Statistics, Data Analysis, and Simulation SS 2013

Transcription

1 Statistics, Data Analysis, and Simulation SS Statistik, Datenanalyse und Simulation Dr. Michael O. Distler Mainz, 23. April 213

2 What we ve learned so far Fundamental concepts random variable, probability frequentist vs. bayesian interpretation probability mass function, probability density function, cumulative distribution function expectation values and moments

3 Definitions probability mass function (pmf) probability density function (pdf) of a measured value (=random variable) f(n) n f(x) f (n) discrete f (x) continuous Normalization: f (n) f (n) = 1 f (x) f (x) dx = 1 Probability: n p(n 1 n n 2 ) = n 2 x n 1 f (n) p(x 1 x x 2 ) = x2 x 1 f (x)dx

4 Expectation values and moments Mean: A random variable X takes on the values X 1, X 2,..., X n with probability p(x i ), then the expected value of X ( mean ) is X = X = n X i p(x i ) i=1 The expected value of an arbitrary function h(x) for a continuous random variable is: E[h(x)] = The mean ist the expected value of x: E[x] = x = h(x) f (x)dx x f (x)dx

5 Expectation values and moments standard deviation = {mean (deviation from x) 2 } 1/2 σ 2 = (x x) 2 = = (x x) 2 f (x)dx (x 2 2x x + x 2 ) f (x)dx = x 2 2 x x + x 2 = x 2 x 2 σ 2 = Variance, σ = Standard deviation Discrete distributions: ( x 2 ( x) 2 ) σ 2 = 1 N N Attention: This is the definition of the variance! To get a bias free estimation of the variance, 1 1 N will be replaced by N 1.

6 Expectation values and moments Moments are the expected value of x n and of (x x ) n. They are called nth algebraic moment µ n and nth central moment µ n, respectivly. Skewness v(x) is a measure of the asymmetry of the probability distribution of a random variable x: v = µ 3 σ 3 = E[(x E[x])3 ] σ 3 Kurtosis is a measure of the peakedness of the probability distribution of a random variable x. β 2 = µ 4 σ 4 = E[(x E[x])4 ] σ 4 γ 2 = β 2 3 (excess kurtosis)

7 Binomial distribution The binomial distribution is the discrete probability distribution of the number of successes r in a sequence of n independent yes/no experiments, each of which yields success with probability p (Bernoulli experiment). P(r) = ( n r ) p r (1 p) n r P(r) is normalized. Proof: Binomial theorem with q = 1 p. The mean of r is: n r = E[r] = rp(r)= np The variance σ 2 is V [r] = E[(r r ) 2 ] = r= n (r r ) 2 P(r)= np(1 p) r=

8 Example: How big is the chance to get with n = 6 throws of a dice exactly zero times the 6, exactly twice the 6, and at least once the 6? For a correct dice is p = 1/6 and ( ) 1 ( ) 5 6 ( ) 6 P() = = 33.5% 6 6 ( ) 1 2 ( ) 5 4 ( ) 6 P(2) = = 2.1% P( 1) = (1 P()) = 66.5%

9 press any key

10 1.2 Special discrete distributions (Poisson distribution) The Poisson distribution gives the probability of getting exactly r events when the number of trials is very large and the probability of the occurrence of an event in a single trial p is very small, with a finite mean r = µ = np. The Poisson distribution can be derived as a limit of the binomial distribution and has only one parameter, namely the mean µ. The Poisson distribution is given as: P(r) = µr e µ r! The Poisson distribution occurs in many cases where one counts things or events, such as the number of nuclear reactions or particle decays or the number of fish caught in a fishing competition.

11 Poisson distribution.6.6 The Poisson distribution ist given by: The mean is: The variance is: P(r) = µr e µ r! r = µ V [r] = σ 2 = µ.5.4 µ = µ = µ = µ =

12 Death by horse kicks in the Prussian army Since 1898 the number of over a period of 2 years killed cavalrymen in the Prussian army is given in many textbooks. Deaths r Σ Years per corps with r deaths Expected # The total number of deaths is 122, and the mean number of deaths per corps and year is µ = 122/2 =.61. The agreement between the expected and observed numbers is very good - actually too well. More examples: Radioactive decay Printing errors per page in books Simultaneously made scientific discoveries

13 1.3 Special probability densities Uniform distribution: This probability distribution is constant in between the limits x = a and x = b: f (x) = Mean and variance: { 1 b a a x < b otherwise x = E[x] = a + b 2 V [x] = σ 2 = (b a)2 12

14 Gaussian distribution The most important probability distribution - also called normal distribution: f (x) = 1 e (x µ)2 2σ 2 2πσ The Gaussian distribution has two parameters, the mean µ and the variance σ 2. The probability distribution with mean µ = and variance σ 2 = 1 is named standard normal distribution or short N(, 1). The Gaussian distribution can be derived from the binomial distribution for large values of n and r and similarly from the Poisson distribution for large values of µ.

15 Gaussian distribution dx N(, 1) =.6827 = (1.3173) dx N(, 1) =.9545 = (1.455) dx N(, 1) =.9973 = (1.27) FWHM: useful to estimate the standard deviation: FWHM = 2σ 2ln2 = 2.355σ

16 Gaussian distribution Left side: The binomial distribution for n = 1 and p =.6 in comparison to the Gaussian distribution for µ = np = 6 and σ = np(1 p) = 2.4. Right side: The Poisson distribution for µ = 6 and σ = 6 in comparison to the Gaussian distribution.

17 Cumulative Gaussian distribution The cumulative Gaussian distribution Φ(x) = 1 2πσ x e (t µ) 2 2σ 2 dt. cannot be expressed analytically and must be evaluated numerically. F(x) = 1 z e x2 2. 2π However it can be expressed in terms of the Gaussian error function erf(x) which is available on many modern calculators or computer libraries erf(x) = 2 π x Φ(x) = 1 2 e t2 dt. ( ( )) x µ 1 + erf. 2σ

18 Cumulative Gaussian distribution *(1+erf(x/sqrt(2))).4*exp(-.5*x*x)

19 Full moon and accidents Do more accidents happen on days with full moon? To discover such an effect the number of accidents in many German cities are compared. We find that in Hamburg, the average number of accidents on days with full moon 1. with a standard deviation of 1., and on the other days it is 7. with negligible error. This effect is significant?

20 Full moon and accidents Do more accidents happen on days with full moon? To discover such an effect the number of accidents in many German cities are compared. We find that in Hamburg, the average number of accidents on days with full moon 1. with a standard deviation of 1., and on the other days it is 7. with negligible error. This effect is significant? But this doesn t mean anything in reality. If one is conducting this investigation in 2 cities, then the probability that the accident rate differs more than 3 standard deviations from the mean in any one city is: And this probability is not small =.23

21 Chi-square distribution If x 1, x 2,..., x n are independend random variables distributed according to the standard Gaussian distribution with mean and variance 1, then the sum u = χ 2 = n i=1 x 2 i ist distributed according to a χ 2 distribution f n (u) = f n (χ 2 ) where n is called the number of degrees of freedom. f n (u) = ( 1 u ) n/ e u/2 Γ(n/2) The χ 2 distribution has a maximum at (n 2). The mean is found to be n and the variance is 2n.

22 Chi-square distribution pdf(2,x) pdf(3,x) pdf(4,x) pdf(5,x) pdf(6,x) pdf(7,x) pdf(8,x) pdf(9,x)

23 Chi-square cumulative distribution function The probability for χ 2 n to take on a value in the interval [, x]. 1.8 cdf(2,x) cdf(3,x) cdf(4,x) cdf(5,x) cdf(6,x) cdf(7,x) cdf(8,x) cdf(9,x)

24 Chi-square distribution with 5 d.o.f % c.l. [ ]

25 Gamma distribution The goal is to calculate the probability density function of f (t) for the time difference t between two events, when events occur at a mean rate λ. Example: the radioactive decay with a mean decay rate λ. The probability density distribution of the gamma distribution is given by: f (x; k) = x k 1 e x Γ(k) mit Γ(z) = t z 1 e t dt; Γ(z+1) = z! this is the wait time t = x from the first to the kth event of Poisson-distributed process with mean µ = 1 an. The generalization for other values of µ is f (x; k, µ) = x k 1 µ k e µx Γ(k)

26 Gamma distribution *exp(-1.*x)

27 Characteristic function If x is a real random variable with the distribution function F(x) and the probability density function f (x), one referred to the expected value of exp(ıtx) as their characteristic function: ϕ(t) = E[exp(ıtx)] so in the case of continuous variables, a Fourier integral with its well-known transforming properties: ϕ(t) = exp(ıtx) f (x)dx f (x) = 1 2π Especially for the algebraic moments one gets: µ n = E[x n ] = ϕ (n) (t) = d n ϕ(t) dt n = ı n ϕ (n) () = ı n µ n x n f (x)dx x n exp(ıtx) f (x)dx exp( ıtx) ϕ(t)dt

28 1.4 Theorems The law of large numbers The law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed. We perform n independent experiments (Bernoulli trials) where the result j occurs n j times. p j = E[h j ] = E[n j /n] The variance of a Binomial distribution is: V [h j ] = σ 2 (h j ) = σ 2 (n j /n) = 1 n 2 σ2 (n j ) = 1 n 2 np j(1 p j ) From the product p j (1 p j ) which is 1 4, we can deduce the law of large numbers: σ 2 (h j ) < 1/n

29 The central limit theorem The central limit theorem (CLT) states conditions under which the mean of a sufficiently large number of independent random variables, each with finite mean and variance, will be approximately normally distributed. Let x i be a sequence of n independent and identically distributed random variables each having finite values of expectation µ and variance σ 2 >. In the limit n the random variable w = n i=1 x i will be normally distributed with mean w = n x and variance V [w] = nσ 2.

30 Illustration: The central limit theorem.5.5 N=1.4 Gauss.4 N= N=3.5.4 N= The sum of uniformly distributed random variables and the standard normal distribution.