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prob - Exponential Distribution Definition X is said to...

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Unformatted text preview: Exponential Distribution Definition X is said to have an exponential distribution with parameter λ ( λ > 0) if the pdf of X is f ( x ; λ ) = ( λ e- λ x x ≥ otherwise Remark: 1. Usually we use X ∼ EXP( λ ) to denote that the random variable X has an exponential distribution with parameter λ . 2. In some sources, the pdf of exponential distribution is given by f ( x ; θ ) = ( 1 θ e- x θ x ≥ otherwise The difference is that λ → 1 θ . Liang Zhang (UofU) Applied Statistics I June 30, 2008 1 / 20 Exponential Distribution Liang Zhang (UofU) Applied Statistics I June 30, 2008 2 / 20 Exponential Distribution Proposition If X ∼ EXP ( λ ) , then E ( X ) = 1 λ and V ( X ) = 1 λ 2 And the cdf for X is F ( x ; λ ) = ( 1- e- λ x x ≥ x < Liang Zhang (UofU) Applied Statistics I June 30, 2008 3 / 20 Exponential Distribution Proof: E ( X ) = Z ∞ x λ e- λ x dx = 1 λ Z ∞ ( λ x ) e- λ x d ( λ x ) = 1 λ Z ∞ ye- y dy y = λ x = 1 λ [- ye- y | ∞ + Z ∞ e- y dy ] integration by parts: u = y , v =- e- y = 1 λ [0 + (- e- y | ∞ )] = 1 λ Liang Zhang (UofU) Applied Statistics I June 30, 2008 4 / 20 Exponential Distribution Proof (continued): E ( X 2 ) = Z ∞ x 2 λ e- λ x dx = 1 λ 2 Z ∞ ( λ x ) 2 e- λ x d ( λ x ) = 1 λ 2 Z ∞ y 2 e- y dy y = λ x = 1 λ 2 [- y 2 e- y | ∞ + Z ∞ 2 ye- y dy ] integration by parts = 1 λ 2 [0 + 2(- ye- y | ∞ + Z ∞ e- y dy )] integration by parts = 1 λ 2 2[0 + (- ye- y | ∞ )] = 2 λ 2 Liang Zhang (UofU) Applied Statistics I June 30, 2008 5 / 20 Exponential Distribution Proof (continued): V ( X ) = E ( X 2 )- [ E ( X )] 2 = 2 λ 2- ( 1 λ ) 2 = 1 λ 2 F ( x ) = Z x λ e- λ y dy = Z x e- λ y d ( λ y ) = Z x e- z dz z = λ y =- e- z | x = 1- e- x Liang Zhang (UofU) Applied Statistics I June 30, 2008 6 / 20 Exponential Distribution Example (Problem 108) The article “Determination of the MTF of Positive Photoresists Using the...
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