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Class Notes in Statistics and Econometrics Part 4

CHAPTER 7 Chebyshev Inequality, Weak Law of Large Numbers, and Central Limit Theorem. 7.1. Chebyshev Inequality If the random variable y has finite expected value µ and standard deviation σ, and k is some positive number, then the Chebyshev Inequality says | CHAPTER 7 Chebyshev Inequality Weak Law of Large Numbers and Central Limit Theorem 7.1. Chebyshev Inequality If the random variable y has finite expected value y and standard deviation a and k is some positive number then the Chebyshev Inequality says 7.1.1 Pr y -mI M T2 k In words the probability that a given random variable y differs from its expected value by more than k standard deviations is less than 1 k2. Here more than and less than are short forms for more than or equal to and less than or equal 189 T98CHEBYSHEV INEQUALITY WEAK LAW OF LARGE NUMBERS AND CENTRAL LIMIT THEC to. One does not need to know the full distribution of y for that only its expected value and standard deviation. We will give here a proof only if y has a discrete distribution but the inequality is valid in general. Going over to the standardized variable z y we have to show Pr z k k2. Assuming z assumes the values Z1 z2 . with probabilities p z1 p z2 . then 7.1.2 Pr z k P z . i Zi k Now multiply by k2 7.1.3 k2 Pr z k k2p zi 7.1.4 z2 P zi 7.1.5 2 z2P zi var z 1. The Chebyshev inequality is sharp for all k 1. Proof the random variable which takes the value k with probability and the value k with probability 7.1. CHEBYSHEV INEQUALITY 191 212 and 0 with probability 1 2 has expected value 0 and variance 1 and the -sign in 7.1.1 becomes an equal sign. Problem 115. HT83 p. 316 Let y be the number of successes in n trials of a Bernoulli experiment with success probability p. Show that 7.1.6 pr y p 1 n 4ns2 Hint first compute what Chebyshev will tell you about the lefthand side and then you will need still another inequality. ANSWER. E y n p and var y n pq n where q 1 p . Chebyshev says therefore 7.1.7 Pr y pl fcJ -2. n n k2 Setting e ky pq n therefore 1 k2 pq ae1 one can rewerite 7.1.7 as 7.1.8 Pr y p e - L. Now note that pq 1 4 whatever their values are. Problem 116. 2 points For a standard normal variable Pr z 1 is approximately 1 3 please look up the precise value in a table. What does the .