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Unformatted text preview: 1 Expected Values of Random Variables Often it is convenient to describe a random variable using some location measure. The most important location measure is the expected value (a.k.a. the mean or the weighted average). For a discrete random variable X its expected value, denote E [ X ], is given by E [ X ] = X k x k p ( x k ) . For a continuous random variable E [ X ] = Z  xf ( x ) dx. Remark: Strictly speaking E [ X ] exists only if both E [ X + ] and E [ X ] exists and are not both where X + = max( X, 0) and X = max( X, 0). Example: Suppose X takes values { , 1 , 2 , 3 } with probabilities { 1 / 8 , 3 / 8 , 3 / 8 , 1 / 8 } . Then, E [ X ] = 0(1 / 8)+ 1(3 / 8) + 2(3 / 8) + 3(1 / 8) = 12 / 8 = 1 . 5. Remark: E [ X ] need not be a possible outcome of X . Frequency, or long run average interpretation of the expected value: Example: Suppose a random variable X represents the profit associated with the production of some item that can be defective or non defective. Suppose that the profit is 2 when the item is defective and 10 when the item is nondefective. Finally, assume that p ( 2) = 0 . 1 , p (10) = 0 . 9. Then E [ X ] = 2(0 . 1) + 10( . 9) = 8 . 8. Suppose that a very large number n of items are produced and let n ( G ) be the number of good items and n ( D ) is the number of defective items. Then the average profit per items is 2 n ( D ) n + 10 n ( G ) n The frequency interpretation is that n ( G ) /n converges in some sense to be defined later to p (10) = 0 . 9. This convergence is know as the Law of Large Numbers. 1.1 Markovs Inequality Proposition: Suppose X is a nonnegative random variable and c > 0. Then P ( X c ) E [ X ] c . Notice that the inequality is nontrivial if c > EX . Proof (Discrete case): E [ X ] = X k kp ( k ) X k c kp ( k ) c X k c p ( k ) = cP ( X c ) . Alternative format: c = kE [ X ] then P ( X kE [ X ]) 1 k ....
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 Spring '04
 GALLEGO

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