Dr. Hackney STA Solutions pg 80

Dr. Hackney STA Solutions pg 80 - 5-14 Solutions Manual for...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5-14 Solutions Manual for Statistical Inference 5.43 a. P (|Yn - | < ) = P (n)(Yn - ) < (n) . Therefore, (n) = P (|Z| < ) = 1, n lim P (|Yn - | < ) = lim P n (n)(Yn - ) < where Z n(0, 2 ). Thus Yn in probability. b. By Slutsky's Theorem (a), g () n(Yn - ) g ()X where X n(0, 2 ). Therefore n[g(Yn ) - g()] = g () n(Yn - ) n(0, 2 [g ()]2 ). 5.45 We do part (a), the other parts are similar. Using Mathematica, the exact calculation is In[120]:= f1[x_]=PDF[GammaDistribution[4,25],x] p1=Integrate[f1[x],{x,100,\[Infinity]}]//N 1-CDF[BinomialDistribution[300,p1],149] Out[120]= e^(-x/25) x^3/2343750 Out[121]= 0.43347 Out[122]= 0.0119389. The answer can also be simulated in Mathematica or in R. Here is the R code for simulating the same probability p1<-mean(rgamma(10000,4,scale=25)>100) mean(rbinom(10000, 300, p1)>149) In each case 10,000 random variables were simulated. We obtained p1 = 0.438 and a binomial probability of 0.0108. 5.47 a. -2 log(Uj ) exponential(2) 2 . Thus Y is the sum of independent 2 random variables. 2 2 By Lemma 5.3.2(b), Y 2 . 2 b. log(Uj ) exponential(2) gamma(1, ). Thus Y is the sum of independent gamma random variables. By Example 4.6.8, Y gamma(a, ) c. Let V = j=1 log(Uj ) gamma(a, 1). Similarly W = V Exercise 4.24, V +W beta(a, b). 5.49 a. See Example 2.1.4. 1 b. X = g(U ) = - log 1-U . Then g -1 (x) = 1+e-y . Thus U fX (x) = 1 e-y e-y = -y )2 (1 + e (1 + e-y )2 a b j=1 log(Uj ) gamma(b, 1). By - < y < , which is the density of a logistic(0, 1) random variable. 1 c. Let Y logistic(, ) then fY (y) = fZ ( -(y-) ) where fZ is the density of a logistic(0, 1). Then Y = Z + . To generate a logistic(, ) random variable generate (i) generate U U uniform(0, 1), (ii) Set Y = log 1-U + . 5.51 a. For Ui uniform(0, 1), EUi = 1/2, VarUi = 1/12. Then 12 X= i=1 Ui - 6 = 12U - 6 = 12 U -1/2 1/ 12 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online