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elemprob-fall2010-page46

elemprob-fall2010-page46 - 3 Exponential etx ex dx = E etX...

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3. Exponential: E e tX = 0 e tx λe - λx dx = λ λ - t if t < λ and if t λ . 4. N (0 , 1): 1 2 π e tx e - x 2 / 2 dx = e t 2 / 2 1 2 π e - ( x - t ) 2 / 2 dx = e t 2 / 2 . 5. N ( μ, σ 2 ): Write X = μ + σZ . Then E e tX = E e e tσZ = e e ( ) 2 / 2 = e + t 2 σ 2 / 2 . Proposition 18.1 If X and Y are independent, then m X + Y ( t ) = m X ( t ) m Y ( t ) . Proof. By independence, m X + Y ( t ) = E e tX e tY = E e tX E e tY = m X ( t ) m Y ( t ) . Proposition 18.2 m X ( t ) = m Y ( t ) < for all t in an interval, then X and Y have the same distribution. We will not prove this, but this is essentially the uniqueness of the Laplace transform. Note E e tX = e tx f X ( x ) dx . If f X ( x ) = 0 for x < 0, this is 0 e tx
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