Random variables. Contents

Random variables Contents 1 Random Variable 2 1.1 Discrete Random Variable............................ 3 1.2 Continuous Random Variable........................... 5 1.3 Measures of Location............................... 7 1.3.1 Expected Value.............................. 7 1.3.2 Quantile.................................. 8 1.3.3 Mode................................... 9 1.4 Measures of Dispersion.............................. 10 1.5 Measures of Concentration............................ 11 2 Excercises 15 3 Models of Discrete Random Variables 16 3.1 The Poisson Distribution............................. 16 3.2 The Distribution of a Bernoulli Random Variable................ 17 3.3 The Binomial Distribution............................ 18 3.4 The Hypergeometric Distribution......................... 20 4 Excercises 23 5 Models of Continuous Random Variables 24 5.1 The Uniform Distribution............................. 24 5.2 The Exponential distribution........................... 26 5.3 The Normal Distribution............................. 27 5.4 The Log-normal Distribution........................... 29 5.5 The Pearson, the Student and the Fisher-Snedecor Distribution......... 30 6 Excercises 33

1 Random Variable Many random experiments have numerical outcomes. Definition 1.1. A random variable is a real-valued function X(ω) defined on the sample space Ω. The set of possible values of the random variable X is called the range of X. M = {x; X(ω) = x}. We denote random variables by capital letters X, Y,... (eventually X 1, X 2,... ) and their particular values by small letters x, y,.... Using random variables we can describe random events, for example X = x, X x, x 1 < X < x 2 etc. Examples of random variables: the number of dots when a die is rolled, the range is M = {1, 2,... 6}; the number of rolls of a die until the first 6 appears, the range is M = {1, 2,... }; the lifetime of the lightbulb, the range is M = {x; x 0}. According to the range M we separate random variables to discrete with M is finite or countable, and continuous where M is a closed or open interval. Examples of discrete random variables: the number of cars sold at a dealership during a given month, M = {0, 1, 2,... }; the number of houses in a certain block, M = {1, 2,... }; the number of fish caught on a fishing trip, M = {0, 1, 2,... }; the number of heads obtained in three tosses of a coin, M = {0, 1, 2, 3}. Examples of continuous random variables: the height of a person, M = (0, ); the time taken to complete an examination, M = (0, ); the amount of milk in a bottle, M = (0, ). For the description of random variables we will use some functions: a (cumulative) distribution function F (x); a probability function p(x) only for discrete random variables; a probability density function f(x) only for continuous random variables and measures of location, dispersion, and concentration. Definition 1.2. Let X be any random variable. The distribution function F (x) of the random variable X is defined as F (x) = P (X x), x R. 2

Remark. Distribution function = cumulative distribution function. The properties of F (x): for every real x: 0 F (x) 1, F (x) is a non-decreasing, right-continuous function, it has limits lim F (x) = 0, lim F (x) = 1, x x if range of X is M = {x; x (a, b]} then F (a) = 0 a F (b) = 1, for every real numbers x 1 and x 2 : P (x 1 < X x 2 ) = F (x 2 ) F (x 1 ). 1.1 Discrete Random Variable For a discrete random variable X, we are interested in computing probabilities of the type P (X = x k ) for various values x k in range of X. Definition 1.3. Let X be a discrete random variable with range {x 1, x 2,... } (finite or countably infinite). The function p(x) = P (X = x) is called the probability function of X. Remark. Probability function = probability mass function. Properties of p(x) for every real number x, 0 p(x) 1, x M p(x) = 1 for every two real numbers x k and x l (x k x l ): P (x k X x l ) = The probability function p(x) can be described by x l x i =x k p(x i ). the table, X x 1 x 2... x i... p(x) p(x 1 ) p(x 2 )... p(x i )... 1 the graph [x, p(x)], 3

the formula, for example, p(x) = where π is a given probability. { π(1 π) x x = 0, 1, 2,..., 0 otherwise, Example 1.1. The shooter has 3 bullets and shoots at the target until the first hit or until the last bullet. The probability that the shooter hits the target after one shot is 0.6. The random variable X is the number of the fired bullets. Find the probability and the distribution function of the given random variable. What is the probability that the number of the fired bullets will not be larger then 2? Solution. Random variable X is discrete with the range M = {1, 2, 3}. The probability function is: p(1) = P (X = 1) = 0.6, p(2) = P (X = 2) = 0.4 0.6 = 0.24, p(3) = P (X = 3) = 0.4 0.4 0.6 + 0.4 0.4 0.4 = 0.4 0.4 = 0.16. The results are summarized in the table x 1 2 3 p(x) 0.6 0.24 0.16 1 The probability function can be described by the formula 0.6 0.4 x 1 x = 1, 2, p(x) = 0.4 2 x = 3, 0 otherwise. We can calculate some values of the distribution function F (x): We can write F (0) = P (X 0) = 0, F (1) = P (X 1) = p(1) = 0.6, F (1.5) = P (X 1.5) = P (X 1) = p(1) = 0.6, F (2) = P (X 2) = p(1) + p(2) = 0.84, F (3) = P (X 3) = p(1) + p(2) + p(3) = 1, F (4) = P (X 4) = p(1) + p(2) + p(3) = 1. 0 x < 1, 0.6 1 x < 2, F (x) = 0.84 2 x < 3, 1 x 3. What is the probability that the number of the fired bullets will not be larger then 2? P (X 2) = P (X = 1) + P (X = 2) = p(1) + p(2) = F (2) = 0.6 + 0.24 = 0.84. 4

Figure 1: The probability and the distribution function 1.2 Continuous Random Variable If the cumulative distribution function is a continuous function, then X is said to be a continuous random variable. Definition 1.4. The probability density function of the random variable X is a non-negative function f(x) such that F (x) = x f(t) dt, x R. Properties of f(x): f(x) dx = M f(x) dx = 1, f(x) = df (x) dx = F (x), where the derivative exists, P (x 1 X x 2 ) = P (x 1 < X < x 2 ) = P (x 1 < X x 2 ) = P (x 1 X < x 2 ) = x 2 F (x 2 ) F (x 1 ) = f(x) dx x 1 Remark. If X is a continuous random variable, then P (X = x) = 0. The function f(x) we can describe by a formula or a graph, for example { 1 x 2 e 5 for x > 2, 5 f(x) = 0 for x 2. 5

Figure 2: The probability density function and the distribution function Example 1.2. The random variable X has the probability density function { cx 2 (1 x) 0 < x < 1, f(x) = 0 otherwise. Determine a constant c in order that f(x) is a probability density function. Find a distribution function of the random variable X. Calculate the probability P (0.2 < X < 0.8). Solution. The probability density function has to fulfil f(x) dx = 1, therefore, M 1 0 cx 2 (1 x) dx = c = c 1 0 [ x (x 2 x 3 3 ) dx = c 3 x4 4 [ 1 3 1 4 ] = c 12 = 1, we get c = 12. The distribution function can be calculated by the definition of the probability density function. We can write for 0 < x < 1 x x [ ] t F (x) = 12t 2 (1 t) dt = 12 (t 2 t 3 3 x ) dt = 12 3 t4 4 0 0 0 [ ] x 3 = 12 3 x4 = 4x 3 3x 4. 4 0 x 0, F (x) = x 3 (4 3x) 0 < x < 1, 1 x 1. Using the probability density function we can calculate P (0.2 < X < 0.8) = 0.8 0.2 ] 1 12x 2 (1 x) dx = [ 4x 3 3x 4] 0.8 0.2 = 0.792. If the distribution function is known, we can do simpler calculation P (0.2 < X < 0.8) = F (0.8) F (0.2) = 0.8 3 (4 3 0.8) 0.2 3 (4 3 0.2) = 0.792. 0 6

Example 1.3. A random variable X is described by the distribution function { 0 x 0, F (x) = 1 e x x > 0. Find a probability density function. Solution. Using the mentioned formula f(x) = get 1.3 Measures of Location f(x) = df (x) dx { 0 x 0, e x x > 0. and the fact that d dx (1 e x ) = e x we The distribution function (the probability function or the probability density function) gives us the complete information about the random variable. Sometimes it is useful to know a simpler and concentrated formulation of this information such as measures of location, dispersion and concentration. The best known measures of location are a mean (an expected value), quantiles (a median, upper and lower quartile,... ) and a mode. 1.3.1 Expected Value Definition 1.5. The mean (the expected value) E(X) of the random variable X (sometimes denoted as µ) is the value that is expected to occur per repetition, if an experiment is repeated a large number of times. For the discrete random variable is defined as E(X) = M x i p(x i ), for the continuous random variable as E(X) = M xf(x) dx if the given sequence or integral absolutely converges. We mention some properties of the mean: the mean of the constant c is equal to this constant E(c) = c, the mean of the product of the constant c and the random variable X is equal to the product of the given constant c and the mean of X: E(cX) = ce(x), the mean of the sum of random variables X 1, X 2,..., X n is equal to the sum of the mean of the given random variables, E(X 1 +X 2 + +X n ) = E(X 1 )+E(X 2 )+ +E(X n ), if X 1, X 2,..., X n are independent, then the mean of their product is equal to the product of their means E(X 1 X 2 X n ) = E(X 1 )E(X 2 ) E(X n ). 7

Figure 3: Quantile x P Definition 1.6. The random variables X 1, X 2,..., X n are independent if and only if for any numbers x 1, x 2,..., x n R is P (X 1 x 1, X 2 x 2,..., X n x n ) = P (X 1 x 1 ) P (X 2 x 2 ) P (X n x n ). Let us have the random vector X = (X 1, X 2,..., X n ), which components X 1, X 2,..., X n are the random variables. F (x) = F (x 1, x 2,..., x n ) = P (X 1 x 1, X 2 x 2,..., X n x n ) is the distribution function of the vector X and F (x 1 ), F (x 2 ),..., F (x n ) are the distribution functions of the random variables X 1, X 2,..., X n. The random variables X 1, X 2,..., X n are independent if and only if F (x 1, x 2,..., x n ) = F (x 1 ) F (x 2 ) F (x n ). If X is the random vector which components are the discrete random variables, the function p(x) = p(x 1, x 2,..., x n ) = P (X 1 = x 1, X 2 = x 2,..., X n = x n ) is the probability function of the vector X, p(x 1 ), p(x 2 ),..., p(x n ) are the probability functions of X 1, X 2,..., X n, then X 1, X 2,..., X n are independent if and only if p(x 1, x 2,..., x n ) = p(x 1 ) p(x 2 ) p(x n ). If X is the random vector which components are the continuous random variables, the function f(x) = f(x 1, x 2,..., x n ) is the probability density function of the vector X, f(x 1 ), f(x 2 ),..., f(x n ) are the probability density functions of X 1, X 2,..., X n, then X 1, X 2,..., X n are independent if and only if f(x 1, x 2,..., x n ) = f(x 1 ) f(x 2 ) f(x n ). 1.3.2 Quantile Definition 1.7. The 100P% quantile x P of the random variable with the increasing distribution function is a such value of the random variable that P (X x P ) = F (x P ) = P, 0 < P < 1. The quantile x 0.50 we call median Me(X), it fulfils P (X Me(X)) = P (X Me(X)) = 0.50. The quantile x 0.25 is called the lower quartile, the quantile x 0.75 is called the upper quartile. The selected quantiles of some important distributions are tabulated. 8

1.3.3 Mode Definition 1.8. The mode Mo(X) is the value of the random variable with the highest probability (for the discrete random variable), or the value, where the function f(x) reaches its maximum (for the continuous random variable). Example 1.4. Find the mean (the expected value) and the mode of the random variable defined as the number of fired bullets (see Ex. 1.1). The probability function is 0.6 0.4 x 1 x = 1, 2, p(x) = 0.4 2 x = 3, 0 otherwise. Solution. The mean (the expected value) we get using the formula from the definition of E(X) E(X) = 3 x i p(x i ) = 1 0.6 0.4 0 + 2 0.6 0.4 1 + 3 0.4 2 = 1.56. i=1 The mode is the value of the given random variable with the highest probability which is Mo(X) = 1, because p(1) = 0.6. Example 1.5. The random variable X is described by the probability density function { 12x 2 (1 x) 0 < x < 1, f(x) = 0 otherwise. Find the mean (the expected value) and the mode. Solution. We calculate the mean using the definition formula 1 1 [ x E(X) = xf(x) dx = x 12x 2 4 (1 x) dx = 12 4 x5 5 0 0 ] 1 0 = 3 5 = 0.6. The mode is the maximum of the probability density function. We have to find the maximum of f(x) on the interval 0 < x < 1, d ( 12x 2 (1 x) ) = 12(2x 3x 2 ) = 0, x(2 3x) = 0 dx we get x = 0 or x = 2/3. The maximum of f(x) is in x = 2/3 Mo(X) = 2/3. Example 1.6. Find the median, the upper and lower quartile of the random variable X with the distribution function { 1 1 x > 1, x F (x) = 3 0 x 1. Solution. The quantile is defined by the formula F (x P ) = P, thus, 1 1 x 3 P = P x P = 1 3 1 P. median 1 x 0.50 = 3 1 0.50 = 1.260, lower quartile 1 x 0.25 = 3 1 0.25 = 1.101, upper quartile 1 x 0.75 = 3 1 0.75 = 1.587. 9

1.4 Measures of Dispersion The elementary and widely-used measures of dispersion are the variance and the standard deviation. Definition 1.9. The variance D(X) of the random variable X (sometimes denoted as σ 2 ) is defined by the formula D(X) = E { [X E(X)] 2}. The variance of the discrete random variable is given by D(X) = M [x i E(X)] 2 p(x i ), the variance of the continuous random variable by D(X) = [x E(X)] 2 f(x)dx. M Some properties of the variance: D(k) = 0, where k is a constant, D(kX) = k 2 D(X), D(X + Y ) = D(X) + D(Y ), if X and Y are independent, D(X) 0 for every random variable, D(X) = E(X 2 ) E(X) 2, D(X) = E[X E(X)] 2 = E[X 2 2XE(X) + E(X) 2 ] = E(X 2 ) E[2XE(X)] + E[E(X) 2 ] = E(X 2 ) 2E(X)E(X) + E(X) 2 = E(X 2 ) E(X) 2. Namely, for the discrete random variable D(X) = M x 2 i p(x i ) E(X) 2, for the continuous random variable D(X) = x 2 f(x)dx E(X) 2. M Definition 1.10. The standard deviation σ(x) of the random variable X is defined as the square root of the variance σ(x) = D(X). Remark. The standard deviation has the same unit as the random variable X. Example 1.7. Find the variance and the standard deviation of the random variable defined as the number of fired bullets (see Example 1.1). 10

Figure 4: Relation between the mean and the standard deviation Solution. The mean is E(X) = 1.56 (see Ex. 1.4). For the purpose of calculation of the variance we use the formula D(X) = E(X 2 ) E(X) 2 E(X 2 ) = M x 2 i p(x i ) = 3 x 2 i p(x i ) = 1 2 0.6 + 2 2 0.24 + 3 2 0.16 = 3, i=1 then D(X) = 3 1.56 2 = 0.566. The standard deviation is the square root of the variance σ(x) = D(X) = 0.753. Example 1.8. Find the variance and the standard deviation of the random variable X with the probability density function { 12x 2 (1 x) 0 < x < 1, f(x) = 0 otherwise. Solution. The mean is E(X) = 3/5 (see Ex. 1.5) and E(X 2 ) = M x 2 f(x) dx = = 2 5 = 0.4, 1 0 [ ] x x 2 12x 2 5 1 (1 x) dx = 12 5 x6 6 0 D(X) = 2 ( ) 3 2 5 5 = 1 = 0.04. The standard deviation is σ(x) = D(X) = 1 = 0.2. 25 5 1.5 Measures of Concentration We will focus on the measures describing the shape of random variables distribution (skewness and kurtosis). These measures are defined by moments. Definition 1.11. The r th moment µ r of the random variable X is defined by the formula µ r(x) = E(X r ) for r = 1, 2,.... The r th moment of the discrete random variable is given by µ r(x) = M x r i p(x i ), the r th moment of the continuous random variable is µ r(x) = x r f(x)dx. M 11

Figure 5: The skewness Definition 1.12. The r th central moment µ r of the random variable X is defined by the formula µ r (X) = E[X E(X)] r for r = 2, 3,.... The r th central moment of the discrete random variable is given by µ r (X) = M [x i E(X)] r p(x i ), the r th central moment of the continuous random variable is µ r (X) = [x E(X)] r f(x)dx. M Remark. From the above definitions, it is obvious that the mean is the first moment and the variance is the second central moment. Definition 1.13. The skewness α 3 (X) is defined by the formula α 3 (X) = µ 3(X) σ(x) 3. According to the values of the skewness we can tell whether distribution is symmetric or asymmetric. α 3 = 0, distribution is symmetric, α 3 < 0, distribution is skewed to the right, α 3 > 0, distribution is skewed to the left. Definition 1.14. The kurtosis α 4 (X) is defined by the formula α 4 (X) = µ 4(X) σ(x) 4 3. 12

Figure 6: The kurtosis Kurtosis is a measure of how outlier-prone a distribution is. The kurtosis of the normal distribution is 0. Distributions that are more outlier-prone than the normal distribution have kurtosis greater than 0; distributions that are less outlier-prone have kurtosis less than 0. Example 1.9. Calculate the skewness and the kurtosis of the random variable defined as the number of fired bullets (see the previous examples). Solution. First of all we have to calculate the 3 rd and 4 th central moment. The mean of the given random variable is E(X) = 1.56, the standard deviation is σ(x) = 0.753. µ 3 = 3 [x i E(X)] 3 p(x i ) i=1 = (1 1.56) 3 0.6 + (2 1.56) 3 0.24 + (3 1.56) 3 0.16 = 0.393 3 µ 4 = [x i E(X)] 4 p(x i ) i=1 = (1 1.56) 4 0.6 + (2 1.56) 4 0.24 + (3 1.56) 4 0.16 = 0.756 The skewness and kurtosis are equal to α 3 = µ 3 σ = 0.922, α 3 4 = µ 4 σ 3 = 0.644. 4 Example 1.10. Calculate the skewness and the kurtosis of the random X with the probability density function { 12x 2 (1 x) 0 < x < 1, f(x) = 0 otherwise. Solution. The mean of the given random variable is E(X) = 3/5 and the standard deviation is 1/5. First of all we calculate the 3 rd and 4 th central moment. µ 3 = µ 4 = 1 0 1 0 [x 0.6] 3 12x 2 (1 x) dx = 2 875 = 0.00229, [x 0.6] 4 12x 2 (1 x) dx = 33 8750 = 0.00377. 13

The skewness and kurtosis are equal to α 3 = µ 3 σ 3 = 2 7 = 0.286, α 4 = µ 4 σ 4 3 = 9 14 = 0.643. 14

2 Excercises 1. The probability function of a random variable X is given in table x 0 1 2 3 4 5 p(x) 0.10 0.15 0.2 0.15 0.25 0.15 1 (a) Determine the distribution function of the random variable and display both functions graphically. (b) Calculate the probability P (X 3), P (X > 4), P (1 < X 4). (c) Determine the mean, variance, standard deviation, mode, coefficients of skewness and kurtosis of the random variable. 2. Consider the distribution function of a continuous random variable X 0 x 0, F (x) = C(1 cos x) 0 < x < π, 1 x π. (a) Determine constant C R. (b) Determine the probability density function. (c) Calculate the probabilities P (0 < X < π 4 ), P ( π 4 < X < π 2 ), P ( π 2 < X < π). (d) Determine the mean, median, mode, variance, standard deviation, coefficient of skewness and kurtosis of the random variable. 15

Table 1: Characteristics of Poisson distribution E(X) D(X) α 3 (X) α 4 (X) Mo(X) λ λ 1 λ 1 λ λ 1 Mo(X) λ 3 Models of Discrete Random Variables 3.1 The Poisson Distribution The Poisson distribution tends to arise when we count the number of occurrences of some unpredictable event over a period of time. Typical examples are earthquakes, car accidents, incoming phone call etc. The Poisson distribution is also possible to use for description of appearance of some elements in the given geometrical area (for example misprints in a newspaper). To apply Poisson distribution the occurrences must be random and independent. Definition 3.1. If X has the probability function { λ x x! p(x) = e λ x = 0, 1, 2,..., 0 otherwise. it is said to have a Poisson distribution with parameter λ > 0, and we write X Po(λ). The parameter λ is the mean number of occurrences. The table 1 summarizes basic information about the Poisson distribution Examples of random variables following the Poisson distribution: the number of telemarketing phone calls received by a household during a given day, the number accidents that occur on a given highway during a one-week period, the number of customers entering the grocery store during a one-hour interval, the number of defects in a five-foot-long iron rod etc. Example 3.1. The secretary received at average 6 phone calls during one hour. We would like to analyse work-load of the secretary during 20-minutes intervals. Describe the random variable the number of received phone calls during 20 minutes by a probability and a distribution function. Find the probability that the secretary receives during 20 minutes a) one phone call at least, b) at most two phone calls, c) one or two phone calls. Calculate the mean, the variance, the standard deviation, the mode, the skewness and the kurtosis of the given random variable. Solution. The range of random variable is M = {0, 1, 2,... } We assume we can use the Poisson distribution. The parameter λ denoted the mean of the random variable is equal to 2 (during one hour we can expect 6 phone calls, during 20 minutes then 2). The probability function is p(x) = X P o(2) { 2 x x! e 2 x = 0, 1, 2,..., 0 otherwise We can calculate probabilities that the secretary receives 16

Table 2: Selected values of the probability and the distribution function P o(2) x 0 1 2 3 4 5 6 p(x) 0.1353 0.2707 0.2707 0.1804 0.0902 0.0361 0.0120 F (x) 0.1353 0.4060 0.6767 0.8571 0.9473 0.9834 0.9955 Figure 7: The probability and the distribution function P o(2) a) at least one phone call P (X 1) = 1 P (X < 1) = 1 P (X = 0) = 1 p(0) = 1 0.1353. = 0.865, b) at most two phone calls P (X 2) = P (X = 0) + P (X = 1) + P (X = 2) = p(0) + p(1) + p(2) c) one or two phone calls = 0.1253 + 0.2707 + 0.2707 = F (2). = 0.677, P (X = 1 X = 2) = P (X = 1)+P (X = 2) = p(1)+p(2) = 0.2707+0.2707 = 0.541. We calculate selected measures: the expected value (the mean) E(X) = λ = 2, the variance D(X) = λ = 2, the standard deviation σ = D(X) = λ = 2 =. 1.414., the mode λ 1 Mo(X) λ, so 2 1 Mo(X) 2, Mo(X) = 1 and 2 (see the table of the probability function), the skewness α 3 = 1 λ = 1. 2 = 0.707, the kurtosis α4 = 1 = 1 = 0.5. λ 2 3.2 The Distribution of a Bernoulli Random Variable Some random experiments can have only 2 possible outcomes: success or failure. The random variable that denotes the number of success in one experiment we call a Bernoulli random variable. If the probability of the success is π (0 < π < 1), then probability function of the Bernoulli random variable is { π x (1 π) 1 x x = 0, 1, p(x) = 0 otherwise. The table 3 summarizes basic information about the distribution of the Bernoulli variable Example 3.2. Find the expected value (the mean) and the variation of the Bernoulli random variable. 17

Table 3: Characteristics of Bernoulli distribution E(X) D(X) α 3 (X) α 4 (X) 1 2π π π(1 π) π(1 π) 1 6π(1 π) π(1 π) Solution. E(X) = M D(X) = M x i p(x i ) = 0 (1 π) + 1 π = π, [x i E(X)] 2 p(x i ) = (0 π) 2 (1 π) + (1 π) 2 π = π 2 (1 π) + (1 π) 2 π = π(1 π)(π + 1 π) = π(1 π). 3.3 The Binomial Distribution Consider an experiment where we are interested in a particular event which occurs with the probability π (0 < π < 1). Suppose that we repeat the experiment independently n times and count the number of success (the event occurs). Denote this number by X, which is then a discrete random variable with the range 0, 1,..., n. Definition 3.2. If X has the probability function {( n p(x) = x) π x (1 π) n x x = 0, 1,..., n, 0 otherwise, it is said to have a binomial distribution with parameters n and π, and we write X B(n, π). The table 4 summarizes basic information about the binomial distribution Table 4: Characteristics of binomial distribution E(X) D(X) α 3 (X) α 4 (X) Mo(X) 1 2π nπ nπ(1 π) nπ(1 π) 1 6π(1 π) nπ(1 π) (n+1)π 1 Mo(X) (n+1)π Examples of variables with the binomial distribution: the number of heads in 10 tosses of a coin, the number of imperfect products in the set of 100 products if the probability that the product is not good is 0.005, the number of defective DVD players in selected 5 ones if it is known that five percent of all DVD players are defective, etc. Let X 1,..., X n are independent Bernoulli random variables with the parameter π, then the random variable M = X 1 + X 2 + + X n has the binomial distribution B(n, π). The Poisson distribution can be used as an approximation to the binomial distribution. If n and π 0, then nπ λ ( n )π x (1 π) n x λx x x! e λ, where λ = nπ. The approximation is good when n > 30, π < 0.1. 18

Figure 8: squares (green) The Poisson distribution, stars (blue) the binomial distribution Example 3.3. The probability that the born child is a boy is 0.51. What is the probability that among five born children are a) exactly 3 girls, b) at most 3 boys? Find the probability and the distribution function of random variable the number of boys among five born children. What is the most probable number of born boy? Calculate the mean, the variation and the standard deviation of the given random variable. Solution. The range of the random variable X is 0, 1, 2,..., 5. The distribution of X can be described by the binomial distribution with parameters n = 5 and π = 0.51, X B(5, 0.51) We get the probability function {( 5 p(x) = x) 0.51 x 0.49 5 x x = 0, 1,..., 5, 0 otherwise. Table 5: The probability function and selected values of the distribution function B(5; 0.51) x 0 1 2 3 4 5 p(x) 0.0282 0.1470 0.3060 0.3185 0.1657 0.0345 F (x) 0.0282 0.1752 0.4813 0.7998 0.9655 1.0000 Now we calculate the probability that among five born children are a) exactly 3 girls, which means just 2 boys P (X = 2) = p(2). = 0.306, b) at most 3 boys P (X 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = p(0) + p(1) + p(2) + p(3) = 0.0282 + 0.147 + 0.3060 + 0.3185 = F (3). = 0.800. 19

Figure 9: The probability function and the distribution function The most probable number of born boys is determined by the mode which we can obtain from (n + 1)π 1 Mo(X) (n + 1)π, (5 + 1) 0.51 1 Mo(X) (5 + 1) 0.51, 2.06 Mo(X) 3.06 we get Mo(X) = 3. The mode is, of course, possible to find in the table of the probability function. The mean is E(X) = nπ = 5 0.51 = 2.55, the variance is D(X) = nπ(1 π) = 5 0.51 (1 0.51). = 1.250 and the standard deviation is σ = D(X). = 1.119. 3.4 The Hypergeometric Distribution Consider a set of N objects, M of which are of a special type. Suppose that we choose n objects, without replacement and without regard to their order. What is the probability that we get exactly x of the special objects? Denote the number of selected special objects by X which is a discrete random variable. Definition 3.3. If X has the probability function ( M x )( N M n x ) max{0, n N + M} x min{n, M}, p(x) = ( N n) 0 otherwise. it is said to have a hypergeometric distribution with parameters N, M and n, written X Hg(N, M, n). The table 6 summarizes basic information about the hypergeometric distribution Table 6: Characteristics of hypergeometric distribution E(X) D(X) α 3 (X) Mo(X) note nπ nπ(1 π) N n N 1 (1 2π)(N 2n) a 1 Mo(X) a π= M (M+1)(n+1), a= (N 2)σ N N+2 Examples of hypergeometric random variables: the number of the defective products among n randomly chosen products from daily output, lotteries, etc. 20

The fraction n denotes so called sample ratio. If the sample ratio is smaller than 0.05, we N can approximate the hypergemetric distribution by the binomial distribution with parameters n and π = M, thus N ( M )( N M ) ( ) x n x n ( N π n) x (1 π) n x. x Whether N is large and n relatively small, there is no significant difference between sampling without replacement (the distribution Hg(N, M, n)) and with replacement (the distribution B(n, π)). If π = M < 0.1 and n > 30, we can use another approximation (the Poisson distribution N λ = n M N ) ( M x )( N M n x ( N n ) ) λx x! e λ. Figure 10: squares (blue) the hypergeometric distribution, stars (green) the binomial distribution Example 3.4. The product is supplied in a set of 100 pieces. The output control checks 5 randomly chosen products from each set and accepts it if the is no defective product. We expect 4 % of defective products in each set. Determine a probability and a distribution function of the random variable the number of defective products in the sample. What is the probability that the the set of the products will be rejected (not accepted)? Find the mean and the standard deviation of the given random variable. Is it possible to use a binomial distribution as an approximation? Solution. The probability that the set of products will not be accepted is P (X 1) = 1 P (X < 1) = 1 P (X = 0) = 1 p(0). = 0.188. The mean of the hypergeometric distribution is E(X) = n M N σ = ( ) D(X) = n M N 1 M N n. = 0.429. N N 1 = 0.2, the standard deviation is 21

Table 7: The probability and the distribution function Hg(100, 4, 5) x 0 1 2 3 4 p(x) 0.8119 0.1765 0.0114 0.0002 1.3 10 6 F (x) 0.8119 0.9884 0.9998 1 1 Figure 11: The probability and the distribution function Hg(100, 4, 5) The sample ratio is n = 0.05 which means that we can approximate the hypergeometric N distribution by the binomial distribution B(5; 0.04). Using this approximation we get ( ) 5 P (X 1) = 1 P (X < 1) = 1 P (X = 0) = 1 0.04 0 0.96 5. = 0.185. 0 22

4 Excercises 1. During the proofreading of a new book there were found an average of 40 errors on 100 pages. (a) What is the probability that on a randomly selected 20 pages of the book will be more than 5 errors, not more than 10 errors, from 5 to 10 errors? (b) Determine the mean value of errors and the most probable number of errors on these 20 pages. 2. We plant 10 seeds of certain plants and suppose that each seed can healthy grow with probability 0.8. (a) What is the most probable number of healthy plants and what is the probability that this number will be plant? (b) Determine the probability that the number of healthy plants will be at least 5, not more than 9, from 4 to 8. 3. A certain type of components is supplied in production series of 200 pieces. During the acceptance inspection are from each series are randomly selected 5 products that the test destroyed. The series is accepted if all controlled products are not defective. Suppose that in each series are 10 defective products. (a) Describe the random variable which indicates the number of defective products among 5 randomly chosen products by a probability and distribution function and display them graphically. (b) What is the probability that series will be accepted? (c) Determine the mean, variance, standard deviation and mode of defective products in the selection. (d) Check whether the conditions for approximating the distribution of random variable by other type of distribution are met. 23

5 Models of Continuous Random Variables 5.1 The Uniform Distribution Definition 5.1. If the probability density function of X is { 1 α < x < β, β α f(x) = 0 otherwise, where α, β R, α < β, then X is said to have a uniform distribution on (α, β), written X R(α, β) The distribution function can be calculated as We obtain F (x) = x f(t) dt = x α 1 β α dt = x α β α 0 x α, x α F (x) = α < x < β, β α 1 x β. for α < x < β. Figure 12: The probability density and the distribution function R(α, β) The table summarizes basic information about the uniform distribution. Table 8: Characteristics of the uniform distribution E(X) D(X) α 3 (X) α 4 (X) quantiles x P Me(X) α+β 2 1 (β 12 α)2 0 1.2 α + P (β α) α+β 2 Examples of random variables following the uniform distribution: a time we wait for a bus (buses go regularly every 10 minutes), a time we wait for a supply of bread in a grocery store (supplies are regular), calculation rounding mistakes,... Using: P (X x 0 ) = F (x 0 ) = x 0 α for x β α 0 (α, β) P (x 1 X x 2 ) = F (x 2 ) F (x 1 ) = x 2 α x 1 α for x β α β α 1, x 2 (α, β) Example 5.1. Trams go regularly every 10 minutes. The passenger comes to the tram-stop at the arbitrary time. The random variable X is the time he/she has to wait for a tram. 24

a) Find the probability density function and the distribution function of X. b) What is the probability that the passenger will wait at most 3 minutes, at least 5 minutes, exactly 7 minutes. c) Calculate the mean, the median, the variance, the standard deviation and the 90% quantile. Solution. The random variable we can describe by the uniform distribution X R(0, 10). The probability density function is { 1 0 < x < 10, 10 f(x) = 0 otherwise, the distribution function is 0 x 0, x F (x) = 0 < x < 10, 10 1 x 10. Figure 13: The probability density and the distribution function R(0, 10) The probability that the passenger will wait at most 3 minutes, P (X 3) = 3 0 1 10 dx = 1 10 [x]3 0 = 0.3 using the distribution function P (X 3) = F (3) = 3 10 = 0.3, at least 5 minutes, P (X 5) = 10 5 1 10 dx = 1 10 [x]10 5 = 0.5 P (X 5) = 1 P (X < 5) = 1 P (X 5) = 1 F (5) = 1 5 10 = 0.5 exactly 7 minutes, P (X = 7) = 0. The mean E(X) = α+β = 10 2 2 (β 12 α)2 = 1 α+β = 5, the median Me(X) = = 10 = 5, the variance 2 2 = 8.333, the standard variation σ = D(X) = 12 D(X) = 1 (10 12 0)2 = 100 100 = 2.887, the 90% quantile x 12 0.90 = α + 0.90(β α) = 0.9 10 = 9. 25

5.2 The Exponential distribution Definition 5.2. If the probability density function of X is { 1 x α e δ x > α, δ f(x) = 0 x α, where α R, δ > 0, then X is said to have an exponential distribution with parameters α and δ, written X Ex(α, δ). The distribution function is F (x) = {1 e x α δ x > α, 0 x α. Figure 14: The probability density and the distribution function Ex(α, δ) The table summarizes some basic information about the exponential distribution. Table 9: Characteristics of the exponential distribution E(X) D(X) α 3 (X) α 4 (X) quantiles x P Me(X) α + δ δ 2 2 6 α δ ln(1 P ) α + δ ln 2 Examples of exponential models: the queuing theory, the reliability theory, the renewal theory, a time we wait for service, a product lifetime,... Using: P (X x 0 ) = F (x 0 ) = 1 e x 0 α δ for x 0 > α P (x 1 X x 2 ) = F (x 2 ) F (x 1 ) = e x 1 α δ e x 2 α δ for x 1, x 2 > α Example 5.2. It has been found out the time we have to wait for a waiter is a random variable which has an exponential distribution with the mean 5 minutes and the standard deviation 2 minutes. Plot the probability density function and the distribution function. What is the probability that we will wait at most 5 minutes? Solution. The mean and the variance of the exponential distribution are E(X) = α + δ and D(X) = δ 2, thus α + δ = 5 α = 3, δ = 2 X Ex(3, 2) δ = 2 The probability that we will wait at most 5 minutes is P (X 5) = F (5) = 1 e 5 3 2 = 1 e 1 = 0.632. 26

Figure 15: The probability density and the distribution function Ex(α, δ) 5.3 The Normal Distribution Definition 5.3. If X has the probability density function f(x) = 1 σ (x µ) 2 2π e 2σ 2 for x R where µ R, σ 2 > 0, it is said to have a normal distribution with parameters µ and σ 2, written X N(µ, σ 2 ). The distribution function is F (x) = x f(t) dt = 1 σ 2π x e (t µ)2 2σ 2 dt for x R Figure 16: The probability density and the distribution function N(µ, σ 2 ) The table summarizes some basic information about the normal distribution. Table 10: Characteristics of the normal distribution E(X) D(X) α 3 (X) α 4 (X) quantiles x P Me(X) Mo(X) µ σ 2 0 0 µ + σu P 1 µ µ The normally distributed random variable fulfils: P (µ σ < X < µ + σ) = 0.683 P (µ 2σ < X < µ + 2σ) = 0.954 P (µ 3σ < X < µ + 3σ) = 0.997 1 quantile of the standard normal distribution N(0, 1) 27

Let us have the random variable X N(µ, σ 2 ). The transformed random variable U U = X µ σ has the normal distribution with the mean 0 and the variance 1 (the standard normal distribution U N(0, 1)). The probability density function is ϕ(u) = 1 2π e u2 2 for u R, the distribution function is Φ(u) = u φ(t) dt = 1 2π u e t2 2 dt for u R Figure 17: The probability density and the distribution function N(0, 1) The table summarizes some basic information about the standard normal distribution. Table 11: Characteristics of the standard normal distribution E(X) D(X) α 3 (X) α 4 (X) quantiles x P Me(X) Mo(X) 0 1 0 0 u P 2 0 0 The values of the distribution function for positive values are tabulated, for negative values we can write Φ( u) = 1 Φ(u). If X N(µ, σ 2 ), U N(0, 1), then the distribution function of the random variable X we can obtain using distribution function of U. ( X µ F (x 0 ) = P (X x 0 ) = P (X µ x 0 µ) = P σ ( = P U x ) ( ) 0 µ x0 µ = Φ σ σ Quantiles of X (quantiles of U are tabulated): ( ) xp µ F (x P ) = Φ = Φ(u P ) σ 2 the values are tabulated, for P < 0.5 is u P = u 1 P x 0 µ σ ) 28

thus u P = x P µ x P = µ + σu P. σ Using: P (X x 0 ) = F (x 0 ) = Φ ( x 0 ) µ σ P (x 1 X x 2 ) = F (x 2 ) F (x 1 ) = Φ ( x 2 ) ( µ σ Φ x1 ) µ σ Example 5.3. During quality control we say that the component is acceptable if its size is within the limits 26 27 mm. The size of the component has a normal distribution with the mean µ = 26.4 mm and the standard deviation σ = 0.2 mm. What is the probability that the size of the component is within the given limits? Solution. The random variable X N(26.4; 0.2 2 ). ( ) ( ) 27 26.4 26 26.4 P (26 X 27) = F (27) F (26) = Φ Φ 0.2 0.2 = Φ(3) Φ( 2) = Φ(3) (1 Φ(2)) = 0.99865 (1 0.97725) = 0.9759 Figure 18: The probability density and the distribution function N(26.4; 0.04) 5.4 The Log-normal Distribution Let us assume that X is a non-negative random variable. If a random variable ln X has a normal distribution N(µ, σ 2 ), then X has a log-normal distribution LN (µ, σ 2 ). Definition 5.4. If the probability density function of X f(x) = { 1 xσ (ln x µ) 2 2π e 2σ 2 x > 0, 0 x 0, where µ 0, σ > 0, then X is said to have a log-normal distribution with parameters µ and σ 2, written X LN (µ, σ 2 ). The table summarizes some basic information about the log-normal distribution. where ω = e σ2. Examples of variables with log-normal distribution: model of entry and wages distributions, a time of renewals, repairs, the theory of non-coherent particles,... 29

Table 12: Characteristics of the log-normal distribution E(X) D(X) α 3 (X) α 4 (X) quantiles x P Mo(X) e µ+σ2 /2 e 2µ ω(ω 1) ω 1(ω+2) ω 4 +2ω 3 +3ω 2 6 e µ+σu P e µ σ 2 If the random variable X has the log-normal distribution X LN(µ, σ), then the transformed random variable U = ln X µ σ has the standard normal distribution U N(0, 1). We can write ( ) ln x0 µ F (x 0 ) = Φ = Φ(u), σ where Φ(u) is the distribution function N(0, 1). Using: P (X x 0 ) = F (x 0 ) = Φ ( ln x 0 ) µ σ P (x 1 X x 2 ) = F (x 2 ) F (x 1 ) = Φ ( ln x 2 µ σ ) ( Φ ln x1 ) µ σ Example 5.4. We suppose that distance between vehicles on the highway (in seconds) is a random variable which is possible to describe by the log-normal distribution with parameters µ = 1.27 a σ 2 = 0.49. What is the probability that the distance will be from 4 to 5 seconds? Figure 19: The probability density and the distribution function LN (1.27; 0.7) Solution. The probability is ( ) ( ) ln 5 1.27 ln 4 1.27 P (4 X 5) = F (5) F (4) = Φ Φ 0.7 0.7 = 0.68613 0.56597 = 0.12016 5.5 The Pearson, the Student and the Fisher-Snedecor Distribution Definition 5.5. Let us assume that U 1, U 2,..., U ν are independent normally distributed random variables (N(0, 1)). The random variable χ 2 = U 2 1 + U 2 2 + + U 2 ν, has a Pearson χ 2 -distribution with ν degrees of freedom. 30

Figure 20: The probability density and the distribution function χ 2 (5) and χ 2 (16) The parameter ν (the number of freedom) usually represents the number of independent observation reduced by the number of linear conditions. In the future the quantiles of the χ 2 distribution will be useful. They are usually tabulated for various values P and degrees of freedom ν 30. For ν > 30 it is possible to use an approximation where u P is the quantile of N(0, 1). χ 2 P (ν) 1 2 ( 2ν 1 + up ) 2, Definition 5.6. If a random variable U has a standard normal distribution U N(0, 1), a random variable χ 2 has a Pearson distribution χ 2 χ 2 (ν) and if U and χ 2 are independent, then a random variable t = U χ 2 ν has a Student distribution with ν degrees of freedom, written t t(ν). Figure 21: The probability density and the distribution function t(2) and t(20) The probability density function is symmetric with the mean E(t) = 0. Quantiles of the Student distribution are tabulated for ν 30 and P > 0.5, for P < 0.5 is Whether ν > 30, we can use an approximation t P = t 1 P. t p u P. 31

Definition 5.7. If a random variable χ 2 1 has χ 2 1 χ 2 (ν 1 ) with ν 1 degrees of freedom and a random variable χ 2 2 has χ 2 2 χ 2 (ν 2 ) with ν 2 degrees of freedom and they are independent, then a random variable F = χ2 1 ν 1 : χ2 2 ν 2 has a Fisher-Snedecor distribution with ν 1 and ν 2 degrees of freedom, written F F (ν 1, ν 2 ). Figure 22: The probability density and the distribution function F (30, 20) and F (3, 50) The Fischer-Snedecor distribution is asymmetric. Quantiles of F distribution are tabulated for P > 0.5, for P < 0.5 we can use the formula F P (ν 1, ν 2 ) = 1 F 1 P (ν 2, ν 1 ). 32

6 Excercises 1. Buses leave at regular intervals of 15 minutes. Traveler comes to a stop at any time. Consider the random variable which represents the waiting time for the bus. (a) Describe this random variable using the density and distribution function, express these functions mathematically and also graphically. (b) Determine the probability that the passenger will wait for a bus not more than 5 minutes, exactly 10 minutes, at least 3 minutes, 3 10 minutes. (c) Determine the mean, median, variance, standard deviation, and 90% quantile of waiting time for the bus. 2. Assume that the time between arrivals of trucks with concrete mixtures is a random variable which has an exponential distribution. The minimum time between arrivals of individual vehicles is 5 minutes, the average time is 10 minutes. (a) Describe this random variable using the density and distribution function. (b) What is the probability that the time between arrivals of each vehicle will be lower than 7 minutes, more than 11 minutes, 7 11 minutes? (c) Determine the mean, standard deviation, median, and 20% quantile of the time between arrivals of trucks. 3. Butter is cutting and packing by the machine. During the long-term observation, it was found that the production line produces packages of butter with an average weight of 246 grams and a standard deviation of 8 grams. Assume that the weight of butter is a random variable with a normal distribution. (a) What is the probability that a randomly chosen package of butter will weigh less than 250 grams? (b) Determine the probability that a randomly chosen package of butter will weigh more than 240 grams. (c) What is the proportion of packages in the production of butter which will undergo a final inspection when the permitted tolerance from the specified weight 250 grams ± is 10 grams? 4. The random variable X has a log-normal distribution LN (3.5; 0.36). (a) Calculate the value of the distribution function F (x) at point x = 16, median, standard deviation, mode, 5% quantile and the coefficient of skewness of this random variable. (b) Determine the probability that the random variable acquires value less than 20, greater than 30, from 20 to 30. What is true for the sum of these probabilities and why? 5. The random variable t has Student s distribution t(24). (a) Determine 2.5% a 97.5% quantiles of a random variable t. (b) Determine the probability P (t > 2.064). 33

6. The random variable χ 2 has Pearson s distribution χ 2 (15). (a) Determine 5% a 95% quantiles of a random variable χ 2. (b) Determine the probability P (χ 2 < 7.26). 7. The random variable F has Fisher s distribution F (12, 7). (a) Determine 5% a 95% quantiles of a random variable F. (b) Determine the probability P (F < 4.666). 34