg with the normal distribution.

10 4 CHAPTER. PROBABILITY DISTRIBUTIONS Some useful integrals, which are often used when working with the Gaussian function: Z + Z + 0 Z + Z + 0 Z + e ax dx = p /a xe ax dx = a x e ax dx = p /a a x n+ e ax dx = n! a n+ x n+ e ax dx = 0, for all odd values of n Normalverteilung, CDF Normaldichte, PDF Figure..5: The standardized Gaussian distribution. On the top graph the cumulative distribution function is shown, on the lower graph its probability distribution function. Here are some numbers for the integrated Gaussian distribution: 68.7% of the area lies within ± around the mean µ % lies within ± % lies within ±3.

11 Gaussian or Normal Distribution 5 90% of the area lies within ± % lies within ± % lies within ± % lies within ±3.90. (x) can also be expressed by the so-called error func- The integrated Gaussian function tion erf(x): (x) = p Z x e (t µ) / dt (..5) erf(x) = p Z x e t dt (..6) 0 => (x) = +erf( x µ p ) (..7) The Full Width Half Maximum (FWHM) is very useful to get a quick estimate for the width of a distribution and for the specific case of the Gaussian we have: FWHM = p ln=.355. (..8) The gaussian distribution is the limiting case for several other p.d.f. s (see Fig...6). This is actually a consequence of the central limit theorem (CLT) (discussed in section.4). Figure..6: Limiting cases. N-dimensional Gaussian Distribution The N-dimensional Gaussian distribution is defined by f(x; µ, V )= ( ) N/ exp V / (x µ)t V (x µ). (..9)

12 6 CHAPTER. PROBABILITY DISTRIBUTIONS Here, x and µ are column vectors with the components x,...,x N and µ,...,µ N,respectively. The transposed vectors x T and µ T are the corresponding row vectors and V is the determinant of the symmetric N N covariance matrix V. The expectation values and the covariances are given by: hx i i = µ i V (x i )=V ii cov(x i,x j )=V ij In the simplified case of a two-dimensional Gaussian distribution we can write f(x,x ; µ µ,,, ) = p exp ( ) " x µ x µ # x µ x µ exp. We will come back to the specific case of the gaussian distribution in multiple dimensions in Ch.3 when talking about the error matrix...3 Distribution Assume that x,x,,x n are independent random variables, which obey a (standardized) Gaussian distribution with mean 0 and variance. Then the joint p.d.f. is: " ny # xi µ f(x; µ, ) = p i exp (..30) i= i i " nx # xi µ i Y n = exp p (..3) i i= i i= Then the variable (n) defined as:a (n) = nx i= xi i µ i (..3) being a function of random variables is itself a random variable distributed as a distribution with n degrees of freedom. The probability density is given by (see Fig...7): / (n) =f( ; n) = ( ) n/ e (n/) n/. (..33) The (n) p.d.f. has the properties: mean = n variance = n mode = n- for n and 0 for n apple

13 Distribution 7 Figure..7: The distribution for di erent degrees of freedom. reproductive property: n +n = n + n = (n + n ) Since the expectation of (n) isn, the expectation of (n)/n is. The quantity (n)/n is called reduced. For n!, (n)! N( ; n, n) becomes a standardized normal distribution. When using the -distribution in practice, the approximation by a normal distribution is already su - cient for n 30. To understand the notation take the particular case of the for degree of freedom. Let z =(x µ)/ so that the p.d.f. for z is N(z;0, ) and the probability that z apple Z apple z + dz is: f(z)dz = p e z dz (..34) Let Q = Z. (We use Q here instead of to emphasize that this is the variable.) This is not a one-to-one transformation because both +Z and Z go into +Q. The probability that Q is between q and q + dq is the sum of the probability that Z is between z and z + dz around z = p q, and the probability that Z is between z and z dz around z = p q. The Jacobian is: J ± = d(±z) dq = ± p q (..35) so f(q)dq = p e q ( J + + J )dq = p e q ( dq p q + dq p q )dq = p q e q dq (..36) Replacing for Q we have: () = p e (..37) (careful, the same symbol is used for the random variable and the p.d.f.).

14 8 CHAPTER. PROBABILITY DISTRIBUTIONS Figure..8: The log-normal distribution for di erent values of the defining parameters...4 Log-Normal Distribution If y obeys a normal distribution with mean µ and standard deviation, then it follows that x = e y obeys a log-normal distribution. This means that ln(x) is normal distributed (see Fig...8): f(x; µ, )= p x e (ln x µ) /. (..38) The expectation value and the variance are given by: <x> = e (µ+ ) (..39) Var(x) = e (µ+ ) (e ) (..40) The log-normal distribution is typically used when the resolution of a measurement apparatus is composed by di erent sources, each contributing a (multiplicative) amount to the overall resolution. As the sum of many small contributions of any random distribution converges by the central limit theorem to a Gaussian distribution, so the product of many small contributions is distributed according to a log-normal distribution. Example Consider the signal of a photomultiplier (PMT), which converts light signals into electric signals. Each photon hitting the photo-cathode emits an electron, which gets accelerated by an electric field generated by an electrode (dynode) behind. The electron hits the dynode and emits other secondary electrons which gets accelerated to the next dynode. This process if repeated several times (as many as the number of dynodes in the PMT). At every stage the number of secondary electrons emitted depends on the voltage applied. If the amplification per step is a i, then the number of electrons after the k th step, n k = k i=0 a i,is approximately log-normal distributed.

15 Exponential Distribution 9 Figure..9: The exponential distribution for di erent values of (right: linear, left: logarithmic scale)...5 Exponential Distribution The exponential distribution (see Fig...9) is defined for a continuous variable t (0 apple t apple) by: f(t, ) = e t/. (..4) The probability density is characterized by one single parameter. The expectation value is h i = Z 0 te t/ dt =. (..4) The variance is Var(t) =. An example for the application of the exponential distribution is the description of the properdecay-time decay (t) of an unstable particles. The parameter corresponds in this case to the mean lifetime of the particle...6 Gamma Distribution The gamma distribution (see Fig...0) is given by: f(x; k, )= k xk e x. (..43) (k) The expectation value is <x>= k/. The variance is = k/. Special cases:

16 30 CHAPTER. PROBABILITY DISTRIBUTIONS Figure..0: The gamma distribution for di erent values of the parameters. for =/ and k = n/ the gamma distribution has the form of a distribution with n degrees of freedom. The Gamma distribution describes the distribution of the time t = x between the first and the k th event in a Poisson process with mean. The parameter k influences the shape of the distribution, whereas is just a scale parameter...7 Student s t-distribution The Student s t distribution is used when the standard deviation of the parent distribution is unknown and the one evaluated from the sample s is used. Suppose that x is a random variable distributed normally (mean µ, variance ), and a measurement of x yields the sample mean x and the sample variance s. Then the variable t is defined as t = ( x µ)/ s/ = x µ. (..44) s You can see that we have canceled our ignorance about the standard deviation of the parent distribution by taking the ratio. The net e ect is to substitute with the estimated s. The quantity t follows the p.d.f. f n (t) withn degrees of freedom (see Fig...): f n (t) = p (n+)/ ((n + )/) + t (..45) n (n/) n If the true mean µ is known, then n = N with N being the number of measurements, if the true mean is unknown then n = N because one degree of freedom is used for the sample mean x. The distribution depends only on the sample mean x and the sample variance s.

17 F Distribution 3 Figure..: The Student s distribution for di erent values of n. In general, the variable: T = Z p S /n (..46) is governed by the Student s t distribution for n degrees of freedom if Z and S are two independent random variables following respectively a normal distribution N (0, ) and the distribution (n) with n degrees of freedom...8 F Distribution Consider two random variables, and, distributed as with and degrees of freedom, respectively. We define a new random variable F as: F = / / (..47) The random variable F follows the distribution (see Fig...): f(f )= n / n n ((n + n )/) (n /) (n /) F (n )/ + n (n +n )/ F. (..48) n This distribution is known by many names: Fisher-Snedecor distribution, Fisher distribution, Snedecor distribution, variance ratio distribution, and F -distribution. By convention, one usually puts the larger value on top so that F. The F distribution is used to test the statistical compatibility between the variances of two di erent samples, which are obtained form the same underlying distribution (more in section 8.6).

18 3 CHAPTER. PROBABILITY DISTRIBUTIONS Figure..: The F-distribution distribution for di erent values of the parameters...9 Weibull Distribution The Weibull distribution (see Fig...3) was originally invented to describe the rate of failures of light bulbs as they are getting older: P (x;, )= ( x) e ( x). (..49) with t 0 and, > 0. The parameter is just a scale factor and describes the width of the maximum. The exponential distribution is a special case ( = ), when the probability of failure at time t is independent of t. The Weibull distribution is very useful to describe the reliability and to predict failure rates. The expectation value of the Weibull distribution is ( + ) and the variance is Cauchy (Breit-Wigner) Distribution The Cauchy probability density function is: f(x) = +x. (..50) For large values of x it decreases only slowly. Neither the mean nor the variance are defined, because the corresponding integrals are divergent. The particular Cauchy distribution of the form f(m; M, ) = (m M) +( /) (..5) is also called Breit-Wigner function (see Fig...4), and is used in particle physics to describe cross sections near a resonance with mass M and width 4. The Breit-Wigner 4 Even if the mean does not exist, noting that the distribution is symmetric, we can define it to be M; is the FWHM. In actual physical problems the distribution is truncated, e.g., by energy conservation, and the resulting distribution is well-behaved.

19 Cauchy (Breit-Wigner) Distribution 33 Figure..3: The Weibull distribution for di erent values of the parameters. Figure..4: The Breit-Wigner distribution for di erent values of the parameters. comes as the Fourier transformation of the wave function of an unstable particle: (t) / e ie it/ h e t/ (..5) (!) / Z 0 (t)e i!t i dt = (!! 0 + i ) (..53)

20 34 CHAPTER. PROBABILITY DISTRIBUTIONS Figure..5: Straggling functions in silicon for 500 MeV pions, normalized to unity at the most probable value p/x. Thewidthw is the full width at half maximum.[4] which squared gives: (!) = (!! 0 ) + 4 (..54).. Landau Distribution The Landau distribution (see Fig...5) is used to describe the distribution of the energy loss x (dimensionless quantity proportional to de) of a charged particle (by ionisation) passing through a thin layer of matter: p(x) = i Z c+i c i exp(s log s + xs)ds (c >0). (..55) The long tail towards large energies models the large energy loss fluctuations in thin layers. The mean and the variance of the distribution are not defined. For numerical purposes the distribution can be rewritten as: p(x) = Z 0 exp( t log t xt)sin( t)dt (..56).3 Characteristic Function The characteristic function (t) is defined as the expectation value of e itx for a p.d.f. f(x): Z (t) :=<e itx >= e itx f(x)dx (.3.57) i.e (t) is the Fourier integral of f(x). The characteristic function completely determines the p.d.f., since by inverting the Fourier transformation we regain f(x): f(x) = Z e itx (t)dt (.3.58)

21 .4. THE CENTRAL LIMIT THEOREM 35 The characteristic function as well as its first and second derivative are readily calculated for the special case where t = 0: (0) = (.3.59) d (0) dt = i<x> (.3.60) d (0) dt = ( + <x> ) (.3.6) What do we need characteristic functions for? They may be useful when performing calculations with probability densities, for example if the convolution of two probability densities f and f for two random variables x and x should be calculated. A convolution of f and f yields a new probability density g(y), according to which the sum of the random variable y = x + x is distributed: Z Z Z Z g(y) = f (x )f (x ) (y x x )dx dx = f (x )f (y x )dx = f (x )f (y x )dx (.3.6) The convolution integral can now be transformed with the help of the characteristic functions: g(t) = f (t) f (t) (.3.63) In words: The characteristic function of the convolution of two variables is obtained by the product of their characteristic functions. Thus it can be easily shown that the convolution of two Gaussian distributions with µ, and, is again a Gaussian distribution with µ = µ +µ and = +. Furthermore, the convolution of two Poisson distributions is again a Poisson distribution. The characteristic functions of some probability densities are given in table.3.. Table.3.: Characteristic functions of some probability densities. Distribution Characteristic Function Binomial (t) =(pe it + q) n Poisson (t) =e (eit ) Gauss (t) =e iµt t / (t) =( it) n/ Uniform (from a to b) (t) =(e ibt e iat )/(b a) it Breit-Wigner (t) =e ie 0t ( /) t Gamma (t) =( it/µ).4 The Central Limit Theorem The Central Limit Theorem (CLT) is probably the most important theorem in statistics and it is the reason why the gaussian distribution is so important. Take n independent variables x i, distributed according to p.d.f. s f i having mean µ i and variance i, then the p.d.f. of the sum of the x i, S = P x i, has mean P µ i and variance

22 36 CHAPTER. PROBABILITY DISTRIBUTIONS Figure.4.6: Sum of random variables from a uniform distribution in [, ] after the iterations to 4. P i and it approaches the normal p.d.f. N(S; P µ i, P i ) as n!. The CLT holds under pretty general conditions: both mean and variance have to exist for each of the random variables in the sum Lindeberg criteria: y k = x k, if x k µ k apple k k y k =0, if x k µ k > k k. Here, k is an arbitrary number. If the variance (y + y + y n )/ y! for n!, then this condition is fulfilled for the CLT. In plain English: The Lindeberg criteria ensures that fluctuations of a single variable does not dominate its sum. An example of convergence for the CLT is given in Fig..4.6 where a uniform distribution is used for 4 iterations. When performing measurements the value obtained is usually a ected by a large number of (hopefully) small uncertainties. If this number of small contributions is large the C.L.T. tells us that their total sum is Gaussian distributed. This is often the case and is the reason resolution functions are usually Gaussian. But if there are only a few contributions, or if a few of the contributions are much larger than the rest, the C.L.T. is not applicable, and the sum is not necessarily Gaussian..5 References Pretty much every book about statistics and probability will cover the material of this chapter. Here a few examples:

23 .5. REFERENCES 37 L. Lyons, Statistics for Nuclear and Particle Physicist : Ch. 3 W. Metzger, Statistical Methods in Data Analysis : Ch. PDG, Passage of particles through matter