1 Chapter 4 Mathematical Expectation  4.1 Mean of Random Variables  4.2 Variance and Covariance  4.3 Means and Variances of Linear Combinations of Random variables  4.4 Chebyshev’s Theorem

2 4.1 Mean of a Random Variables Example Consider the example of tossing two coins 16 times, what is the average of heads observed per toss? Let X be the number of heads that occur per toss, then the values of X can be 0, 1, and 2. Suppose that the experiment yields no head, one head, and two heads a total of 4, 7, and 5 times, respectively. The average-number of heads per toss of the two coins over 16 tosses is: Notice that this average value 1.06 is not a possible outcome of X. Rewrite the above computation as The fractions of the total tosses resulting in 0,1 and 2 heads respectively =1.06

3 Definition 4.1 Let X be a random variable with probability distribution f(x). The mean or expected value of X is if X is discrete, and if X is continuous. Remark: The mean  of a random variable X can be thought of as a measure of the “center of location” in the sense that it indicates where the “center” of the density line.

4 Example 4.1, page 89  The probability distribution of a random variable X is given by x=0,1,2,3. f(0)=1/35 f(1)=12/35 f(2)=18/35 f(3)=4/35

5 Example  The probability distribution of a random variable X is given by ( )

6 Example 4.2, page 90 In a gambling game a man is paid $5 if he gets all heads or all tails when three coins are tossed, and he will pay out $3 if either one or two heads show. What is his expected gain? Let Y be the amount of gain per bet. The possible values are 5 and –3 dollars. Let X be the number of heads that occur in tossing three coins. The possible values of X are 0, 1, 2, and 3. Solution: P(Y = 5) = P(X = 0 or X = 3) = 1/8 + 1/8 = ¼ P(Y = -3) = P(X =1 or X = 2) = 6/8 = ¾  = (5)(1/4) + (-3)(3/4) = –1 Interpretation: Over the long run, the gambler will, on average, lose $1 per bet. Most likely, the more the gambler play the games, the more he would lose.

7 Notice that in the preceding example, there are two random variables, X and Y; and Y is a function of X, for example if we let E(Y) = E(g(X)) = (5)P(Y = 5) + (-3)P(Y = -3) = (5)[P(X = 0) + P(X = 3)] + (-3)[P(X = 1) + P(X = 2)] = (5)P(X = 0) + (5)P(X = 3) + (-3)P(X = 1) + (-3)P(X = 2) = g(0)P(X = 0) + g(3)P(X = 3) + g(1)P(X = 1) + g(2)P(X = 2) =

8 Theorem 4.1 Let X be a random variable with probability distribution f(x). The mean or expected value of random variable g(X) is if X is discrete, and if X is continuous.

9 Example Let X denote the length in minutes of a long- distance telephone conversation. Assume that the density for X is given by Find E(X) and E(2X+3) Solution: E(X ) = = = 10 E(2X+3)= = 2(10) + 3 = 23.

10 Definition 4.2 Let X and Y be random variables with joint probability distribution f(x, y). The mean or expected value of the random variable g(X, Y) is if X and Y are discrete, and if X and Y are continuous. Extension

11 Example  Example: Suppose two dice are rolled, one red and one white. Let X be the number on the top face of the red die, and Y be the number on the top face of the white one. Find E(X+Y). E[X + Y] = = = = = 7

12  Example 4.7, page 93. Find E[Y/X] for the density Solution: = =

13 In general If X and Y are two random variables, f(x, y) is the joint density function, then:  E(X)= = (discrete case)  E(X) = (continuous case)  E(Y) = = (discrete case)  E(Y) = (continuous case) g(x) and h(y) are marginal probability distributions of X and Y, respectively.