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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 1 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: (0)(4) (1)(7) (2)(5) =1.06 16 Notice that this average value 1.06 is not a possible outcome of X. Rewrite the above computation as 4 7 5 (0) (1) (2) 16 16 16 =1.06 The fractions of the total tosses resulting in 0,1 and 2 heads respectively 2 Definition 4.1 Let X be a random variable with probability distribution f(x). The mean or expected value of X is E ( X ) xf ( x) x if X is discrete, and E ( X ) xf ( x)dx 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. 3 Example 4.1, page 89 The probability distribution of a random variable X is given by f ( x) f(0)=1/35 4 3 x 3 x 7 3 f(1)=12/35 f(2)=18/35 x=0,1,2,3. f(3)=4/35 18 4 12 (0)( 351 ) (1)( 12 ) ( 2 )( ) ( 3 )( ) 35 35 35 7 4 Example The probability distribution of a random variable X is given by e x x 0 f ( x) 0 elsewhere E( X ) xe dx xe 0 x x 0 ( e 0 x 0) 1 dx 5 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 6 play the games, the more he would lose. 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 X 0,3 5 Y g( X ) 3 X 1,2 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) = g ( x) f ( x) x 7 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 g ( X ) E[ g ( X )] g ( x) f ( x) x if X is discrete, and g ( X ) E[ g ( X )] g ( x) f ( x)dx if X is continuous. 8 Example Let X denote the length in minutes of a longdistance telephone conversation. Assume that the density for X is given by f ( x) 1 10 e x /10 x0 Find E(X) and E(2X+3) Solution: x f ( x)dx E(X ) = E(2X+3)= (2 x 3) 0 1 x / 10 = x e dx = 10 0 10 1 x /10 e dx 10 = 2(10) + 3 = 23 . 9 Extension 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 g ( X ,Y ) E[ g ( X , Y )] g ( x, y) f ( x, y) ( x, y ) if X and Y are discrete, and g ( X ,Y ) E[ g ( X , Y )] g ( x, y ) f ( x, y )dxdy if X and Y are continuous. 10 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). 6 6 E[X + Y] =( x, y )( x y) P( X x, Y y) = y1 x1( x y ) P( X x, Y y ) 6 6 6 6 6 6 = y1 x1 xP( X x, Y y ) y1 x1 yP( X x, Y y ) 6 6 = y1 x1 x(1 / 36) y1 x1 y (1 / 36) = 3.5 + 3.5 = 7 11 Example 4.7, page 93. Find E[Y/X] for the density x3 y3 , f ( x, y ) 16 0, Solution: 0 x 2, 0 y 2 elsewhere g ( X ,Y ) E[ g ( X , Y )] g ( x, y ) f ( x, y )dxdy 3 3 2 2 x2 y4 y x y = dxdy dxdy = 0 0 16 0 0 x 16 2 2 12 In general If X and Y are two random variables, f(x, y) is the joint density function, then: E(X)= xf ( x, y ) = x y xg ( x) x E(X) = xf ( x, y )dxdy xg ( x)dx E(Y) = yf ( x, y ) = yh( y ) y x (discrete case) y (continuous case) (discrete case) E(Y) = yf ( x, y )dxdy yh( y )dy (continuous case) g(x) and h(y) are marginal probability distributions of X and Y,13 respectively.