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ST 522-001: Statistical Theory II

Solution to Lab Exercises - 4

Prepared by Chen-Yen Lin

Feb. 09, 2011

6.24; (i) When   λ  = 0, random variable   X   degenerates to 0. It follows immediately thatthis family is not complete. (ii) When   λ   = 1, to show this family is not complete,we can let  g(0) =  −e  and  g(X ) = 1,   ∀x >   0. Then it can be shown that  Eg(X ) =

g(0)e−1 + e−1∞

x=1h(x)x!

  = −1 + 1 = 0.

6.26: (a) If P 0 consists two distributions N (θ1, 1) and N (θ0, 1). By   Theorem 6.6.5 , a minimalsufficient statistic for  P   is

T (x) = pθ1 pθ0

=  e−

nθ212

e−nθ202

exp

  ni=1

xi(θ1 − θ0)

which is equivalent to  X̄ (b) Similarly, consider a case in which P 0 consists only two distributions gamma(α, β 1)and gamma(α, β 0). By   Theorem 6.6.25 , a minimal sufficient statistic for  P 0   is

T (x) = β α0β α1

exp

−

ni=1

xi

 1

β 0−

  1

β 1

which is equivalent to n

i=1 X i.(c) Consider a case in which  P 0  consists only two distributions U(0,θ1) and U(0,θ0).By   Theorem 6.6.5 , the minimal sufficient statistic for  P 0   is

T (X ) =θ−n1   I (X (n),∞)(θ1)

θ−n0   I (X (n),∞)(θ0)

which is X (n)

6.31(a) (i) When  σ2 is known, it can be shown that  X̄   is complete sufficient statistic for  µ,whereas the distribution of (n −  1)S 2/σ2 does not depend on   µ  and therefore   σ2 isancillary for  µ. By   Basu Theorem ,  X̄  ⊥ S 2.(ii) When σ2 is unknown but fixed, first observe that  S 2 = (n− 1)−1

ni=1(X i−

 X̄ )2 =(n −  1)−1

ni=1(Z i  −

 Z̄ )2, where   Z i   =   X i  −  µ   is   N (0, σ2). Therefore,   S 2 is still an

ancillary statistic. By Basu Theorem,  X̄  ⊥ S 2 for unknown µ  and fixed σ2. Since σ2 isarbitrary, this result can be extended to general unknown  µ  and σ2. (More discussioncan be found in Example 6.2.27.)

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6.31(c) (i) Given that X/Y   and Y  are independent, we have

EX k =   E Y  X Y 

k

EX k =   EY kE 

X 

Y 

kEX k

EY k  =   E 

X 

Y 

k

(ii) Since   α   is known, it can be argued that   T   = n

i=1 X i   is complete sufficient forβ . Moreover, since   X i  comes from a scale family with standard variable   Z i   whose

distribution does not depend on  β . It follows that  X (i)

n

i=1 X iis ancillary for β . Therefore

E (X (i)|T ) = E 

X (i)T 

  T |T 

= E 

X (i)T   |T 

E (T |T ) = E 

X (i)T 

T   =  E (X (i))

ET   T 

The last equality holds by using the result from (i).

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