Statistical Inference Test Set 3
1. Let X X1, 2,...,Xn be a random sample from a population with density function
2
( ) exp , 0, 0.
2
x x
f x x θ
θ θ
= − > >
Find FRC lower bound for the variance of an unbiased estimator of θ. Hence derive a UMVUE for θ.
2. Let X X1, 2,...,Xnbe a random sample from a discrete population with mass function
1 1
( 1) , ( 0) , ( 1) , 0 1.
2 2 2
P X = − = −θ P X = = P X = =θ < <θ
Find FRC lower bound for the variance of an unbiased estimator of θ. Show that the variance of the unbiased estimator 1
X+2 is more than or equal to this bound.
3. Let X X1, 2,...,Xn be a random sample from a population with density function
(1 )
( ) (1 ) , 0, 0.
f x =θ +x − +θ x> θ > Find FRC lower bound for the variance of an unbiased estimator of 1/θ. Hence derive a UMVUE for 1/θ.
4. Let X X1, 2,...,Xn be a random sample from a Pareto population with density
( ) 1 , , 0, 2.
fX x x
x
β β
βα+ α α β
= > > > Find a sufficient statistics when (i) α is known, (ii) when βis known and (iii) when both α β, are unknown.
5. Let X X1, 2,...,Xnbe a random sample from a Gamma ( , )p λ population. Find a sufficient statistics when (i) p is known, (ii) when λ is known and (iii) when both p,λ are unknown.
6. Let X X1, 2,...,Xnbe a random sample from a Beta ( , )λ µ population. Find a sufficient statistics when (i) µ is known, (ii) when λ is known and (iii) when both λ µ, are unknown.
7. Let X X1, 2,...,Xn be a random sample from a continuous population with density function
( ) 1 , 0, 0.
(1 )
f x x
x θ
θ + θ
= > >
+ Find a minimal sufficient statistic.
8. Let X X1, 2,...,Xnbe a random sample from a double exponential population with the density ( ) 1 | |, , .
2
f x = e− −xθ x∈ θ∈ Find a minimal sufficient statistic.
9. Let X X1, 2,...,Xnbe a random sample from a discrete uniform population with pmf ( ) 1, 1, , ,
p x x θ
=θ = where θis a positive integer. Find a minimal sufficient statistic.
10. Let X X1, 2,...,Xnbe a random sample from a geometric population with pmf
1 ,
( ) (1 )x , 1, 2
f x = −p − p x= , 0< <p 1. Find a minimal sufficient statistic.
11. Let X have a N(0,σ2) distribution. Show that X is not complete, but X2is complete.
12. Let X X1, 2,...,Xnbe a random sample from an exponential population with the density
( ) x, 0, 0.
f x =λe−λ x> λ> Show that
1 n
i i
Y X
=
=
∑
is complete.13. Let X X1, 2,...,Xnbe a random sample from an exponential population with the density
( ) x, , .
f x =eµ− x>µ µ∈ Show that Y =X(1)is complete.
14. Let X X1, 2,...,Xnbe a random sample from a N( ,θ θ2) population. Show that ( ,X S2) is minimal sufficient but not complete.
Hints and Solutions 1.
2
log ( | ) log log 2 f x θ x θ x
= − − θ .
2 2 2 4 2 2
2 4 3 2 4 3 2 2
lo g 1 ( ) ( ) 1 8 2 1 1
2 4 4
f X E X E X
E E θ θ
θ θ θ θ θ θ θ θ θ θ
∂ = − = − + = − + =
∂
.
So FRC lower bound for the variance of an unbiased estimator ofθis
2
n θ .
Further 1 2
2 i
T X
= n
∑
is unbiased forθand2
( ) Var T
n
=θ . Hence Tis UMVUE for θ.
2. 1
( ) 2
E X = −θ . So 1
T = X +2is unbiased for θ.
1 4 4 2
( ) 4
Var T
n
θ θ
+ −
= .
1
1
( 1) , 1
log ( | ) 0, 0
, 1
x
f x x
x θ
θ θ
θ
−
−
− = −
∂ = =
∂ =
So we get
log 2 1
2 (1 ) E f
θ θ θ
∂ =
∂ −
The FRC lower bound for the variance of an unbiased estimator ofθis 2 (1 ) n θ −θ
. It can be seen that
2 (1 ) (1 2 )2
( ) 0.
Var T 4
n n
θ −θ − θ
− = ≥
3. We have 1
log ( | )f x θ log(1 x)
θ θ
∂ = − +
∂ . Further, E{log(1+X)}=θ−1, and
2 2
{log(1 )} 2
E +X = θ− . So
2 2
log f 1
E θ θ
∂ =
∂
, and FRC lower bound for the variance of an unbiased estimator of θ−1is
2
n
θ . Now 1
log(1 i)
T X
=n
∑
+ is unbiased for θ−1and2
( ) Var T
n
=θ .
4. Using Factorization Theorem, we get sufficient statistics in each case as below (i)
1 n
i i
X
∏
= ,(ii) X(1) (iii) (1) 1,
n i i
X X
=
∏
5. Using Factorization Theorem, we get sufficient statistics in each case as below (i)
1 n
i i
X
∑
= ,(ii) 1 ni i
X
∏
= (iii)1 1
,
n n
i i
i i
X X
= =
∑ ∏
6. Using Factorization Theorem, we get sufficient statistics in each case as below (i)
1 n
i i
X
∏
= ,(ii) 1(1 )
n
i i
X
=
∏
− (iii)1 1
, (1 )
n n
i i
i i
X X
= =
−
∏ ∏
7. Using Lehmann-Scheffe Theorem, a minimal sufficient statistic is
1
(1 )
n
i i
X
=
∏
+ .8. Using Lehmann-Scheffe Theorem, a minimal sufficient statistic is the order statistics
(1) ( )
(X ,, X n ).
9. Using Lehmann-Scheffe Theorem, a minimal sufficient statistic is the largest order statisticsX( )n .
10. Using Lehmann-Scheffe Theorem, a minimal sufficient statistic is
1 n
i i
X
∑
= .11. Since E X( )=0 for all σ >0 , but but (P X =0)=1 for all σ >0 . Hence X is not complete. Let W =X2. The pdf of Wis 1 / 2 2
( ) , 0
2
w
fW w e w
w
σ
σ π
= − > .
Now E g Wσ ( )=0 for allσ >0 1/ 2 / 2 2
0
( ) w 0 for all 0
g w w e σ dw σ
∞
− −
⇒
∫
= > . Uniquenessof the Laplace transform implies g w( )=0 . .a e Hence X2is complete.
12. Note that
1 n
i i
Y X
=
=
∑
has a Gamma ( , )n λ distribution. Now proceeding as in Problem 11, it can be proved that Yis complete.13. Note that the density of Y = X(1)is given by fY( )y =n en(µ−x),x>µ µ, ∈. ( ) 0 for all
Eg Y = µ∈
( )
( ) n y 0 for all n g y e µ dy
µ
∞ µ
⇒
∫
− = ∈( ) ny 0 for all g y e dy
µ
∞ µ
⇒
∫
− = ∈Using Lebesgue integration theory we conclude that g y( )=0 a.e. Hence Yis complete.
14. Minimal sufficiency can be proved using Lehmann-Scheffe theorem. To see that ( ,X S2) is not complete, note that 2 2 0 for all 0.
1
E n X S
n θ
− = >
+
However, 2 2 0
1
P n X S
n
= =
+
.
