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Statistics 510: Notes 8 Reading: Sections 4.1-4.4 I. Random Variables So far, we have been defining probability functions in terms of the elementary outcomes making up an experiment’s sample space. Thus, if two fair dice were tossed, a probability was assigned to each of the 36 possible pairs of upturned faces,: P((3,2))=1/36, P((2,3))=1/36, P((4,6))=1/36 and so on. We have seen that in certain situations some attribute of an outcome may hold more interest for the experimenter than the outcome itself. A craps player, for example, may be concerned only that he throws a 7, not whether the 7 was the result of a 5 and a 2, a 4 and a 3 or a 6 and a 1. That, being the case, it makes sense to replace the 36member sample space of (x,y) pairs with the more relevant (and simpler) 11-member set of all possible two-dice sums, S {x y : x y 2,3, ,12} . This redefinition of the sample space not only changes the number of outcomes in the space (from 36 to 11) but also changes the probability structure. In the original sample space, all 36 outcomes are equally likely. In the revised sample space, the 11 outcomes are not equally likely. The probability of getting a sum equal to 2 is 1/36[=P((1,1))], but the probability of getting a sum equal to 3 is 2/36[=P((1,2))+P((2,1))]. In general, rules for redefining sample spaces – like going from (x,y)’s to (x+y)’s – are called random variables. A random variable is simply a function that is defined on the sample space of the experiment and that assigns a numerical variable to each possible outcome of the experiment. We denote random variables by uppercase letters, often X, Y or Z. A random variable that can take on a finite or at most countably infinite number of values is said to be discrete; a random variable that can take on values in an interval of real numbers, bounded or unbounded, is said to be continuous. We will focus on discrete random variables in Chapter 4 and consider continuous random variables in Chapter 5. Associated with each discrete random variable X is a probability mass function (pmf) p ( a ) that gives the probability that X equals a: p(a) P{ X a} P({s S | X ( s) a}) . Example 1: Suppose two fair dice are tossed. Let X be the random variable that is the sum of the two upturned faces. X is a discrete random variable since it has finitely many possible values (the 11 integers 2, 3, ..., 12). The probability mass function of X is P(X=2)=1/36 P(X=3)=2/36 P(X=4)=3/36 P(X=5)=4/36 P(X=6)=5/36 P(X=7)=6/36 P(X=8)=5/36 P(X=9)=4/36 P(X=10)=3/36 P(X=11)=2/36 P(X=12)=1/36 It is often instructive to present the probability mass function in a graphical format plotting p ( xi ) on the y-axis against xi on the x-axis. Example 2: Three balls are to be randomly selected without replacement from an urn containing balls numbered 1 through 20. Let X denote the largest number selected. X is a random variable taking on values 3, 4, ..., 20. Since we select the balls randomly, each of the 20 combinations of the balls is equally likely to be 3 chosen. The probability mass function is i 1 2 P{ X i} , i 3, , 20 20 3 This equation follows because the number of selections that result in the event { X i} is just the number of selections that result in the ball numbered i and two of the balls numbered 1 through i-1 being chosen. Suppose the random variable X can take on values x1 , x2 , Since the probability mass function is a probability function on the redefined sample space that considers values of X, we have that P( X x ) 1 . i i 1 [This follows from 1 P( S ) P( i 1 { X xi }) P( X xi ) ] i 1 Example 3: Independent trials, consisting of the flipping of a coin having probability p of coming up heads, are continually performed until either a head occurs or a total of n flips is made. Let X be the random variable that denotes the number of times the coin is flipped. The probability mass function for X is P{ X 1} P{H } p P{ X 2} P{(T , H )} (1 p) p P{ X 3} P{(T , T , H )} (1 p) 2 p P{ X n 1} P{(T , T , , T , H )} (1 p) n 2 p n2 P{ X n} P{(T , T , , T , T ), (T , T , n 1 , T , H )} (1 p) n 1 n 1 As a check, note that n n 1 i 1 i 1 P{ X i} p(1 p) i 1 (1 p) n 1 1 (1 p) n 1 n 1 p (1 p) 1 (1 p) 1 (1 p) n 1 (1 p) n 1 1 II. Expected Value Probability mass functions provide a global overview of a random variable’s behavior. Detail that explicit, though, is not always necessary – or even helpful. Often times, we want to focus the information contained in the pmf by summarizing certain of its features with single numbers. The first feature of a pmf that we will examine is central tendency, a term referring to the “average” value of a random variable. The most frequently used measure for describing central tendency is the expected value. Motivation for expected value: Let X 1 , , X n be the sum of the two dice in n independent throws of two dice. The mean of X 1 , , X n is 12 # of throws in which X = i i * n i 2 From the frequentist definition of probability, as n becomes # of throws in which X i large, becomes close to p (i ) n so that the mean of X 1 , , X n becomes close to 12 i * p(i) . i2 12 This last quantity i * p(i) is called the i2 expected value of X -- it is what the mean value of X over many repeated experiments converges to. Generally, for a discrete random variable, the expected value of a random variable X is a weighted average of the possible values X can take on, each value being weighted by the probability that X assumes it: E[ X ] xp ( x ) . x: p ( x ) 0 Example 1 continued: The expected value of the random variable X is E[ X ] 2*(1/ 36) 3*(2 / 36) 4*(3 / 36) 5*(4 / 36) 6*(5 / 36) 7*(6/36)+8*(5/36)+9*(4/36)+10*(3/36)+11*(2/36)+12*(1/36)=7 Example 2 continued: The expected value of the random variable X is i 1 20 2 E( X ) i 15.75 20 i 3 3 IV. Expectation of Function of a Random Variable (Chapter 4.4) Suppose we are given a discrete random variable X along with its pmf and that we want to compute the expected value of some function of X, say g(X). One approach is to directly determine the pmf of g(X). Example 3: Let X denote a random variable that takes on the values -1, 0, 1 with respective probabilities P{X=-1}=.2, P{X=0}=.5, P{X=1}=.3 2 Compute E ( X ) . Although the procedure we used in Example 3 will always enable us to compute the expected value of g(X) from knowledge of the pmf of X, there is another way of thinking about E[ g ( X )] . Noting that g(X) will equal g(x) whenever X is equal to x, it seems reasonable that E[ g ( X )] should just be a weighted average of the values g(x) with g(x) being weighted by the probability that X is equal to x. Proposition 4.1: If X is a discrete random variable that takes on one of the values xi , i 1 with respective probabilities p ( xi ) , then for any real valued function g, E[ g ( X )] g ( xi ) p( xi ) . i Applying the proposition to Example 3, E ( X 2 ) (1)2 (.2) 02 (.5) 12 (.3) .5 Proof: g ( xi ) p( xi ) i j i:g ( xi ) y j yj j g ( xi ) p ( xi ) i:g ( xi ) y j p ( xi ) y j P{g ( X ) y j } j E[ g ( X )] A corollary of Proposition 4.1 is: Corollary 4.1: If a and b are constants, then E[aX b] aE[ X ] b . Proof: E[aX b] (ax b) p ( x) x: p ( x ) 0 a xp( x) b x: p ( x ) 0 aE[ X ] b x: p ( x ) 0 p ( x)