# Hw02ans - STAT 410 Fall 2008 Homework#2(due Friday...

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Unformatted text preview: STAT 410 Fall 2008 Homework #2 (due Friday, September 12, by 3:00 p.m.) 1. ( ~ 1.9.19 ) Let X be a nonnegative continuous random variable with p.d.f. f ( x ) and c.d.f. F ( x ). Show that E ( X ) = ( ) ( ) & ∞-1 x x d F . E ( X ) = ( ) & ∞ x x f x d = ( ) & & ∞ ± ± ² ³ ´ ´ µ ¶ x x f y d x d = ( ) & & ∞ ± ± ² ³ ´ ´ µ ¶ x y x f d x d → ( ) & & ∞ ± ± ² ³ ´ ´ µ ¶ x y x f d x d = ( ) & & ∞ ± ± ± ² ³ ´ ´ ´ µ ¶ ∞ y x x f d y d · E ( X ) = ( ) & & ∞ ± ± ± ² ³ ´ ´ ´ µ ¶ ∞ y x x f d y d = ( ) & ∞ > X P y y d = ( ) ( ) & ∞-1 y y d F . OR ( ) ( ) & ∞-1 x x d F = by parts: u = 1 – F ( x ) dv = dx du = – f ( x ) dx v = x = ( ) ( ) ( ) ( ) & ∞--∞-1 x x f x x x d F = ( ) & ∞ x x f x d = E ( X ). 2. Suppose that X follows a uniform distribution on the interval [ – π / 2 , π / 2 ] . Find the c.d.f. and the p.d.f. of Y = tan X. f X ( x ) = & & ± & & ² ³ < <-o.w. 2 2 1 π x F X ( x ) = & & & ± & & & ² ³ ≥ < ≤-+-< 2 1 2 2 2 1 2 x x x x F Y ( y ) = P ( Y ≤ y ) = P ( tan X ≤ y ) = P ( X ≤ arctan ( y ) ) = ( ) 2 1 1 arctan + y , – ∞ < y < ∞ . f Y ( y ) = ( ) 2 1 1 y + , – ∞ < y < ∞ . ( Standard ) Cauchy distribution. OR g ( x ) = tan x g – 1 ( y ) = arctan ( y ) d x / d y = 2 1 1 y + f Y ( y ) = f X ( g – 1 ( y ) ) y x d d = ´ ´ µ ¶ · · ¸ ¹ + ´ µ ¶ · ¸ ¹ 2 1 1 1 y = ( ) 2 1 1 y + , – ∞ < y < ∞ . F Y ( y ) = ( ) º ∞ +-y du u 2 1 1 = ( ) 2 1 1 arctan + y , – ∞ < y < ∞ . 3. An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder’s loss, Y, follows a distribution with density function: f ( y ) = & & ± & & ² ³ > otherwise 1 if 2 3 y y What is the expected value and the variance of the benefit paid under the insurance policy? The benefit paid under the insurance policy = & ± & ² ³ ≥ ≤ < 10 for 10 10 1 for y y y E ( Benefit Paid ) = ´ ´ ∞ + ⋅ ⋅ 10 3 10 1 3 2 10 2 dy y dy y y = 10 2 1 10 10 2 ∞--y y = 1.9. E ( Benefit Paid 2 ) = ´ ´ ∞ + ⋅ ⋅ 10 3 2 10 1 3 2 2 10 2 dy y dy y y = 10 2 1 10 100 2 ln ∞--y y = 2 ln 10 + 1. Var ( Benefit Paid ) = 2 ln 10 + 1 – 1.9 2 = 2 ln 10 – 2.61 ≈ 1.99517. 4. The time, T, that a manufacturing system is out of operation has cumulative distribution function F ( t ) = & & ± & & ² ³ > ´ µ ¶ · ¸ ¹-otherwise 2 if 2 1...
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## This note was uploaded on 10/22/2009 for the course STAT 410 taught by Professor Alexeistepanov during the Fall '08 term at University of Illinois at Urbana–Champaign.

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Hw02ans - STAT 410 Fall 2008 Homework#2(due Friday...

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