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# SolvedProblems - 5.5 SOLVED PROBLEMS 81 Example 5.5.10 Let...

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5.5. SOLVED PROBLEMS 81 Example 5.5.10. Let X be uniformly distributed in [0 , 2 π ] and Y = sin( X ) . Calculate the p.d.f. f Y of Y . Since Y = g ( X ), we know that f Y ( y ) = X 1 | g 0 ( x n ) | f X ( x n ) where the sum is over all the x n such that g ( x n ) = y . For each y ( - 1 , 1), there are two values of x n in [0 , 2 π ] such that g ( x n ) = sin( x n ) = y . For those values, we find that | g 0 ( x n ) | = | cos( x n ) | = q 1 - sin 2 ( x n ) = p 1 - y 2 , and f X ( x n ) = 1 2 π . Hence, f Y ( y ) = 2 1 p 1 - y 2 1 2 π = 1 π p 1 - y 2 . Example 5.5.11. Let { X, Y } be independent random variables with X exponentially dis- tributed with mean 1 and Y uniformly distributed in [0 , 1] . Calculate E (max { X, Y } ) . Let Z = max { X, Y } . Then P ( Z z ) = P ( X z, Y z ) = P ( X z ) P ( Y z ) = z (1 - e - z ) , for z [0 , 1] 1 - e - z , for z 1 . Hence, f Z ( z ) = 1 - e - z + ze - z , for z [0 , 1] e - z , for z 1 .

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82 CHAPTER 5. RANDOM VARIABLES Accordingly, E ( Z ) = Z 0 zf Z ( z ) dz = Z 1 0 z (1 - e - z + ze - z ) dz + Z 1 ze - z dz To do the calculation we note that Z 1 0 zdz = [ z 2 / 2] 1 0 = 1 / 2 , Z 1 0 ze - z dz = - Z 1 0 zde - z = - [ ze - z ] 1 0 + Z 1 0 e - z dz = - e - 1 - [ e - z ] 1 0 = 1 - 2 e - 1 . Z 1 0 z 2 e - z dz = - Z 1 0 z 2 de - z = - [ z 2 e - z ] 1 0 + Z 1 0 2 ze - z dz = - e - 1 + 2(1 - 2 e - 1 ) = 2 - 5 e - 1 . Z 1 ze - z dz = 1 - Z 1 0 ze - z dz = 2 e - 1 . Collecting the pieces, we find that E ( Z ) = 1 2 - (1 - 2 e - 1 ) + (2 - 5 e - 1 ) + 2 e - 1 = 3 - 5 e - 1 1 . 16 . Example 5.5.12. Let { X n , n 1 } be i.i.d. with E ( X n ) = μ and var ( X n ) = σ 2 . Use Chebyshev’s inequality to get a bound on α := P ( | X 1 + · · · + X n n - μ | ≥ ² ) . Chebyshev’s inequality (4.8.1) states that α 1 ² 2 var( X 1 + · · · + X n n ) = 1 ² 2 n var( X 1 ) n 2 = σ 2 2 . This calculation shows that the sample mean gets closer and closer to the mean: the variance of the error decreases like 1 /n .
5.5. SOLVED PROBLEMS 83 Example 5.5.13. Let X = D P ( λ ) . You pick X white balls. You color the balls indepen- dently, each red with probability p and blue with probability 1 - p . Let Y be the number of red balls and Z the number of blue balls. Show that Y and Z are independent and that Y = D P ( λp ) and Z = D P ( λ (1 - p )) . We find P ( Y = m, Z = n ) = P ( X = m + n ) m + n m p m (1 - p ) n = λ m + n ( m + n )! m + n m p m (1 - p ) n = λ m + n ( m + n )! × ( m + n )! m ! n ! p m (1 - p ) n = [ ( λp ) m m ! e - λp ] × [ ( λ (1 - p )) n n ! e - λ (1 - p ) ] , which proves the result.

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6.7. SOLVED PROBLEMS 95 6.7 Solved Problems Example 6.7.1. Let ( X, Y ) be a point picked uniformly in the quarter circle { ( x, y ) | x 0 , y 0 , x 2 + y 2 1 } . Find E [ X | Y ] . Given Y = y , X is uniformly distributed in [0 , p 1 - y 2 ]. Hence E [ X | Y ] = 1 2 p 1 - Y 2 . Example 6.7.2. A customer entering a store is served by clerk i with probability p i , i = 1 , 2 , . . . , n . The time taken by clerk i to service a customer is an exponentially distributed random variable with parameter α i . a. Find the pdf of T , the time taken to service a customer. b. Find E [ T ] . c. Find V ar [ T ] . Designate by X the clerk who serves the customer. a. f T ( t ) = n i =1 p i f T | X [ t | i ] = n i =1 p i α i e - α i t b. E [ T ] = E ( E [ T | X ]) = E ( 1 α X ) = n i =1 p i 1 α i . c. We first find E [ T 2 ] = E ( E [ T 2 | X ]) = E ( 1 α 2 i ) = n i =1 p i 2 α 2 i . Hence, var( T ) = E ( T 2 ) - ( E ( T )) 2 = n i =1 p i 2 α 2 i - ( n i =1 p i 1 α i ) 2 .
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