ContRVs - Outline The Density Function The Normal Random...

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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations Continuous Random Variables Michael Akritas Michael Akritas Continuous Random Variables
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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations The Density Function The Normal Random Variable Normal Approximation to the Binomial The Exponential Distribution Transformations Michael Akritas Continuous Random Variables
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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations Definition X is continuous if there exists a 0 function f ( x ) such that P ( X A ) = Z A f ( x ) dx . This function is called the probability density function , or pdf. Properties: I R -∞ f ( x ) dx = 1 (= P ( -∞ < X < )) . I P ( a X b ) = P ( a < X < b ) = R b a f ( x ) dx . I P ( X = a ) = R a a f ( x ) dx = 0. Read Example 1a, p. 187. Michael Akritas Continuous Random Variables
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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations Example X = life time of an indicator light, has pdf f ( x ) = 100 x I ( x > 100) 1. Find P ( X < 150) 2. What is the probability that two of 5 such indicator lights will have to be replaced within the first 150 hours? Definition F ( x ) = P ( X x ) = R x -∞ f ( t ) dt is called the cumulative distribution function. I By the Fundamental Theorem of Calculus, d dx F ( x ) = f ( x ) Michael Akritas Continuous Random Variables
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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations Example X f X ( x ). Find the pdf of Y = 2 X . I μ X = E ( X ) = R -∞ xf X ( x ) dx (Definition) I E ( g ( X )) = R -∞ g ( x ) f ( x ) dx (To be shown – see next page) I E ( ag 1 ( X ) + bg 2 ( X )) = aE ( g 1 ( X )) + bE ( g 2 ( X )) Example X f X ( x ) = I (0 x 1). Find E ( e X ). Solution: a) Apply above formula with g ( x ) = e x b) Find the pdf, f Y ( y ), of Y = e X and use the definition of expected value, i.e. E ( Y ) = R -∞ yf Y ( y ) dy . Michael Akritas Continuous Random Variables
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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations Lemma If Y is a 0 r.v., then E ( Y ) = R 0 P ( Y > y ) dy Proposition E ( g ( X )) = R -∞ g ( x ) f ( x ) dx. Proof: Assume first that g ( x ) 0, for all x. Then, E ( g ( X )) = Z 0 P ( g ( X ) > y ) dy = Z 0 Z x : g ( x ) > y f ( x ) dxdy = Z -∞ Z g ( x ) 0 dyf ( x ) dx = Z -∞ g ( x ) f ( x ) dx For general g write g ( x ) = g ( x ) I ( g ( x ) 0) + g ( x ) I ( g ( x ) < 0) = g + ( x ) - g - ( x ), and use E ( g ( X )) = E ( g + ( X )) - E ( g - ( X )) Michael Akritas Continuous Random Variables
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Outline The Density Function The Normal Random Variable The Exponential Distribution Transformations I σ 2 X = Var( X ) = E ( X - μ X ) 2 = E ( X 2 ) - μ 2 X .
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This note was uploaded on 01/23/2011 for the course STAT 418 at Pennsylvania State University, University Park.

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ContRVs - Outline The Density Function The Normal Random...

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