Dr. Hackney STA Solutions pg 73

# Dr. Hackney STA Solutions pg 73 - Second Edition 5-7 5.18...

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Second Edition 5-7 5.18 If X t p , then X = Z/ p V/p where Z n(0 , 1), V χ 2 p and Z and V are independent. a. E X = E Z/ p V/p = (E Z )(E1 / p V/p ) = 0, since E Z = 0, as long as the other expectation is ﬁnite. This is so if p > 1. From part b), X 2 F 1 ,p . Thus Var X = E X 2 = p/ ( p - 2), if p > 2 (from Exercise 5.17b). b. X 2 = Z 2 / ( V/p ). Z 2 χ 2 1 , so the ratio is distributed F 1 ,p . c. The pdf of X is f X ( x ) = " Γ( p +1 2 ) Γ( p/ 2) # 1 (1 + x 2 /p ) ( p +1) / 2 . Denote the quantity in square brackets by C p . From an extension of Stirling’s formula (Exercise 1.28) we have lim p →∞ C p = lim p →∞ 2 π ( p - 1 2 ) p - 1 2 + 1 2 e - p - 1 2 2 π ( p - 2 2 ) p - 2 2 + 1 2 e - p - 2 2 1 = e - 1 / 2 π lim p →∞ ( p - 1 2 ) p - 1 2 + 1 2 ( p - 2 2 ) p - 2 2 + 1 2 p = e - 1 / 2 π e 1 / 2 2 , by an application of Lemma 2.3.14. Applying the lemma again shows that for each x lim p →∞ ( 1+ x 2 /p ) ( p +1) / 2 = e
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