MA 324 Mid-semester Examination

Statistical Inference and Multivariate Analysis 14:00–16:00 IST

IIT GUWAHATI March 01, 2022

Model Answers of Mid-semester Examination

i.i.d.
1. (2 points) Let X1 , X2 ∼ P oi(λ), where λ > 0 is unknown parameter. Is the family of
distributions induced by the statistic T = (X1 , X2 ) complete?

Solution: Note that E (X1 − X2 ) = 0 for all λ > 0. Now,

P (X1 ̸= X2 ) ≥ P (X1 = 0, X2 = 1) = λe−2λ > 0 =⇒ P (X1 − X2 = 0) = P (X1 = X2 ) < 1.

Thus, the family of distributions induced by the statistics (X1 , X2 ) is not complete. [2 points ]

2. (5 points) Let X1 , X2 , . . . , X9 be a random sample of size 9 form population having U (θ1 , θ2 )

distribution, where both θ1 and θ2 are unknown and −∞ < θ1 < θ2 < ∞. Derive the
estimators of θ1 and θ2 using method of moment.

θ2 +θ θ +θ2
Solution: Here E(X1 ) = θ1 +θ
2
2
and E (X12 ) = 1 13 2 2 . [1 point]
Let M1 = 91 9i=1 Xi , M2 = 19 9i=1 Xi2 and S 2 = 19 9i=1 (Xi − M1 )2 . The method of moment
P P P
estimators can be found by solving the following equations:

θ1 + θ2 =2m1 and θ12 + θ1 θ2 + θ22 =3m2 . [1 point]

√ √
√ √

The solutions for (θ1 , θ2 ) are m1 − 3s, m1 + 3s and m1 + 3s, m1 − 3s , where S is
√
the positive√square root of S 2 . As θ1 < θ2 , the estimator for θ1 and θ2 are θb1 = M1 − 3S and
θb2 = M1 + 3S, respectively. [3 points ]

3. Let X1 , X2 , . . . , Xn be a random sample of size n (≥ 2) from a population having probability

density function
( h 2i
2
x exp − xθ if x > 0
f (x, θ) = θ
0 otherwise,
1
where θ > 0 is a unknown parameter. Consider the problem of estimation of τ (θ) = √ .
θ
(a) (5 points) Derive minimum variance unbiased estimator of τ (θ).
Solution: Note that f (·, θ) belongs to a full rank
P exponential family. Thus, using prop-
erty of exponential family of distributions, T = ni=1 Xi2 is complete and sufficient statis-
tic for θ. [2 points ]
Notice that Xi2 follows exponential distribution with mean θ. Therefore,
 
1
T ∼ Gamma n, . [1 point]
θ

Thus, for k > −n,

Z ∞  
k 1 k+n−1 − θt Γ(n + k) k Γ(n)
Tk = θk .


E T = n t e dt = θ =⇒ E
θ Γ(n) 0 Γ(n) Γ(n + k)

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2021-22 Even Mid-semester Examination MA 324

Therefore, using Lehmann-Scheffee theorem and taking k = − 12 , we have

n
!− 12
Γ(n) X
τb = Xi2
Γ(n − 12 ) i=1

1
is the UMVUE of τ (θ) = θ− 2 . [2 points ]

(b) (3 points) Show that the estimator that you obtained in (a) is consistent. You may use
√ 1
Stirling’s approximation for Γ(n): Γ(n) ∼ 2π (n − 1)n− 2 e−n+1 .
Solution: Here E (X12 ) = θ. Therefore, using WLLN,
n
1X 2
X −→ θ
n i=1 i

in probability. [1 point]
Now,

n
!− 12
Γ(n) X
τbn = Xi2
Γ(n − 21 ) i=1
n
!− 12
Γ(n) 1X 2
=√ X
nΓ(n − 21 ) n i=1 i
  12 n
!− 12
1 1 1 1X 2
∼ e− 2 1− X −→ τ (θ)
n i=1 i
n−1
n

1
1− 2(n−1)

in probability. Therefore θb is a consistent estimator of τ (θ). [2 points ]

(c) (3 points) Compute Cramer-Rao lower bound of an unbiased estimator of τ (θ).

3
Solution: Here τ ′ (θ) = − 12 θ− 2 . The Fisher information present in X1 is
 2 
d
IX1 (θ) = −E ln f (X, θ)
dθ2
 2  
d 2X1 − X12
= −E ln e θ
dθ2 θ
 2 
X12

d
= −E ln 2 − ln θ + ln X1 −
dθ2 θ
2X 2
 
1
= −E 2 − 31
θ θ
1
= 2 . [1 point]
θ
Therefore, CRLB is

(τ ′ (θ))2 1
= . [2 points ]
nIX1 (θ) 4nθ

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2021-22 Even Mid-semester Examination MA 324

4. (5 points) Let X1 , X2 , . . . , Xn be a random sample from a Bernoulli distribution with success

1
probability p = 1+e θ , where θ ∈ R. Find the maximum likelihood estimator of θ. [Hint:
Investigate the existence and non-existence of maximum likelihood estimator.]

Solution: The likelihood function of θ is

 m  n−m
1 1
L(θ) = 1− ,
1 + eθ 1 + eθ
Pn
where m is the realized value of i=1 Xi . [1 point]
Now, consider the following cases.
Case I: m = 0. In this case, the likelihood function of θ is
 n
1
L(θ) = 1 − ,
1 + eθ

which is an increasing function in θ ∈ R. Therefore, in this case the MLE of θ does not exist.
[1 point]
Case II: m = n. In this case the likelihood function of θ is
 n
1
L(θ) = ,
1 + eθ

which is a decreasing function in θ ∈ R. Thus, the MLE of θ does not exist in this case also.
[1 point]
Case III: m = 1, 2, . . . , n − 1. In this case, the log-likelihood function is

l(θ) = −n ln 1 + eθ + (n − m)θ.

Taking first derivative with respect to θ and equate it to zero, we obtain

neθ n 
− + n − m = 0 =⇒ θ = ln −1 .
1 + eθ m
Moreover,

d2 neθ
l(θ) = − <0
dθ2 (1 + eθ )2
n


for all θ ∈ R. Therefore, l(θ) attains
 it’s maximum
 at θ = ln m
− 1 . Thus, MLE of θ exists
in this case and the MLE is θb = ln Pnn Xi − 1 . [2 points ]
i=1

5. (7 points) Let X1 , X2 , . . . , Xn be a random sample of size n (≥ 2) from a population with

probability density function
(
1 − xθ
e if x > 0
f (x; θ) = θ
0 otherwise,

where θ > 0 is unknown. With preassigned α ∈ (0, 1), derive a level α likelihood ratio test for
H0 : θ = θ0 (> 0) against H1 : θ ̸= θ0 .

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2021-22 Even Mid-semester Examination MA 324

Solution: Here Θ0 = {θ0 } and Θ1 = (0, ∞) \ Θ0 . The likelihood function is

" n
#
1 1X
L(θ) = n exp − xi . [1 point]
θ θ i=1

Therefore,
" n
#
1 1 X
sup L(θ) = L(θ0 ) = n exp − xi .
θ∈Θ0 θ0 θ0 i=1

To find supθ∈Θ0 ∪Θ1 L(θ), we need to find MLE of θ > 0. Now, standard calculation shows that
MLE of θ is X = n1 ni=1 Xi . Therefore,
P

" n
#
1 1X 1
sup L(θ) = sup L(θ) = L (x) = n exp − xi = n e−n . [1 point]
θ∈Θ0 ∪Θ1 θ∈(0, ∞) x x i=1 x

The likelihood ratio test statistics is

 n   
x x
Λ= exp −n − 1 . [1 point]
θ0 θ0

A LRT rejects null hypothesis if

Λ < k [1 point]
 n  
x nx
⇐⇒ exp − < k.
θ0 θ0

Here k is used as an generic constant. Now consider the function

f (y) = y n e−ny for y > 0.

It is easy to see that f has unique maximum at y = 1, f is strictly increasing for 0 < y < 1
and strictly decreasing for y > 1. Moreover, f (0) = 0 and limy→∞ f (y) = 0. Therefore,
 n  
x nx nx nx
Λ < k ⇐⇒ exp − < k ⇐⇒ < k1 or > k2 , [2 points ]
θ0 θ0 θ0 θ0

for k1 < k2 . Now, under null hypothesis,

nX
∼ Gamma(n, 1).
θ0
Thus, the test function of level α LRT is
(
1 if nx
θ0
< G1− α2 or nx
θ0
> G α2
ψ(x) =
0 otherwise,

where Gα is upper α-point of a Gamma(n, 1) distribution. [1 point]

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