ma316f03hw11 - N (0 , 1) by example 2 from section 5.3 so...

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MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #11 SOLUTIONS: SECTIONS 5.2 AND 5.3 FALL 2003 5.9: If W n is a random variable with mean μ and variance b/n p , to show that W n converges in probability to μ we need to show that for arbitrary ² > 0 we have lim n →∞ P ( | W n - μ | ≥ ² ) = 0 . By Chebyshev’s inequality we know that in general for ² = we have P ( | W n - μ | ≥ ² ) = P ( | W n - μ | ) ) 1 /k 2 so in particular with k = ²/σ = ²/ p b/n p we have lim n →∞ P ( | W n - μ | ≥ ² ) = lim n →∞ P ( | W n - μ | ) ) lim n →∞ 1 /k 2 = lim n →∞ b n p ² 2 = 0 since p > 0 ,² > 0, and b is a constant that is independent of n . 5.11: If X n is a random variable with gamma distribution with parameters α = n and β and Y n = X n /n , then the limiting distribution of Y n is degenerate normal at β since the moment generating function of Y n is M ( t ) = lim n →∞ E ( e tX n /n ) = lim n →∞ ± 1 - βt n - n = e βt . 5.13: If X is χ 2 (50) then Y = ( X - n ) / ( 2 n ) = ( X - 50) / 100 = ( X - 50) / 10 is approximately
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Unformatted text preview: N (0 , 1) by example 2 from section 5.3 so the desired probability is P (40 < X < 60) = P 40-50 10 < X-50 10 < 60-50 10 = (1)-(-1) = 2(1)-1 = 0 . 682 . 5.15: If Z n has a Poisson distribution with parameter = n then the limiting distribution of Y n = ( Z n-n ) / n is normal with mean zero and variance 1 since lim n E h e t ( Z n-n ) / n i = lim n n e-t n exp h n ( e t/ n-1) io = lim n exp -t n + n t/ n + t 2 2 n + t 3 6 n 3 / 2- = lim n exp t 2 2 + t 3 6 n 1 / 2- = exp( t 2 / 2) which is the moment generating function of N (0 , 1) . 1...
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This note was uploaded on 09/10/2009 for the course STATS 517 taught by Professor Song during the Fall '07 term at Purdue University-West Lafayette.

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