PostLecture02EN.pdf - Lecture 2 Chapter 2.4-2.5(till...

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Lecture 2: Chapter 2.4-2.5 (till Factorial moments). Plan for lecture 2 1. Expected value of a function of a random variable E ( u ( X )) 2. Properties of the expected value 3. Variance and properties of the variance 4. Moments of a random variable E ( X k ) 5. The inequalities of Markov and Chebychev
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The expected value of a function of a random variable Example 1: Let X be a discrete random variable with the probability mass function P ( X = 1 ) = 1 2 , P ( X = 2 ) = 1 4 , P ( X = 3 ) = 1 4 . I Calculate E ( X 2 ) E ( X 2 ) = 1 2 P ( X = 1 ) + 2 2 P ( X = 2 ) + 3 2 P ( X = 3 ) = 15 4 I Calculate E ( max ( X , 3 2 )) . E ( max ( X , 3 2 )) = max ( 1 , 3 2 ) P ( X = 1 ) + max ( 2 , 3 2 ) P ( X = 2 ) + max ( 3 , 3 2 ) P ( X = 3 ) = 3 2 * 1 2 + 2 * 1 4 + 3 * 1 4 = 2 . The expected value of a function of a random variable Theorem ( Th. 2.4.1 ) I Discrete variable The expected value of a function u ( X ) of a discrete random variable X : S 7→ R with the probability mass function P [ · ] is E [ u ( X )] = X x u ( x ) P ( X = x ) . I Continuous variable The expected value of a function u ( X ) of a continuous random variable X with the probability density function f ( x ) is E ( u ( X )) = Z + -∞ u ( x ) f ( x ) dx , if the integral exists. Pay attention to the boundaries of the integral!
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Example 2 Example 2 Choose a number X between 0 and 1. What is the expected value of e X if the cumulative distribution function of X is given by: I F ( x ) = 0 x 0 x 0 x 1 1 otherwise J (3) I E ( e X ) = R + -∞ e x f ( x ) dx . I f ( x ) = 1 0 < x < 1 0 otherwise I E ( e X ) = R 1 0 e x dx = e - 1 = 1 , 718 . I PLEASE NOTE! E ( e X ) 6 = e E ( X ) !!!!!!!!!!!!!! Properties of the expected value Theorem Let X be a random variable with probability density function f ( x ) , a , b R and g and h two functions defined on the codomain of X. Then E [ ag ( X ) + bh ( X )] = aE [ g ( X )] + bE [ h ( X )] .
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  • Fall '19
  • Variance, Probability theory, probability density function

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