Math 362 Problem Set #9 â€“ Solution 27 April 2012 - Faculty web ...

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$Math 361 Problem Set #5 Due 5 October 2012 - Faculty web pages$

For large values of ν, the Student t distribution is approximately the same as thestandard normal distribution.Furthermore, since the normalizing constants have to match up, we have shown indirectlythatΓ ( )ν+121√ (πν Γν= √ .2)2πlimν→∞5. We showed in a previous problem set that the mean of n independent exponential(1)random variables has a Gamma distribution with parameters r = n and λ = n.(a) Show that the moment-generating function for a random variable Y with a Gammadistribution with parameters r and λ is given by(m Y (t) = 1 −λ) t −r.<strong>Solution</strong>: We haveE(e tY ) = λrΓ(r)= λrΓ(r)∫ ∞0∫ ∞0e ty y r−1 e −λy dyy r−1 e −(λ−t)y dyAs long as λ − t > 0, this integral converges. In fact, since y r−1 e −(λ−t)y is the“core” of the probability density function for a Gamma distribution, we knowthatIt follows thatas required.∫ ∞0y r−1 e −(λ−t)y dy =Γ(r)(λ − t) r .m Y (t) = λrΓ(r) · Γ(r)(λ − t)( ) rr λ=λ − t(= 1 − t ) −rλ
(b) Now suppose Y follows a Gamma distribution with parameters r = n and λ = n.Let U = √ n(Y − 1), and find the limit of m U (t) as n → ∞.What does this tell you about the mean of a large number of exponential(1)random variables?(m Y −1 (t) = e −t m Y (t) = e −t 1 − t ) −nnand so(m U (t) = e −t√ n1 − √ t ) −n= n(e t √ n(1 − t √ n)) −n.We’ll rewrite this limit in terms of a new variable s =s → 0, so our limit becomeslims→0 (es (1 − s))) − t2s 2 =[]lims→0 (es (1 − s)) 1 −t 2s 2 .t √ n. As n → ∞, we haveThe limit inside the brackets has the form 1 ∞ , and we can attack it using l’Hôpital’srule. We getThus lim(e s (1 − s)) 1s 2 = e − 1 2 ands→0ln lim(e s (1 − s)) 1 ln(e s (1 − s))s 2 = lims→0 s→0 s 2lim m U(t) = limn→∞s→0[e s (1 − s) − e s= lims→0 2se s (1 − s)= lims→0= − 1 2 .−12(1 − s)lims→0 (es (1 − s)) 1s 2 ] −t 2= e t2 2which is the moment generating function for the standard normal random variable.This says that for large n, √ n(Y − 1) has approximately the standard normaldistribution. Working backward, we can conclude that the distribution of Y − 1is approximately normal with mean zero and variance 1 , and so Y , the samplen
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Math 362 Problem Set #9 â€“ Solution 27 April 2012 - Faculty web ...

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