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Unformatted text preview: Vinh Nguyen Stat 120a (Fall 2007): HW3 Solutions October 23, 2007 Text: Probabilty and Statistics , by Degroot and Schervish (3 rd ed) Problems: 4.1.1, 4.1.2, 4.1.3, 4.1.11, 4.1.13, 4.2.1, 4.2.4, 4.2.7, 4.3.1, 4.3.5, and 4.3.10 4.1.1 Suppose X ∼ uniform [ a,b ]. Then E [ X ] = Z b a xf ( x ) dx = Z b a x 1 b a dx = x 2 2( b a ) b a = a + b 2 4.1.2 We would like to know the expected value of an integer chosen between 1 and 100 at random. Note that we have a discrete uniform random variable with P ( X = i ) = 1 100 for i ∈ { 1 , 2 ,..., 100 } . Thus, E [ X ] = 100 X i =1 i P ( X = i ) = 1 100 100 X i =1 i = 1 100 100(100 + 1) 2 = 50 . 5 . 4.1.3 Based on the given information we have the following probabilities: Age i n i P (Age i ) 18 20 20/50 19 22 22/50 20 4 4/50 21 3 3/50 25 1 1/50 E [Age i ] = 18 × 20 50 + 19 × 22 50 + 20 × 4 50 + 21 × 3 50 + 25 × 1 50 = 18 . 92 4.1.11 Suppose X 1 ,...,X n ∼ iid uniform [0 , 1]. Let Y 1 = min { X 1 ,...,X n } and Y n = max { X 1 ,...,X n } . We would like to find E ( Y 1 ) and E ( Y n ). For any continuous random variable Y , we know (by definition) E ( Y ) = R ∞∞ yf Y ( y ) dy and F Y ( y ) = f Y ( y ). To get E ( Y ), we need to get the pdf of Y by first deriving the cdf of Y . To get he cdf of Y 1 and Y n , we will need the cdf of a uniform [0 , 1] (will be used later, so we’ll just derive it first). Let U ∼ uniform [0 , 1]. Then F U ( u ) = P ( U ≤ u ) = Z u∞ f U ( t ) dt = Z u 1 dt = t  u = u for 0 ≤ u ≤ 1. Now, 1 F Y 1 ( y ) = P ( Y 1 ≤ y ) (definition) = 1 P ( Y 1 > y ) (compliments) = 1 P ( min { X 1 ,...,X n } > y ) = 1 P ( X 1 > y,...,X n > y ) = 1 n Y i =1 P ( X i > y ) (independence) = 1 n Y i =1 [1 P ( X i ≤ y )] (compliments) = 1 n Y i =1 [1 F X i ( y )] (definition) = 1 n Y i =1 [1 y ] plug in cdf of uniform [0 , 1] = 1 (1 y ) n for 0 ≤ y ≤...
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This note was uploaded on 01/21/2010 for the course STATS 37850 taught by Professor Phelan during the Spring '09 term at UC Irvine.
 Spring '09
 PHELAN
 Statistics

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