Department of Statistics, Yale University s1

Department of Statistics, Yale University STAT242b Theory of Statistics

Suggested Solutions to Homework 4

Compiled by Marco Pistagnesi

Problem 9.3c For this problem, I will use the delta method, which is faster and easier. 2 X1, , X n iid ~N (m , s ) . Following the book notation (and results), we have;  2 0  1  1  1  2  g( , ) 1.64 J n  I n   ; we then notice          g    ; n  0   1 . 6 4   2 

ˆ 2 0  1  1  1  ˆ 2  ˆ ˆ T ˆ ˆ  2  ˆ 2 2 s e (  )  ( g ) J n ( g )  1 1 . 6 4  ˆ      ( 1 . 6 4 )  n  0  1 . 6 4  n  2   2  2 where ö2  n1 X  X  i  I do not report the numeric calculations.

Problem 9.4 ˆ We take Y= max{ X1 , , X n} and we know that Y = qMLE . Consider: P{Y-q > e} = P{ q - Y > e} = P { Y < q - e} (1) We consider for any fixed c that n P{Y< c} = P{max{ X1 , , Xn} < c} = P{ X 1 < c}创 ⋯ P { X n < c} = ( c q ) (2) Thus we use (2) to evaluate (1): n P {Y

ˆ 1 1 ˆ 1 X 1 , K , X n : N ( 0 , 1 )  I 1 (  )  1 , hence s e (  )    s e (  )   ( X ) . n I 1 (  ) n n Thus the approximate 95% confidence interval is (ö  1.96 * se(ö)) . c) By the weak law of large numbers,  is consistence. d) Because Y 1 , K , Y n : B e r ( p ) , p    (  ) , we can apply the CLT to state :   1   ö ö %   ()1 ()   Y : approxN  ,  , hence: sµe ö  .  n    n s e ( ˆ )   ( X ) 1 In part b) we computed n . (ö) Hence AREö,% . (Plot omitted) (ö) 1 (ö)   e) From before, we have ö   X . Hence, by the WLLN and its property oif   preservation under continuous transformations, we have:

. X  p   ( X )  ˆ  p (  )

But if the data are not normal, (  )  p ( X i  0 )   .

Hence, ˆ  p  .

Problem 5 a) The MLE of p for a binomial distribution is pˆ MLE = X n .From that we get: ö  pö  pö  X n  X n . MLE 1 MLE 2 MLE 1 1 2 2 b) We write the likelihood function: n- x n - x L p, p f x , p f x , p px1 1 p1 1 p x 2 1 p 2 2 ( 1 2) =( 1( 1 1))( 1( 1 1)) =( 1( - 1) )( 2( - 2 ) ) We take the log-likelihood function:

lnL ( p1 , p 2) = x 1 ln p 1 +( n 1 - x 1) ln( 1 - p 1) + x 2 ln p 2 +( n 2 - x 2) ln( 1 - p 2 ) . From this we can compute:

 xi ni  xi lnL p1, p2   , i=1,2 p p 1 p i i i 2 2 We now take the second derivatives. Note: mixed derivatives 抖 p1 p 2 and 抖 p2 p 1 are zero. We have: 2  xi ni  xi 2 lnL p1, p2   2  2 , for i=1,2 pi pi 1 p  i  next step: 骣抖2lnLn p n- n p n 骣 2 ln L n -E琪 =1 1 + 1 1 1 = 1 and also - E 琪 = 2 p2 p 22 p1 p p 2 p 1 p 桫�1 1(1- p1 ) 1( � 1) 桫 2 2( 2 ) Thus the Fisher information matrix is:

轾n1 p 1(1- p 1 ) 0 I( p1, p 2 ) = 犏臌 0n2 p 2( 1- p 2 ) c) We ultimately want to use the formula: T ˆ se(yˆ ) = (蜒g) Jn ( g)

To find the standard error we notice that y =g( p1, p 2) = p 1 - p 2 . We take the gradient:

轾抖g p 轾�( p p) p 轾1 T �g=1 =1 2 1 垩 ( = g) - [ 1 1] 犏抖犏犏臌g p2 臌�( p1 p 2) p 2 臌-1

We also know that Jn is the inverse of the fisher information matrix (7) 2-1 2 轾n1 p 1(1- p 1) 0 轾 p 1( 1 - p 1) n 1 0 Jn =犏2 = 犏 2 臌0n2 p 2( 1- p 2) 臌 0 p 2( 1 - p 2) n 2 ˆ Taking Jn instead of Jn means we insert the estimators of the parameters instead of the real values. Thus, 轾pˆ(1- p ˆ ) n 0 轾1 se(yˆ ) =[ 1 - 1] 1 1 1 = 犏 ˆ ˆ 犏臌 0p2( 1- p 2) n 2 臌-1 pˆ(1-p ˆ) p ˆ( 1 - p ˆ ) se(yˆ ) =1 1 + 2 2 n1 n 2 d) from before we have: ö  X n  X n  160  148 200 = 1.54 1 1 2 2  

pˆ1(1-p ˆ 1) p ˆ 2( 1 - pˆ 2 ) 200 160骣 160 148 骣 148 se(yˆ ) = + =琪 1 - + 琪 1 - = 0.04197 n1 n 2 200 200桫 200 200 桫 200 It is thus simple to find a 90-percent confidence interval: 1.54  1.25 0.04197  1.487,1.593 .    