5.3 EXPECTED VALUE AND VARIANCE def: The expected value of a random - PDF document

5.3.1 Section 5.3 Expected Value and Variance 5.3 EXPECTED VALUE AND VARIANCE def: The expected value of a random variable X on a probability space ( S, p ) is the sum � E ( X ) = X ( s ) p ( s ) s ∈ S The expected outcome of a Example 5.3.1: fair die is 1 · 1 6 + 2 · 1 6 + 3 · 1 6 + 4 · 1 6 + 5 · 1 6 + 6 · 1 6 = 21 6 = 7 2 The expected outcome of the Example 5.3.2: standard loaded die is 1 · 1 21 + 2 · 2 21 + · · · + 6 · 6 21 = 91 21 = 13 3 Over a non-uniform probability space the expected value of a random variable is the weighted mean. Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.2 Chapter 5 DISCRETE PROBABILITY Flip a fair coin three times. Example 5.3.3: The expected number of heads is 0 · 1 8 + 1 · 3 8 + 2 · 3 8 + 3 · 1 8 = 12 8 = 3 2 This calculation is based on the binomial distribution. Flip a standard loaded coin Example 5.3.4: three times. The expected number of heads is 125 + 1 · 12 1 125 + 2 · 48 125 + 3 · 64 0 · 125 = 0 + 12 + 96 + 192 = 300 125 = 12 125 5 This calculation too is based on the binomial distribution. Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.3 Section 5.3 Expected Value and Variance SUMMING RANDOM VARIABLES Theorem 5.3.1. Let X 1 and X 2 be random variables on a probability space ( S, p ) . Then E ( X 1 + X 2 ) = E ( X 1 ) + E ( X 2 ) Proof: � E ( X 1 + X 2 ) = ( X 1 ( s ) + X 2 ( s )) p ( s ) s ∈ S � = X 1 ( s ) p ( s ) + X 2 ( s ) p ( s ) s ∈ S � � = X 1 ( s ) p ( s ) + X 2 ( s ) p ( s ) s ∈ S s ∈ S = E ( X 1 ) + E ( X 2 ) When two fair dice are rolled, Example 5.3.5: here are both calculations: E ( X 1 ) + E ( X 2 ) = 7 2 + 7 2 = 7 and 6 6 E ( X 1 + X 2 ) = 1 ( j + k ) = 252 � � 36 = 7 36 j =1 k =1 Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.4 Chapter 5 DISCRETE PROBABILITY Theorem 5.3.2. Let X 1 , . . . , X n be random variables on a probability space ( S, p ) . Then E ( X 1 + · · · + X n ) = E ( X 1 ) + · · · + E ( X n ) By induction on n , using Thm 5.3.1. ♦ Proof: When 100 fair coins are Example 5.3.6: tossed, the expected number of heads is 1 2 · 100 = 50 When 100 standard loaded Example 5.3.7: coins are tossed, the expected number of heads is 0 . 8 · 100 = 80 Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.5 Section 5.3 Expected Value and Variance GEOMETRIC DISTRIBUTION def: The geometric distribution on the positive integers is pr ( k ) = (1 − p ) k − 1 p A coin with p ( H ) = p is Example 5.3.8: tossed until the first occurrence of heads. Then the probability of requiring exactly k tosses is (1 − p ) k − 1 p . We observe that ∞ ∞ p (1 − p ) k − 1 = � � (1 − p ) k − 1 p = p 1 − (1 − p ) = 1 k =1 k =1 It is proved in the text that ∞ � (1 − p ) k − 1 pk E ( X ) = k =1 x =1 − p = 1 = p d dx (1 − x ) − 1 � � p � Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.6 Chapter 5 DISCRETE PROBABILITY INDEPENDENT RANDOM VARIABLES def: The random variables X and Y on the probability space ( S, p ) are independent if for all real numbers r 1 and r 2 p ( X = r 1 ∧ Y = r 2 ) = p ( X = r 1 ) · p ( Y = r 2 ) Suppose that X is the sum of Example 5.3.9: two fair dice and Y is the product. Then p ( X = 2) = 1 p ( Y = 5) = 1 and 36 18 However, p ( X = 2 ∧ Y = 5) = 0 � = 1 36 · 1 18 Thus X and Y are not independent. Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.7 Section 5.3 Expected Value and Variance VARIANCE and STANDARD DEVIATION def: The variance of a random variable X on a probability space ( S, p ) is the sum � 2 p ( s ) � σ 2 ( X ) = V ( X ) = � X ( s ) − E ( X ) s ∈ U def: The standard deviation of a random variable X on a probability space ( U, p ) is � σ ( X ) = V ( X ) Flip a fair coin three times. Example 5.3.10: The variance of the number of heads is 0 − 3 · 1 1 − 3 · 3 2 − 3 · 3 3 − 3 · 1 � 2 � 2 � 2 � 2 � � � � 8 + 8 + 8 + 2 2 2 2 8 = 9 4 · 1 8 + 1 4 · 3 8 + 1 4 · 3 8 + 9 4 · 1 8 = 24 32 = 3 4 The standard deviation is √ � 3 3 4 = 2 Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.

5.3.8 Chapter 5 DISCRETE PROBABILITY Flip a standard loaded coin Example 5.3.11: three times. The variance of the number of heads is 0 − 12 1 1 − 12 · 12 � 2 � 2 � � · 125 + 5 5 125 2 − 12 · 48 3 − 12 · 64 � 2 � 2 � � + 125 + 5 5 125 = 144 + 49 · 12 + 4 · 48 + 9 · 64 = 1500 = 12 5 5 5 5 25 The standard deviation is √ � 12 25 = 2 3 5 CHEBYSHEV INEQUALITY Chebyshev Inequality. Let X be a random variable on any probability space that has a mean and a variance. Then ≤ 1 � � | X ( s ) − E ( X ) | ≥ kσ ( X ) p k 2 Coursenotes by Prof. Jonathan L. Gross for use with Rosen: Discrete Math and Its Applic., 5th Ed.