Statistical Inference Test Set 1 1. Let λ Find unbiased ... - Nptel

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Statistical Inference Test Set 1

1. Let X ~ ( ).Ρ λ Find unbiased estimators of ( )i λ³, ( )ii e⁻^λcos ,λ (iii)sinλ. (iv) Show that there does not exist unbiased estimators of 1 / , and exp{ 1 / }.λ − λ

2. Let X X₁, ₂,...,X_nbe a random sample from a N( ,µ σ²)population. Find unbiased and consistent estimators of the signal to noise ration µ

σ and quantile µ+^bσ , where b is any given real.

3. Let X X₁, ₂,...,X_nbe a random sample from a U(−θ θ, 2 )population. Find an unbiased and consistent estimator of θ.

4. Let X X₁, ₂be a random sample from an exponential population with mean 1 /λ. Let

1 2

1 , 2 1 2

2 X X

T = + T = X X . Show that T₁ is unbiased and T₂is biased. Further, prove that

2 1

( ) ( )

MSE T ≤Var T .

5. Let T₁and T₂ be unbiased estimators of θ with respective variances σ₁² and σ₂² and

1 2 12

cov( ,T T )=σ (assumed to be known). Consider T =αT₁+ −(1 α) , 0T₂ ≤ ≤α 1. Show that Tis unbiased and find value of αfor which Var T( )is minimized.

6. Let X X₁, ₂,...,X_nbe a random sample from anExp( , )µ σ population. Find the method of moment estimators (MMEs) of µ and σ .

7. Let X X₁, ₂,...,X_n be a random sample from a Pareto population with density

( ) 1 , , 0, 2.

fX x x

x

β β

βα₊ α α β

= > > > Find the method of moments estimators of α β, . 8. Let X X₁, ₂,...,X_nbe a random sample from a U(−θ θ, )population. Find the MME of θ. 9. Let X X₁, ₂,...,X_nbe a random sample from a lognormal population with density

2 2

1 1

( ) exp (log ) , 0.

2 2

X e

f x x x

x µ

σ π σ

 

= − −  >

  Find the MMEs of µ and σ².

10. Let X X₁, ₂,...,X_nbe a random sample from a double exponential( , )µ σ population. Find the MMEs of µ and σ.

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Hints and Solutions 1. (i) E X X{ ( −1)(X −2)}=λ³

(ii) For this we solve estimating equation. Let T X( )be unbiased for e⁻^λcosλ. Then

ET X( )=e⁻^λcosλfor all >0λ .

0

( ) cos for all >0

!

x

T x e e

x

λλ λ λ λ

∞ −

−

=

⇒

∑

=

2 4

0

( ) 1 for all >0

! 2! 4!

x

T x x

λ λ λ λ

∞

=

⇒

∑

= − + −^

As the two power series are identical on an open interval, equating coefficients of powers of λon both sides gives

( ) 0, if 2 1, 1, if 4 ,

1, if 4 2, 0,1, 2,

T x x m

x m

x m m

= = +

= =

= − = + = 

(iii) For this we have to solve estimating equation. However, we use Euler’s identity to solve it.

Let U X( )be unbiased for sinλ. Then

( ) 1 ( ) for all 0

! 2

0

1 (1 ) (1 )

( ) for all 0

2

1 (1 ) (1 )

for all 0.

2 0 ! 0 !

x i i

U x e e e

x i

x

i i

e e

i

k k k k

i i

i k k k k

λ λ λ λ λ

λ λ λ

∞ −

⇒ ∑ = − >

=

+ −

= − >

 ∞ + ∞ − 

 

= ∑ − ∑ >

 = = 

 

Applying De-Moivre’s Theorem on the two terms inside the parentheses, we get

0

( ) 1 ( 2) cos sin ( 2) cos sin

! 2 ! 4 4 4 4

0

1 ( 2) cos sin ( 2) cos sin

2 ! 4 4 4 4

( 2) sin

! 4

k k k

k k

k

k k

k

x

U x i i

x i k

x

k k k k

i i

i k

k k

λ λ π π π π

λ π π π π

λ π

∞

=

∞

=

∞

=

∞ =   +  −  − + −  

∑=         

       

=   + −  − + − 

= 

∑

^^{for all}^λ ^>⁰



Equating the coefficients of powers of λon both sides gives ( ) ( 2) sin , 0,1, 2,

4

x x

U x _π _ x

=   =

  

In Parts (iv) and (v) , we can show in a similar way that estimating equations do not have any solutions.

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2. Let ² ²

1 1

, and ( )

1

n n

i i

X X S X X

n = n =

= = −

∑

−

∑

^.

Then

2 2

2 2 1

( 1)

~ , , and W= n S ~ _n

X N n

µ σ χ

σ ⁻

  −

 

  . It can be seen that

1/ 2

2 2

( )

1 2

n E W = n

− and ^{1/ 2}

2 ( ) 2

2 1 2 n

E W n

−

= − . Using these, we get unbiased estimators

of σ and ¹

σ as ₁ ₂

1 1

1 2 and 2 2 1

2 1 2

2 2

n n

T n S T

n S

n n

− −

= − =

− − respectively. As and 2

X S are independently distributed, U₁ =X T₂ is unbiased for µ.

σ Further,

2 1

U =X +bT is unbiased for µ+bσ. As X and S² are consistent for µ and σ² respectively, U₁ and U₂are also consistent for µ

σ and µ+bσ respectively.

3. As ₁ 3 2

2 , 3

T X

µ′ = θ = is unbiased for θ. T is also consistent for θ.

4. As E X( _i) ¹,T₁

=λ is unbiased. Also X₁and X₂are independent. So

( ) ( )

² ²

2 1 2 1

( ) ( ) 1 .

2 4

E T E X X E X π π

λ λ

 

= = =  = ¹ ²

( ) 1 Var T 2

= λ .

( )

2

2 1 2 1 2 1 2 2

2

1 2 1

( ) ( )

2 1 4

MS ET E X X E X X E X X

λ λ λ

π λ

 

=  −  = − +

 

 

=  − 

 

5. The minimizing choice of α is obtained as

2

2 12

2 2

1 2 2 12

σ σ

σ σ σ

−

+ − .

6. ( ) ¹ exp x , , 0.

f x µ x µ σ

σ σ

 − 

= −  > > µ1′= +µ σ µ, 2′ =

(

µ σ+

)

²+σ².

So µ µ= ₁′− µ₂′−µ σ′², = µ₂′−µ′² . The method of moments estimators for and

µ σ are therefore given by

2 2

1 1

ˆ ( ) , an ˆ d ( )

n n

MM i MM i

i i

X X X X X

n n

µ σ

= =

= −

∑

− =

∑

− ^.

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7.

2

1 , 2

1 2

βα βα

µ µ

β β

′= ′ =

− − . So ¹ ² ² ₂

2

2 1

, 1

µ µ µ

α β

µ µ µ µ

′ ′ ′

= = +

′− ′

The method of moments estimators for α and βare therefore given by

2 2

1 1

2 2 2

1 1 1

ˆ , ˆ 1

( )

n n

i i

MM n n MM n

i i i

X X X

X X

X X X

α ⁼ β ⁼

= = =

= = +

− + −

∑ ∑

∑ ∑ ∑

.

8. Since µ′ =₁ 0, we consider

2

2 3

µ′ =θ . So ²

1

ˆ 3 ⁿ

MM i

i

n X θ

=

∑

9. µ₁′=e^{µ σ}⁺ ²^{/ 2},µ₂′ =e²^{µ σ}⁺² ² . So

2

1 2 2

2 2 1

log µ , log µ

µ σ

µ µ

 ′   ′ 

=  ′  =  ′  and the method of moments estimators for µ and σ²are therefore given by

2 2

2 1

2 2

1

ˆ log , ˆ log

1

n i i

MM n MM

i i

X n X

X X n

µ σ ⁼

=

   

   

=   =  

   

   

 

∑

.

10. 1

( ) exp , , , 0.

2

f x x µ x µ σ

σ σ

 − 

= −  ∈ ∈ >

    µ1′=µ µ, 2′ =µ²+2σ².

So ₁ 1 ₂ ²

, ( )

µ µ σ= ′ = 2 µ′ −µ′ . The method of moments estimators for µ and σ are therefore given by

2 1

ˆ , and ˆ 1 ( )

2

n

MM MM i

i

X X X

µ σ n

=

= =

∑

− ^.