University of California, Los AngelesDepartment of StatisticsStatistics 100A Instructor: Nicolas ChristouRandom variables• Discrete random variables.• Continuous random variables.• Discrete random variables. Denote a discrete random variable with X:It is a variable that takes values with some probability.Examples:a. Roll a die. Let X be the number observed.b. Draw 2 cards with replacement. Let X be the number of aces among the 2 cards.c. Roll 2 dice. Let X be the sum of the 2 numbers observed.d. Toss a coin 5 times. Let X be the number of tails among the 5 tosses.e. Randomly select a US household. Let X be the number of people live in thishousehold.• Probability distribution of a discrete random variable XIt is the list of all possible values of X with the corresponding probabilities. It can berepresented by a table, a graph, or a function.Examples:a. Roll a die. Let X be the number observed. The probability distribution of X is:X P (X = x)116216316416516616XP(X=x)1 2 3 4 5 60.00 0.05 0.10 0.151b. Roll two dice. Let X be the sum of the two numbers observed. The probabilitydistribution of X is:X P (X = x)21363236433654366536763685369436103361123612136XP(X=x)2 3 4 5 6 7 8 9 10 11 120.00 0.05 0.10 0.15We can also represent this distribution with a function: P (X = x) =6−|x−7|36,for x = 2, 3, ···, 12. This is called probability mass function and returns theprobability for each possible value of the random variable X.• Expected value (or mean) of a discrete random variableIt is denoted with E(X) or µ and it is computed as follows:Definition:µ = E(X) =XxxP (X = x)It is a weighted average. The weights are the probabilities.2Example:Roll a die. Let X be the number observed. Find the expected value of X. The E(X)must be somewhere between 1, 6: E(X) = 116+ 216+ 316+ 416+ 516+ 616= 3.5.What does this number mean?Example: Casino roulette. Below you see the standard Nevada roulette table:A player will bet 1$ on four joining numbers. This bet pays 8 : 1. Let X be the player’spayoff. Find the player’s expected payoff.3• Expected value of a sum of random variablesLet X and Y be 2 random varables. The expected value of the sum of these 2 randomvariables is:E(X + Y ) = E(X) + E(Y )Example:Roll 2 dice. Let X be the number observed on the first die and Y be the numberobserved on the second die. Let W be the sum of the 2 dice. Find the expected valueof W . There are 2 ways to solve this problem:a. E(W ) = E(X + Y ) = E(X) + E(Y ) = 3.5 + 3.5 = 7b. Or using the distribution of the sum of the two dice (see page 2):E(W ) =XwwP (W = w) = 2136+ 3236+ ··· + 12136= 7.The expected value of the sum can be extended to more than two random variables:E(X + Y + Z + ···) = E(X) + E(Y ) + E(Z) + ···• Expected value of a function of a random variableLet g(X) a function of a discrete random variable X. Then its expected value iscomputed as follows:E[g(X)] =Xxg(x)P (X = x)4• Variance and standard deviation of a discrete random variableConsider the following 2 probability distributions:X P (X) Y P (Y )012512112-412What do you observe?Definition:V ar(X) = σ2= E(X − µ)2=Xx(x − µ)2)P (X = x) =Xxx2P (X = x) − µ2The standard deviation of a discrete random variable is the square root of the variance:SD(X) =√σ2=sXx(x − µ)2)P (X = x) =Xxx2P (X = x) − µ2It follows that:σ2= EX2− µ2or EX2= σ2+ µ2Example:Roll a die. Let X be the number observed. Find the variance of X.V ar(X) = 1216+ 2216+ 3216+ 4216+ 5216+ 6216− 3.52= 2.917.The standard deviation is:SD(X) =√2.917 = 1.708.• Some properties expectation and varianceLet X, Y random variables and a, b constants.a. E(aX) = aE(X)b. E(aX + b) = aE(X) + bc. V ar(X + a) = V ar(X)d. V ar(aX) = a2V ar(X)e. V ar(aX + b) = a2V ar(X).f. If X, Y are independent then V ar(X + Y ) = V ar(X) + V ar(Y )5Example:An insurance policy costs $100, and will pay policyholders $10000 if they suffer a majorinjury (resulting in hospitalization) or $3000 if they suffer a minor injury (resulting in losttime from work). The company estimates that each year 1 in every 2000 policyholders mayhave a major injury, and 1 in 500 a minor injury.a. Construct the probability distribution for the profit on a policy.b. What is the company’s expected profit on this policy?c. Do you think the standard deviation is large or small. Why?d. Compute the standard deviation.e. Suppose that the company writes (a) 36, (b) 10000 of these policies per year. Whatare the mean and standard deviation of the annual profit for these 2 cases?f. Comment!6Some examplesExample 1The number N of residential homes that a fire company can serve depends on the dis-tance r (in city blocks) that a fire engine can cover in a specified (fixed) period of time.If we assume that N is proportional to the area of a circle r blocks from the firehouse,then N = Cπr2, where C is a constant, π = 3.1416..., and r, a random variable, isthe number of blocks that a fire engine can move in the specified time interval. For aparticular fire company, C = 8, the probability distribution for r is as shown in theaccompanying table, and p(r) = 0 for r ≤ 0 and r ≥ 27.r 21 22 23 24 25 26p(r) .05 .20 .30 .25 .15 .05Find the expected value of N, the number of homes that the fire department can serve.Example 2Suppose that X takes on one of the values 0, 1, 2. If for some constant c, P (X = i) =cP (X = i − 1), i = 1, 2 find E(X) in terms of c.Example 3Two balls are chosen randomly without replacement from and urn containing 8 white,4 black, and 2 orange balls. Suppose that we win $2 for each black ball selected andwe lose $1 for each white ball selected. We neither win nor we lose any money forselecting an orange ball. Let X denote our winnings.a. What is the expected value of X?b. What is the variance of X?c. Given that our winnings are negative, what is the probability that we lost exactly$2?Example 4To determine whether or not they have a certain disease, 100 people are to have theirblood tested. However, rather than testing each individual separately, it has been de-cided first to group the people in groups of 10. The blood samples of the 10 peoplein each group will be pooled and analyzed together. If the test is negative, one testwill suffice for the 10 people. If the test is positive each of the 10 people will alsobe individually tested. Suppose the probability that a person has the disease is 0.10for all people independently from each other. Compute the expected number of testsnecessary for each