10-07 Random Variables (2)

# 10-07 Random Variables (2) - BTRY 4080 STSCI 4080 Fall 2010...

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BTRY 4080 / STSCI 4080 Fall 2010 190 Section 4.3: Expected Value Definition: If X is a discrete random variable having a probability mass function p ( x ) , the expected value of X is defined by E ( X ) = summationdisplay x : p ( x ) > 0 x p ( x ) = summationdisplay i : x i S x x i p ( x i ) . where the second form simply recalls the definition of the support of X (i.e., S x ). The expected value is a probability-weighted average of the possible values that X can take on. The resulting value need not satisfy p ( E ( X )) > 0 !

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BTRY 4080 / STSCI 4080 Fall 2010 191 Example (4.3.3a) Let X be the outcome when we roll one fair die. Then: E ( X ) = 6 summationdisplay i =1 i P ( X = i ) = 6 summationdisplay i =1 parenleftbigg i × 1 6 parenrightbigg = 7 2 Example (see also 4.3.3b) Define the random variable Y = I { A occurs } , where A is some event. Then, Y = 1 if A occurs 0 if A c occurs What is E ( Y )?
BTRY 4080 / STSCI 4080 Fall 2010 192 More Examples: Roll a fair die until a “6” appears. What is the expected number of trials required? Let W be a discrete random variable with pmf P ( W = i ) = e - λ λ i i ! for i = 0 , 1 , 2 ,... , where λ > 0 . (i.e., S w is the set of nonnegative integers). What is E ( W ) ? Let X be a random variable with pmf P ( X = i 2 ) = 6 π 2 i for i = 1 , 2 ,... (i.e., S x is the set of positive integers). What is E ( X ) ?

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BTRY 4080 / STSCI 4080 Fall 2010 193 Example (Intel, Slides 165-167) Define the random variable T to be the division on which the first error occurs. Let the pmf of T be p ( n ) = p (1 p ) n 1 , n 1 . What is E ( T ) = n =1 nP ( T = n ) ? E ( T ) = summationdisplay n =1 np (1 p ) n 1 = p summationdisplay n =1 n (1 p ) n 1 = p (1 (1 p )) 2 = 1 p , the so-called “mean time before failure” (MTBF).
BTRY 4080 / STSCI 4080 Fall 2010 194 Section 4.4: Expectations of Functions of Random Variables Example (4.4.4a) Let X be a random variable with pmf: P ( X = 1) = 0 . 2

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