Bernoulli Distribution Example: Toss of coin De¯ne X = 1 if head comes up and X = 0 if tail comes up. 1 Both realizations are equally likely: (X = 1) = (X = 0) = 2 Examples: Often: Two outcomes which are not equally likely: Success of medical treatment ± Interviewed person is female ± Student passes exam ± Transmittance of a disease ± Bernoulli distribution (with parameter ) X takes two values, 0 and 1, with probabilities p and 1 p ± ¡ Frequency function of X ± x 1 x (1 ) ¡ for x 0; 1 p(x) = ¡ 2 f g ½ 0 otherwise Often: ± 1 if event A has occured X = ½ 0 otherwise Example: A = blood pressure above 140/90 mm HG. Distributions, Jan 30, 2003 - 1 - Bernoulli Distribution Let X1; : : : ; Xn be independent Bernoulli random variables with same parameter . Frequency function of X1; : : : ; Xn x1+:::+xn n x1 ::: xn p(x ; : : : ; xn) = p(x ) p(xn) = (1 ) ¡ ¡ ¡ 1 1 ¢ ¢ ¢ ¡ for xi 0; 1 and i = 1; : : : ; n 2 f g Example: Paired-Sample Sign Test Study success of new elaborate safety program ± Record average weekly losses in hours of labor due to accidents before ± and after installation of the program in 10 industrial plants Plant 1 2 3 4 5 6 7 8 9 10 Before 45 73 46 124 33 57 83 34 26 17 After 36 60 44 119 35 51 77 29 24 11 De¯ne for the ith plant 1 if ¯rst value is greater than the second Xi = ½ 0 otherwise Result: 1 1 1 1 0 1 1 1 1 1 The Xi's are independently Bernoulli distributed with unknown parameter . Distributions, Jan 30, 2003 - 2 - Binomial Distribution Let X1; : : : ; Xn be independent Bernoulli random variables Often only interested in number of successes ± Y = X1 + : : : + Xn Example: Paired Sample Sign Test (contd) De¯ne for the ith plant 1 if ¯rst value is greater than the second Xi = ½ 0 otherwise n Y = Xi iP=1 Y is the number of plants for which the number of lost hours has decreased after the installation of the safety program We know: Xi is Bernoulli distributed with parameter ± Xi's are independent ± What is the distribution of Y ? Probability of realization x ; : : : ; xn with y successes: ± 1 y n y p(x ; : : : ; xn) = (1 ) ¡ 1 ¡ Number of di®erent realizations with y successes: n ± y ¡ ¢ Distributions, Jan 30, 2003 - 3 - Binomial Distribution Binomial distribution (with parameters n and ) Let X1; : : : ; Xn be independent and Bernoulli distributed with pa- rameter and n Y = Xi: iP=1 Y has frequency function n y n y p(y) = (1 ) ¡ for y 0; : : : ; n µy¶ ¡ 2 f g Y is binomially distributed with parameters n and . We write Y Bin(n; ): » Note that the number of trials is ¯xed, ± the probability of success is the same for each trial, and ± the trials are independent. ± Example: Paired Sample Sign Test (contd) Let Y be the number of plants for which the number of lost hours has decreased after the installation of the safety program. Then Y Bin(n; ) » Distributions, Jan 30, 2003 - 4 - Binomial Distribution Binomial distribution for n = 10 0.4 0.4 θ = 0.1 θ = 0.3 0.3 0.3 0.2 0.2 p(x) p(x) 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 x x 0.4 0.4 θ = 0.5 θ = 0.8 0.3 0.3 0.2 0.2 p(x) p(x) 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 x x Distributions, Jan 30, 2003 - 5 - Geometric Distribution Consider a sequence of independent Bernoulli trials. On each trial, a success occurs with probability . ± Let X be the number of trials up to the ¯rst success. ± What is the distribution of X? x 1 Probability of no success in x 1 trials: (1 ) ¡ ± ¡ ¡ Probability of one success in the xth trial: ± The frequency function of X is x 1 p(x) = (1 ) ¡ ; x = 1; 2; 3; : : : ¡ X is geometrically distributed with parameter . Example: 1 Suppose a batter has probability 3 to hit the ball. What is the chance that he misses the ball less than 3 times? The number X of balls up to the ¯rst success is geometrically distributed 1 with parameter 3. Thus 2 1 1 2 1 2 (X 3) = + + = 0:7037: · 3 3 ¢ 3 3³3´ Distributions, Jan 30, 2003 - 6 - Hypergemetric Distribution Example: Quality Control Quality control - sample and examine fraction of produced units N produced units ± M defective units ± n sampled units ± What is the probability that the sample contains x defective units? The frequency function of X is M N M x n¡x p(x) = ¡ ; x = 0; 1; : : : ; n: ¡ ¢¡N ¢ n X is a hypergeometric¡ ¢ random variable with parameters N, M, and n. Example: Suppose that of 100 applicants for a job 50 were women and 50 were men, all equally quali¯ed. If we select 10 applicants at random what is the probability that x of them are female? The number of chosen female applicants is hypergeometrically distributed with parameters 100, 50, and 10. The frequency function is 50 50 x 10 x p(x) = ¡ for x 0; : : : ; n ¡ ¢100¡ ¢ 10 2 f g ¡ ¢ for x = 0; 1; : : : ; 10. Distributions, Jan 30, 2003 - 7 - Poisson Distribution Often we are interested in the number of events which occur in a speci¯c period of time or in a speci¯c area of volume: Number of alpha particles emitted from a radioactive source during a ± given period of time Number of telephone calls coming into an exchange during one unit of ± time Number of diseased trees per acre of a certain woodland ± Number of death claims received per day by an insurance company ± Characteristics Let X be the number of times a certain event occurs during a given unit of time (or in a given area, etc). The probability that the event occurs in a given unit of time is ± the same for all the units. The number of events that occur in one unit of time is inde- ± pendent of the number of events in other units. The mean (or expected) rate is ¸. ± Then X is a Poisson random variable with parameter ¸ and frequency function x ¸ ¸ p(x) = e¡ ; x = 0; 1; 2; : : : x! Distributions, Jan 30, 2003 - 8 - Poisson Approximation The Poisson distribution is often used as an approximation for binomial probabilities when n is large and is small: x n x n x ¸ ¸ p(x) = (1 ) ¡ e¡ µx¶ ¡ ¼ x! with ¸ = n . Example: Fatalities in Prussian cavalry Classical example from von Bortkiewicz (1898). Number of fatalities resulting from being kicked by a horse ± 200 observations (10 corps over a period of 20 years) ± Statistical model: Each soldier is kicked to death by a horse with probability . ± Let Y be the number of such fatalities in one corps. Then ± Y Bin(n; ) » where n is the number of soldiers in one corps. Observation: The data are well approximated by a Poisson distribution with ¸ = 0:61 Deaths per Year Observed Rel. Frequency Poisson Prob. 0 109 0.545 0.543 1 65 0.325 0.331 2 22 0.110 0.101 3 3 0.015 0.021 4 1 0.005 0.003 Distributions, Jan 30, 2003 - 9 - Poisson Approximation Poisson approximation of Bin(40; ) 1.0 1 1.0 1 = 400 ¸ = 10 0.8 0.8 0.6 0.6 p(x) p(x) 0.4 0.4 0.2 0.2 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x 0.5 1 0.5 = 40 ¸ = 1 0.4 0.4 0.3 0.3 p(x) p(x) 0.2 0.2 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x 0.2 1 0.2 = 8 ¸ = 5 0.1 0.1 p(x) p(x) 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x 0.2 1 0.2 = 4 ¸ = 10 0.1 0.1 p(x) p(x) 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x Distributions, Jan 30, 2003 - 10 - Continuous Distributions Uniform distribution U(0; ) 40 U(0, θ) Range (0; 1) 30 1 f(x) = 1(0,)(x) 20 Frequency 10 (X) = 2 0 2 var(X) = −2 −1 0 1 2 3 4 12 X 40 Exponential distribution Exp(¸) Exp(λ) 30 Range [0; ) 1 20 f(x) = ¸ exp( ¸x)1[0; )(x) ¡ 1 Frequency 1 10 (X) = ¸ 1 0 var(X) = −2 −1 0 1 2 3 4 ¸2 X 2 40 Normal distribution (¹; ) 2 N N(µ, σ ) 30 Range ¡ 1 1 f x x ¹ 2 20 ( ) = exp 2 ( ) p ¼2 ¡2 ¡ Frequency 2 ³ ´ (X) = ¹ 10 var(X) = 2 0 −2 −1 0 1 2 3 4 X 6 4 2 0 −2 U(0, θ) Exp(λ) N(µ, σ2) Distributions, Jan 30, 2003 - 11 -.