elemprob-fall2010-page45

# elemprob-fall2010-page45 - ∑ e tx p x and in the...

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One more example. Suppose X 1 ,X 2 ,... is an i.i.d. sequence and each X i has mean 0 and variance 25. How large must n be so that the probability that the absolute value of the average is less than .1 is at least .99? Answer. We want to choose n such that P ±² ² ² S n n ² ² ² < . 1 ³ . 99 . Then . 99 = P (( - . 1) n < S n < ( . 1) n ) = P ± ( - . 1) n 5 < S n - n E X 1 n Var X 1 < ( . 1) n 5 ³ P ± - . 02 n < Z < . 02 n ³ . This means that P ( Z . 02 n ) = . 995, so . 02 n = Φ - 1 ( . 995) = 2 . 576 . Solving for n shows that n must be at least 16,590. 18 Moment generating functions We deﬁne the moment generating function m X by m X ( t ) = E e tX , provided this is ﬁnite. In the discrete case this is equal to
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Unformatted text preview: ∑ e tx p ( x ) and in the continuous case R e tx f ( x ) dx . Let us compute the moment generating function for some of the distribu-tions we have been working with. 1. Binomial: using independence, E e t P X i = E Y e tX i = Y E e tX i = ( pe t + (1-p )) n , where the X i are independent Bernoulli’s. 2. Poisson: E e tX = X e tk e-λ λ k k ! = e-λ X ( λe t ) k k ! = e-λ e λe t = e λ ( e t-1) . 45...
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