Assignment

Assignment Solutions
Authors: Manas Sharma (MB2412)
Advisor/Instructor: Dr. Arijit Chakrabarti, Associate Professor, Applied Statistics Unit
Course: Large Sample Statistical Methods — M. Stat. — Indian Statistical Institute, Kolkata

Question 1 (Supplementary Sufficient Conditions for Fréchet Domain). Let F be a distribution function
with density f , and assume that f (x) > 0 for all sufficiently large x. Suppose further that
xf (x)
lim = ν > 0.
x→∞ 1 − F (x)
Show that under this condition, F lies in the domain of attraction of the Fréchet distribution H1,ν , i.e.,
1 − F (tx)
lim = x−ν , for all x > 0.
t→∞ 1 − F (t)

Proof. Define the survival function S(x) = 1 − F (x). The given condition can then be rewritten as
xf (x) −xS ′ (x)
lim = lim = ν > 0.
x→∞ S(x) x→∞ S(x)
We aim to establish that for all x > 0,
S(tx)
lim = x−ν .
S(t)
t→∞

Consider the function h(y) = log S(ey ). Then its derivative is

d ey S ′ (ey )
h′ (y) = log S(ey ) = .
dy S(ey )
Substituting x = ey , the initial condition becomes
lim −h′ (y) = ν, or equivalently, lim h′ (y) = −ν.
y→∞ y→∞

Now, consider the logarithmic form of the desired limit:

lim (log S(tx) − log S(t)) = −ν log x.
t→∞

Let t = eu , so that u → ∞ as t → ∞. Then

log S(tx) = log S(eu x) = log S(eu+log x ) = h(u + log x),
and similarly log S(t) = h(u). Thus,
lim (h(u + log x) − h(u)) = −ν log x.
u→∞

Let c = log x. By the Mean Value Theorem, for each u, there exists some ξu ∈ (u, u + c) such that
h(u + c) − h(u) = c · h′ (ξu ).
As u → ∞, ξu → ∞, and since limy→∞ h′ (y) = −ν, we have
lim h′ (ξu ) = −ν.
u→∞

Consequently,
lim (h(u + c) − h(u)) = c · (−ν) = −ν log x.
u→∞
Exponentiating both sides, we obtain:
S(eu+log x )
lim = e−ν log x = x−ν .
u→∞ S(eu )
Substituting back t = eu , it follows that
S(tx)
lim = x−ν .
t→∞ S(t)

Hence,
1 − F (tx)
lim = x−ν , for all x > 0,
t→∞1 − F (t)
which confirms that F is in the domain of attraction of the Fréchet distribution H1,ν . ■

10 May 2025 1 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 2. Consider a multinomial distribution with k cells. Let the cell probabilities π1 , . . . , πk be specified
functions π1 (θ), . . . , πk (θ) of m(< k) unknown parameters θ = (θ1 , . . . , θm )⊤ . Assume the following conditions
hold:

(a) Each πi (θ) admits continuous partial derivatives with respect to θj for all j = 1, . . . , m.

(b) The matrix M = M (θ) defined by its (r, s)-th element as

1 ∂πr (θ)
Mrs = p
πr (θ) ∂θs

for r = 1, . . . , k and s = 1, . . . , m, has rank m.

Let I(θ) be the Fisher Information matrix for a single observation from this multinomial distribution. Show
that
I(θ) = M ⊤ M.

Proof. Let X = (X1 , . . . , Xk )⊤ represent the outcome of a single trial from the multinomial distribution with
probabilities π(θ) = (π1 (θ), . . . , πk (θ))⊤ . Here, Xi = 1 if the outcome falls into cell i, and Xi = 0 otherwise.
The log-likelihood function for a single observation is:
k
X
ℓ(θ; X) = Xi log πi (θ).
i=1

The score vector S(θ) has components:

k
∂ℓ(θ; X) X 1 ∂πi (θ)
Sj (θ) = = Xi .
∂θj i=1
π i (θ) ∂θj

The Fisher Information matrix I(θ) is defined as:

I(θ) = Eθ [S(θ)S(θ)⊤ ].

The (j, s)-th element of I(θ) is:

I(θ)js = Eθ [Sj (θ)Ss (θ)].
Substituting the expression for Sj (θ), we have:
k
! k
!
X 1 ∂πi (θ) X 1 ∂πl (θ)
Sj (θ)Ss (θ) = Xi Xl .
i=1
π i (θ) ∂θj πl (θ) ∂θs
l=1

Expanding the product:

k X
k
X 1 ∂πi (θ) ∂πl (θ)
Sj (θ)Ss (θ) = Xi Xl .
i=1 l=1
πi (θ)πl (θ) ∂θj ∂θs

Taking the expectation and using the properties of multinomial indicator variables (E[Xi ] = πi (θ), E[Xi Xl ] =
0 for i ̸= l), we get: (
πi (θ) if i = l,
E[Xi Xl ] =
0 if i ̸= l.
Thus, the expectation simplifies to:
k
X 1 ∂πi (θ) ∂πi (θ)
I(θ)js = πi (θ) .
i=1
πi (θ)2 ∂θj ∂θs

10 May 2025 2 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Simplifying further:
k
X 1 ∂πi (θ) ∂πi (θ)
I(θ)js = .
i=1
πi (θ) ∂θj ∂θs

Now, consider the matrix M with entries:

1 ∂πr (θ)
Mrs = p .
πr (θ) ∂θs

The (j, s)-th element of M ⊤ M is:

k k
X X 1 ∂πr (θ) ∂πr (θ)
(M ⊤ M )js = (M ⊤ )jr Mrs = .
r=1 r=1
πr (θ) ∂θj ∂θs

Comparing this with I(θ)js , we see that:

I(θ)js = (M ⊤ M )js .

Since this holds for all j, s = 1, . . . , m, we conclude:

I(θ) = M ⊤ M. ■

10 May 2025 3 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 3. Let X1 , . . . , Xn be i.i.d. random variables from a distribution depending on a parameter

θ ∈ Θ ⊆ Rp . Let the first k population moments exist and be denoted by µj (θ) = Eθ (X j ) for j = 1, . . . , k.
Pn
Define the vector function e(θ) = (µ1 (θ), . . . , µk (θ))⊤ . Let mj = n1 i=1 Xij be the j-th sample moment, and
m = (m1 , . . . , mk )⊤ .
If e : Θ → Rk is one-to-one, the Method of Moments (MOM) estimator θ bn is defined implicitly by e(θ
bn ) = m,
−1 −1
or explicitly as θ n = e (m) if the inverse function e exists.
b
Assume the following conditions hold:

(a) e(θ) is one-to-one on an open set Θ ⊆ Rp .

(b) e(θ) is continuously differentiable at the true parameter value θ 0 ∈ Θ.
∂e(θ)
(c) The Jacobian matrix J = ∂θ ⊤
evaluated at θ 0 has full column rank p. (If p = k, this means J is
non-singular).
(d) The first 2k population moments exist, i.e., Eθ0 (X 2k ) < ∞.

Then, the MOM estimator θ

bn exists with probability approaching 1 and satisfies
√ d
bn − θ 0 ) −
n(θ → Np (0, Vθ0 ).

Explicitly determine the asymptotic covariance matrix Vθ0 .

Solution.

Step 1: Asymptotic Distribution of Sample Moments Let Yi = (Xi , Xi2 , . . . , Xik )⊤ . Then Y1 , . . . , Yn
are i.i.d. random vectors.P
The population P mean vector is E[Yi ] = (µ1 (θ 0 ), . . . , µk (θ 0 ))⊤ = e(θ 0 ). The sample
mean vector is Yn = ( n1 Xi , . . . , n1 Xik )⊤ = m. The covariance matrix of Yi is Σ, a k × k matrix with
entries
Σrs = CoV(X r , X s ) = Eθ0 (X r+s ) − Eθ0 (X r )Eθ0 (X s ) = µr+s (θ 0 ) − µr (θ 0 )µs (θ 0 ).
The existence of the 2k-th moment ensures that all entries of Σ are finite. By the Multivariate Central Limit
Theorem (CLT, see Question 13):
√ d
n(m − e(θ 0 )) −→ Nk (0, Σ).

Step 2: Applying the Multivariate Delta Method The MOM estimator is θ bn = g(m), where g = e−1
is the inverse function mapping the vector of moments back to the parameter vector. The true parameter
is θ 0 = g(e(θ 0 )). The function g maps from a subset of Rk to Rp . Let G be the Jacobian matrix of the
transformation g evaluated at the point e(θ 0 ). G is a p × k matrix:

∂g(y)
G= .
∂y⊤ y=e(θ 0 )

√ bn − θ 0 ) is given by:
By the Multivariate Delta Method (see Question 13), the asymptotic distribution of n(θ
√ d
→ Np (0, GΣG⊤ ).
n(g(m) − g(e(θ 0 ))) −

Therefore, the asymptotic covariance matrix is Vθ0 = GΣG⊤ .

Step 3: Relating G to the Jacobian of e Let J be the k × p Jacobian matrix of e(θ) evaluated at θ 0 :

∂e(θ)
J= .
∂θ ⊤ θ=θ 0

10 May 2025 4 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Since g = e−1 and we have the conditions that e is continuously differentiable with J having full column rank,
the inverse function theorem ensures that g is continuously differentiable in a neighborhood of e(θ 0 ).
By the chain rule, differentiating g(e(θ)) = θ with respect to θ ⊤ gives:

∂g(y) ∂e(θ)
· = Ip =⇒ GJ = Ip .
∂y⊤ y=e(θ 0 ) ∂θ ⊤ θ=θ 0

Similarly, differentiating e(g(y)) = y with respect to y⊤ gives:

∂e(θ) ∂g(y)
· = Ik =⇒ JG = Ik when θ = θ 0 .
∂θ ⊤ θ=g(y) ∂y⊤ y=e(θ 0 )

These matrix relations lead to two cases:

• If p = k (equal number of parameters and moments), then J is square and non-singular, and G = J −1 .

• If p < k (more moments than parameters), then J is k × p with full column rank p, and G is a left
inverse of J. Specifically, G = (J ⊤ J)−1 J ⊤ , which is the Moore-Penrose pseudo-inverse when J has full
column rank.

Step 4: Final Expression for Vθ0 Substituting the expression for G into Vθ0 = GΣG⊤ :

• For p = k:
Vθ0 = J −1 Σ(J −1 )⊤ = J −1 Σ(J ⊤ )−1 .

• For p < k:
Vθ0 = (J ⊤ J)−1 J ⊤ ΣJ(J ⊤ J)−1 .

∂µr (θ)
The entries of J are Jrs = ∂θs evaluated at θ 0 , and the entries of Σ are Σrs = µr+s (θ 0 ) − µr (θ 0 )µs (θ 0 ).
Therefore, the asymptotic covariance matrix of the MOM estimator is explicitly given by:
(
J −1 Σ(J ⊤ )−1 , if p = k,
Vθ 0 = ⊤ −1 ⊤ ⊤ −1
(J J) J ΣJ(J J) , if p < k.

10 May 2025 5 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 4. Let (n1 , . . . , nk ) be the observed counts from n independent trials, where each trial results in
Pk
b = (n1 /n, . . . , nk /n)⊤
one of k categories with probabilities p = (p1 , . . . , pk )⊤ , such that j=1 pj = 1. Let p
be the vector of sample proportions. Show that
√ d
p − p) −
n(b → Nk (0, Σ),
where Σ = diag(p) − pp⊤ .

Solution. Let Xi = (Xi1 , . . . , Xik )⊤ for i = 1, . . . , n be the outcome vector for the i-th independent trial. The
component Xij is an indicator variable defined as:
(
1 if the i-th trial results in category j
Xij =
0 otherwise
Pk
Since each trial must result in exactly one category, we have j=1 Xij = 1 for each i. The vectors X1 , . . . , Xn
are independent and identically distributed (i.i.d.).

Distribution of a single trial vector Xi Each Xi follows a Multinomial distribution with 1 trial and
category probabilities (p1 , . . . , pk ), denoted as Multinomial(1, p).

Mean of Xi The expectation of the j-th component Xij is:

E[Xij ] = 1 · P(Xij = 1) + 0 · P(Xij = 0) = P(trial i is category j) = pj .
The mean vector of Xi is therefore:
E[Xi ] = (E[Xi1 ], . . . , E[Xik ])⊤ = (p1 , . . . , pk )⊤ = p.

Covariance Matrix of Xi Let Σ be the k × k covariance matrix of Xi . The (j, l)-th entry is Σjl =
CoV(Xij , Xil ) = E[Xij Xil ] − E[Xij ]E[Xil ].

• Diagonal entries (j = l): We need Var(Xij ) = E[Xij

2
]−(E[Xij ])2 . Since Xij is an indicator, Xij
2
= Xij .
2
E[Xij ] = E[Xij ] = pj .
So, the diagonal entries are:
Σjj = Var(Xij ) = pj − p2j = pj (1 − pj ).

• Off-diagonal entries (j ̸= l): We need CoV(Xij , Xil ) = E[Xij Xil ]−E[Xij ]E[Xil ]. The product Xij Xil
is 1 only if the i-th trial results in both category j and category l. Since the categories are mutually
exclusive, this is impossible. Thus, Xij Xil = 0 always for j ̸= l.
E[Xij Xil ] = 0.
So, the off-diagonal entries are:
Σjl = 0 − pj pl = −pj pl .

Combining these, the covariance matrix is:

 
p1 (1 − p1 ) −p1 p2 ... −p1 pk
 
 −p2 p1 p2 (1 − p2 ) . . . −p2 pk 
Σ= .
 
.. .. .. ..

 . . . . 

−pk p1 −pk p2 . . . pk (1 − pk )
This matrix can be written compactly using matrix notation. Let diag(p) be the diagonal matrix with
p1 , . . . , pk on the diagonal. The matrix pp⊤ is the outer product, an k × k matrix with (j, l)-th entry pj pl .
Then, Σ = diag(p) − pp⊤ .

10 May 2025 6 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

b ThePtotal count for category j after n trials is nj = ni=1 Xij . The vector
P
Sample Proportion Vector p
n
of observed counts is (n1 , . . . , nk )⊤ = i=1 Xi . The vector of sample proportions is p b = (n1 /n, . . . , nk /n)⊤ .
We can write this as:
n
1 1X
b = (n1 , . . . , nk )⊤ =
p Xi .
n n i=1

This shows that p

b is the sample mean of the i.i.d. random vectors X1 , . . . , Xn .

Applying the Multivariate Central Limit Theorem (CLT) The Multivariate CLT states that if
Y1 , . . . , Yn are i.i.d. k-dimensional random vectors with mean vector µ and finite covariance matrix Σ, then
as n → ∞: √ d
n(Yn − µ) −→ Nk (0, Σ),
Pn
where Yn = n1 i=1 Yi .
In our case, Yi = Xi , Yn = p b , µ = E[Xi ] = p, and the covariance matrix is Σ = diag(p) − pp⊤ . The
covariance matrix Σ is finite since all pj ∈ [0, 1].
Applying the Multivariate CLT directly gives:
√ d
p − p) −
n(b → Nk (0, Σ),

where Σ = diag(p) − pp⊤ .

This proves the asymptotic
Pk normality
Pkof the sample proportions
Pk vector. Note that the covariance matrix Σ is
singular because j=1 pbj = 1 and j=1 pj = 1, implying j=1 (b pj − pj ) = 0. This linear dependency means
the distribution is concentrated on a (k − 1)-dimensional subspace of Rk .

10 May 2025 7 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 5. Consider Pa k × l contingency table with observed cell counts nij (i = 1, . . . , k; j = 1, . . . , l) and
total sample size n = i,j nij . Let πij be the probability of an observation falling
P into cell (i, j). P We want to
test the hypothesis of independence H0 : πij = πi. π.j for all i, j, where πi. = j πij and π.j = i πij are the
marginal probabilities. Assume the marginal probabilities πi. and π.j are known.
Define the following chi-squared statistics:
k X l 2
X (nij − nπi. π.j )
T =
i=1 j=1
nπi. π.j
k 2
X (ni. − nπi. ) X
T1 = where ni. = nij
i=1
nπi. j
l 2
X (n.j − nπ.j ) X
T2 = where n.j = nij
j=1
nπ.j i

The test statistic often used for testing independence when marginals are known is T3 = T − T1 − T2 . Show
the algebraic simplification of T3 to the form
k X l 2
X (nij − ni. π.j − n.j πi. + nπi. π.j )
T3 =
i=1 j=1
nπi. π.j

and state its asymptotic distribution under H0 .

Solution. We are given the statistics T , T1 , and T2 . We want to simplify T3 = T − T1 − T2 .

Let eij = nπi. π.j , ei. = nπi. , and e.j = nπ.j . We show that

X (nij − eij )2 X (ni. − ei. )2 X (n.j − e.j )2 X (nij − ni. π.j − n.j πi. + nπi. π.j )2
− − = .
i,j
eij i
ei. j
e.j i,j
nπi. π.j

2
We start by expanding the term (nij − eij ) :

nij − eij = nij − nπi. π.j

= (nij − ni. π.j − n.j πi. + nπi. π.j )
+ (ni. π.j − nπi. π.j )
+ (n.j πi. − nπi. π.j )
= (nij − ni. π.j − n.j πi. + eij ) + π.j (ni. − nπi. ) + πi. (n.j − nπ.j )

Let dij = nij − ni. π.j − n.j πi. + eij , di. = ni. − ei. , and d.j = n.j − e.j . Then

nij − eij = dij + π.j di. + πi. d.j .

Now, we square this expression and divide by eij = nπi. π.j :

2 2
(nij − eij ) (dij + π.j di. + πi. d.j )
= .
eij nπi. π.j

We need to show that the cross-product terms sum to zero when summed over i and j.
X 2dij (π.j di. ) X 2di. X
= dij
i,j
nπi. π.j i
nπi. j
X 2dij (πi. d.j ) X 2d.j X
= dij
i,j
nπi. π.j j
nπ.j i

10 May 2025 8 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

! 
X 2 (π.j di. ) (πi. d.j ) X 2di. d.j 2 X X
= = di.  d.j 
i,j
nπi. π.j i,j
n n i j

Let’s check the sums of dij :

X X
dij = (nij − ni. π.j − n.j πi. + nπi. π.j )
j j
       
X X X X
= nij  − ni.  π.j  −  n.j  πi. + nπi.  π.j 
j j j j
 
XX
= ni. − ni. (1) −  ni′ j  πi. + nπi. (1)
j i′
 
XX
= ni. − ni. −  ni′ j  πi. + nπi.
i′ j

= −nπi. + nπi. = 0.
P P P P P
Similarly,
P i dij = 0. Also, i di. = i (ni. − nπi. ) = ( i ni. ) − n ( i πi. ) = n − n(1) = 0. Similarly,
j d.j = 0.
P P P P
Since j dij = 0 and i dij = 0, the first two cross-product sums are zero. Since i di. = 0 and j d.j = 0,
the third cross-product sum is also zero.
Therefore,
X (nij − eij )2 X d2ij + (π.j di. )2 + (πi. d.j )2
=
i,j
eij i,j
nπi. π.j
X d2ij X π.j 2 2
di. X πi.2 d2.j
= + +
i,j
eij i,j
nπi. π.j i,j
nπi. π.j
X d2ij X d2 X X d2.j X
i.
= + π.j + πi.
i,j
eij i
nπi. j j
nπ.j i
X d2ij X d2 X d2.j
i.
= + (1) + (1)
i,j
eij i
nπi. j
nπ.j
X (nij − ni. π.j − n.j πi. + eij )2 X (ni. − ei. )2 X (n.j − e.j )2
= + +
i,j
eij i
ei. j
e.j

Substituting back the definitions of T, T1 , T2 :

X (nij − ni. π.j − n.j πi. + nπi. π.j )2
T = + T1 + T2 .
i,j
nπi. π.j

Rearranging gives the desired result:

k X l 2
X (nij − ni. π.j − n.j πi. + nπi. π.j )
T − T1 − T2 = .
i=1 j=1
nπi. π.j

This confirms the algebraic simplification provided in the question prompt. This result is a specific instance
of Lancaster-Irwin partitioning for contingency tables.

Asymptotic Distribution Under the null hypothesis H0 : πij = πi. π.j with known marginal probabilities
πi. and π.j :

10 May 2025 9 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

d
• T −→ χ2kl−1 (since the cell counts follow a multinomial distribution with kl cells, and probabilities sum
to 1).
d
• T1 − → χ2k−1 (since the row sums ni. follow a multinomial distribution with k cells and known probabilities
πi. ).
d
• T2 −→ χ2l−1 (since the column sums n.j follow a multinomial distribution with l cells and known proba-
bilities π.j ).

By Cochran’s theorem or properties of partitioning chi-squared statistics for multinomial data, the statistic
T3 = T − T1 − T2 measures the interaction (departure from independence) after accounting for the marginal
discrepancies. Its asymptotic distribution under H0 is chi-squared with degrees of freedom equal to the
difference in the degrees of freedom of the components:

df = (kl − 1) − (k − 1) − (l − 1) = kl − 1 − k + 1 − l + 1 = kl − k − l + 1 = (k − 1)(l − 1).

Thus, under H0 , as n → ∞,
d
→ χ2(k−1)(l−1) .
T3 −

10 May 2025 10 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 6. Show that the Lyapunov condition:

n
1 X h
2+δ
i
E |Xk − µk | →0 for some δ > 0,
s2+δ
n k=1

implies the Lindeberg-Feller condition:

n   
1 X 2 |Xk − µk |
E (Xk − µk ) 1 > ϵ →0 for all ϵ > 0,
s2n sn
k=1
Pn
where s2n = k=1 σk2 and σk2 = Var(Xk ).

Proof. Assume the Lyapunov condition holds:

n
1 X h
2+δ
i
E |Xk − µk | →0 for some δ > 0.
s2+δ
n k=1

We need to show that the Lindeberg-Feller condition is satisfied:

n   
1 X 2 |Xk − µk |
E (Xk − µk ) 1 >ϵ →0 for all ϵ > 0.
s2n sn
k=1

For any ϵ > 0, let  

|Xk − µk |
Ank = >ϵ
sn
. Then,
h i h i  1 δ
2 2+δ
E (Xk − µk ) 1 (Ank ) ≤ E |Xk − µk | · ,
ϵsn
 δ
2 2+δ 1
where we used the inequality (Xk − µk ) ≤ |Xk − µk | · |Xk −µ k|
on the set Ank .
Summing over k = 1, . . . , n, we get
n n
1 X h 2
i 1 1 X h 2+δ
i
E (Xk − µk ) 1 (A nk ) ≤ · E |Xk − µk | .
s2n s2n ϵδ sδn
k=1 k=1

By the Lyapunov condition,

n
1 X h
2+δ
i
E |Xk − µk | → 0.
s2+δ
n k=1

Therefore,
n
1 X h 2
i
E (Xk − µk ) 1 (A nk ) →0 as n → ∞.
s2n
k=1

Since this holds for all ϵ > 0, the Lindeberg-Feller condition is satisfied. ■

10 May 2025 11 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 7. Let {Xn } be a sequence of independent random variables such that

(n + 1)i/2
 
n
P Xn = = , i, n ∈ N.
n (n + 1)i

Does the sample mean X n − E(X n ) converge in probability to zero?

Solution. First, we verify that the probability mass function (PMF) is valid for each Xn :
∞ ∞
X n X 1
i
= n
i=1
(n + 1) i=1
(n + 1)i
1 1
=n· · 1
n + 1 1 − n+1
1
= n · = 1.
n

Now compute E[Xn ] for each n ∈ N:

∞
X (n + 1)i/2 n
E[Xn ] = ·
i=1
n (n + 1)i
∞
X 1
=
i=1
(n + 1)i/2
1 1
=√ · 1
n + 1 1 − √n+1
1
=√
n+1−1
√
n+1+1
= .
n

Let µn = E[Xn ]. To apply the weak law of large numbers via Markov’s criterion, we examine whether
n
1X
E[|Xk − µk |] → 0 as n → ∞.
n
k=1

Note that for i = 1, √

(n + 1)1/2 n+1+1
< µn = ,
n n
while for i ≥ 2,
(n + 1)i/2
> µn .
n
Therefore,
∞
X (n + 1)i/2 n
E[|Xn − µn |] = − µn ·
i=2
n (n + 1)i
(n + 1)1/2 n
+ − µn · .
n n+1

Simplifying:
∞ ∞
X 1 X n 1
E[|Xn − µn |] = − µn · i
+
i=2
(n + 1)i/2 i=2
(n + 1) n + 1

10 May 2025 12 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

√
1 1 ( n + 1 + 1) n + 1 1
= · 1 − · +
n + 1 1 − √n+1 n(n + 1) n n+1
2
= .
n+1
Now consider:
n n
1X 1X 2
E[|Xk − µk |] =
n n k+1
k=1 k=1
n
2 X 1
= .
n k+1
k=1
Pn 1
Since k=1 k+1 ∼ ln n, we have
n
1X 2 ln n
E[|Xk − µk |] ∼ → 0.
n n
k=1

Thus, by Markov’s version of the weak law of large numbers,

p
X n − E(X n ) −
→ 0.

Hence, the sample mean converges in probability to its expected value.

Question 8. Let F be a distribution function on Rk . For a = (a1 , . . . , ak )⊤ ∈ Rk , define
k
[
x ∈ Rk : xi = ai , xj ≤ aj ∀ j ̸= i .

S=
i=1

Prove that F is continuous at a if and only if P(S) = 0.

Proof. Let F be a distribution function on Rk , and let a = (a1 , . . . , ak )⊤ ∈ Rk . By definition, F is continuous

at a if
lim F (x) = F (a).
x→a

Since F is a distribution function, we can write F (x) = P(X1 ≤ x1 , . . . , Xk ≤ xk ) for a random vector
X = (X1 , . . . , Xk ).
For any b = (b1 , . . . , bk ) ∈ Rk , let us define
F (b− ) = lim F (x)
x→b
xi <bi ,∀i

(Only if direction): Assume F is continuous at a. Then F (a) = F (a− ). We know that

P(S) = F (a) − F (a− )

This is because S precisely captures the set of points that contribute to the jump of F at a. Since F (a) =
F (a− ) by continuity, we have P(S) = 0.

(If direction): Assume P(S) = 0. We need to show F is continuous at a, i.e., F (a) = F (a− ).
Let Si = {x ∈ Rk : xi = ai , xj ≤ aj ∀ j ̸= i} so that S = ∪ki=1 Si . Then,
F (a) − F (a− ) = P(S) = 0
where the inequality follows from the sub-additivity of probability and the equality to zero follows from our
assumption that P(S) = 0.
Since F (a) − F (a− ) = 0, we have F (a) = F (a− ), implying that F is continuous at a.
Therefore, F is continuous at a if and only if P(S) = 0. ■

10 May 2025 13 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

p
Question 9. Show that if Xn −
→ a and f is differentiable at a, then

f (Xn ) = f (a) + (Xn − a)f ′ (a) + (Xn − a)Zn ,

p
where Zn −
→ 0.

Proof. Since f is differentiable at a, by definition, there exists a function r(x) such that for all x near a, we
have
f (x) = f (a) + (x − a)f ′ (a) + (x − a)r(x),
where limx→a r(x) = 0.
Define a function g : R → R as follows:
′

 f (x) − f (a) − (x − a)f (a)
, if x ̸= a
g(x) = x−a
0, if x = a

The function g is continuous at a and satisfies g(a) = 0, by the differentiability of f .

For each n with Xn ̸= a, we can write:

f (Xn ) = f (a) + (Xn − a)f ′ (a) + (Xn − a)g(Xn ).

p
Define Zn = g(Xn ). Since Xn −
→ a and g is continuous at a, it follows from the Continuous Mapping Theorem
p
that Zn = g(Xn ) −
→ g(a) = 0.
Hence, we have
f (Xn ) = f (a) + (Xn − a)f ′ (a) + (Xn − a)Zn ,
p
where Zn −
→ 0, completing the proof. ■
Question 10 (Compulsory). Let Tn = (T1n , . . . , Tkn )⊤ be a sequence of k-dimensional statistics for n ≥ 1,
with the asymptotic behavior:
√ d
n(Tn − θ) − → Nk (0, Σ),
where θ ∈ Rk and Σ is a positive semi-definite covariance matrix. Let gi : Rk → R, i = 1, . . . , ℓ, be functions
that are differentiable. The asymptotic distribution of
√

n g(Tn ) − g(θ) ,

where g(Tn ) = (g1 (Tn ), . . . , gℓ (Tn ))⊤ , is given by

√
d
n g(Tn ) − g(θ) −→ Nℓ (0, GΣG⊤ ),
 
∂gi
where G = ∂θj is the Jacobian matrix of g evaluated at θ.
ℓ×k

Lemma 10.1. Let g : Rk → Rℓ be differentiable at θ ∈ Rk . Then, for any T ∈ Rk ,

g(T) = g(θ) + G(T − θ) + R,

 
∂gi
where G = ∂θj is the Jacobian of g evaluated at θ, and the remainder term R satisfies
ℓ×k

∥R∥ = o(∥T − θ∥) as T → θ.

√
Lemma 10.2. Let Tn ∈ Rk be a sequence such that n(Tn − θ) is Op (1). If ∥Rn ∥ = o(∥Tn − θ∥), then
√ p
nRn −
→ 0.

10 May 2025 14 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Proof. Since ∥Rn ∥ = o(∥Tn −θ∥), there

√ exists a sequence ϵn → 0 such that ∥Rn ∥ ≤ ϵn ∥Tn −θ∥ for sufficiently
large n. Multiplying both sides by n, we obtain
√ √
n∥Rn ∥ ≤ ϵn n∥Tn − θ∥.
√
Since n(Tn − θ) is bounded in probability, i.e., Op (1).
√ √ p
Therefore, n∥Rn ∥ ≤ ϵn · Op (1) = op (1) as n → ∞, since ϵn → 0. Hence, n∥Rn ∥ −
→ 0, implying
√ p
nRn −→ 0. ■

Proof. Using Lemma 10.1, we can expand g(Tn ) via a first-order Taylor series around θ:

g(Tn ) = g(θ) + G(Tn − θ) + Rn ,

where ∥Rn ∥ = o(∥Tn − θ∥) as n → ∞.

√
Rearranging and multiplying by n, we obtain
√
√ √
n g(Tn ) − g(θ) = G n(Tn − θ) + nRn .

√ d
Given that n(Tn − θ) −
→ Nk (0, Σ), by the continuous mapping theorem, we get:
√ d
→ G · Nk (0, Σ) = Nℓ (0, GΣG⊤ ).
G n(Tn − θ) −
√ √ p
Moreover, by Lemma 10.2, since n(Tn − θ) is convergent in distribution, we have nRn −
→ 0.
Finally, by Slutsky’s theorem, which allows us to combine converging sequences, we conclude:
√ √ √ d
→ Nℓ (0, GΣG⊤ ).


n g(Tn ) − g(θ) = G n(Tn − θ) + nRn −

■
Question 11. Suppose Xi are i.i.d. random variables with a finite fourth moment. Let
σ
γ=
µ
represent the population coefficient of variation. Determine the asymptotic distribution of
√
 
sn
n −γ ,
Xn
Pn Pn
where X n = n1 i=1 Xi and s2n = n1 i=1 (Xi − X n )2 .

Solution. Let µ = E(Xi ) and σ 2 = Var(Xi ). The coefficient of variation is γ = σµ . We aim to find the
√  
asymptotic distribution of n Xsn − γ . We will also denote the third and fourth central moments as
n

µ3 = E[(Xi − µ)3 ] and µ4 = E[(Xi − µ)4 ].

Step 1: Asymptotic Distribution of X n . By the Central Limit Theorem (CLT), for i.i.d. random
variables with finite variance: √
d
→ N 0, σ 2 .


n Xn − µ −

10 May 2025 15 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Step 2: Asymptotic Distribution of s2n . We now analyze the asymptotic behavior of s2n . Expanding s2n :
n
1X
s2n = (Xi − X n )2
n i=1
n
1X
= [(Xi − µ) − (X n − µ)]2
n i=1
n
1X
= (Xi − µ)2 − (X n − µ)2 .
n i=1

Consider each term:

Pn
(i) The first term, n1 i=1 (Xi − µ)2 , is the sample mean of squared deviations from the mean, which has
mean σ 2 and variance µ4 − σ 4 . By the CLT:
n
!
√ 1X d
n (Xi − µ)2 − σ 2 −→ N (0, τ 2 ),
n i=1

where τ 2 = µ4 − σ 4 .
√ √
(ii) The second term, (X n − µ)2 , is of order Op (n−1 ), as n(X n − µ) = Op (1). Therefore, n(X n − µ)2 =
Op (n−1/2 ) = op (1), which means its contribution becomes negligible in the limit.

Thus, combining the results:

√ d
n(s2n − σ 2 ) −
→ N (0, τ 2 ).

Step 3: Joint Asymptotic Distribution. To use the multivariate delta method for the ratio Xsn , we
√ √ n
need the joint asymptotic distribution of n(s2n − σ 2 ) and n(X n − µ). The covariance between these two
quantities is: √ √
CoV n(X n − µ), n(s2n − σ 2 ) = µ3 .

Therefore, the joint asymptotic distribution is:

! ! !!
√ s2n − σ 2 d 0 µ4 − σ 4 µ3
n −
→ N2 , .
Xn − µ 0 µ3 σ2

sn
Step 4: Asymptotic Distribution of Xn
. Applying the multivariate delta method to the function
√
x
g(x, y) = y at (σ 2 , µ), we compute the partial derivatives:

∂g 1 ∂g σ
= , = − 2.
∂x 2σµ ∂y µ
Using the delta method, we find:
√
 
sn σ d
n − −
→ N (0, V ),
Xn µ
where:
1 4 σ4 µ3
V = (µ4 − σ ) + − .
4σ 2 µ2 µ4 2σµ3

10 May 2025 16 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

σ √  sn 
Step 5: Final Result. Since γ = µ, the asymptotic distribution of n X − γ is:
n

√ 1 µ4 − σ 4
    
sn d 4 µ3 γ
n −γ −
→ N 0, 2 + γ − .
Xn µ 4σ 2 2σµ

Question 12. Suppose (Xi , Yi ), for i = 1, . . . , n, is an i.i.d. sample from the bivariate normal distribution:
2
BV N (µX , µY , σX , σY2 , ρ),

where ρ is the population correlation coefficient. Define the sample correlation coefficient as:
Pn
i=1 (Xi − X)(Yi − Y )
rn = qP qP ,
n 2 n 2
i=1 (Xi − X) i=1 (Yi − Y )

1
Pn 1
Pn
with sample means X = n i=1 Xi and Y = n i=1 Yi . Show that:
√ d
→ N (0, (1 − ρ2 )2 ).
n(rn − ρ) −

2
Proof. Let (Xi , Yi ), i = 1, . . . , n, be an i.i.d. sample from BV N (µX , µY , σX , σY2 , ρ). Define:
n n n
1X 1X 1X
SXY = (Xi − X)(Yi − Y ), SXX = (Xi − X)2 , SY Y = (Yi − Y )2 .
n i=1 n i=1 n i=1

Then:
SXY
rn = √ .
SXX SY Y

Step 1: CLT for sample means. By the Central Limit Theorem:

! ! !!
√ 2
X − µX d 0 σX ρσX σY
n −
→ N2 , .
Y − µY 0 ρσX σY σY2

Step 2: Joint asymptotics of SXY , SXX , SY Y . Applying the multivariate Central Limit Theorem and
delta method:     
SXY − ρσX σY 0
√  2
 d   
 SXX − σX  −
n  → N3 0 , Σ ,
  
SY Y − σY2 0
where Σ is given by:  
2 2
σX σY (1 + ρ2 ) 3
ρσX σY ρσX σY3
1 3 4

Σ=  ρσX σY 2σX 2ρ2 σX
2 2 .
σY 
n
ρσX σY3 2ρ2 σX
2 2
σY 2σY4

Step 3: Delta method for rn . Let:

SXY
g(SXY , SXX , SY Y ) = √ .
SXX SY Y
The gradient at the population values is:
 T
1 ρ ρ
∇g = , − 2 , − 2 .
σX σY 2σX 2σY

10 May 2025 17 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Then:

V = ∇g T Σ∇g
ρ2 ρ2
 
1 2 2 2 4
= (1 + ρ ) + + −ρ −ρ +ρ
n 2 2
(1 − ρ2 )2
= .
n

Conclusion. Therefore, by the delta method:

√ d
→ N (0, (1 − ρ2 )2 ).
n(rn − ρ) − ■
 
Question 13. Let X1 , . . . , Xn be i.i.d. random variables such that E(Xi ) = µ, Var(Xi ) = σ 2 , and E Xi2r <
∞ (i.e., the 2r-th raw moment exists). Define the r-th sample raw moment as
n
1X r
Mr,n = X .
n i=1 i

Determine the joint asymptotic distribution of the first r sample raw moments (M1,n , M2,n , . . . , Mr,n ).

Solution. Each Mk,n is the sample mean of the i.i.d. sequence X1k , . . . , Xnk . Let µk = E[Xik ], the population’s
k-th raw moment.
Because E[Xi2r ] < ∞, all lower-order moments (up to 2r) also exist. In particular, for each 1 ≤ k ≤ r,

Var(Xik ) = E[Xi2k ] − µ2k < ∞.

Step 1: Define the vector of sample moments. Let Mn = (M1,n , M2,n , . . . , Mr,n )⊤ . Then:
n
1X
Mn = Yi , where Yi = (Xi , Xi2 , . . . , Xir )⊤ .
n i=1

The Yi are i.i.d. vectors in Rr with finite mean and covariance.

Step 2: Compute mean vector and covariance matrix. Define µ = (µ1 , µ2 , . . . , µr )⊤ and let the
covariance matrix Σ = (σjk )1≤j,k≤r be:

σjk = CoV(Xij , Xik ) = E[Xij+k ] − µj µk .

Since E[Xi2r ] < ∞, all entries of Σ are finite.

Step 3: Apply the multivariate CLT. By the multivariate central limit theorem:
√ d
n(Mn − µ) −
→ Nr (0, Σ).

Conclusion: The joint asymptotic distribution of (M1,n , M2,n , . . . , Mr,n ) is multivariate normal with mean
vector µ and covariance matrix Σ/n.
iid
Question 14. Let X1 , . . . , Xn ∼ N (θ, θ), where θ > 0.

(a) Determine the Maximum Likelihood Estimator (MLE) of θ.

(b) Find a variance-stabilizing transformation g(θbn ), where θbn is the MLE.

10 May 2025 18 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Solution of (a). The PDF of Xi ∼ N (θ, θ) is

(xi − θ)2
 
1
f (xi ; θ) = √ exp − , θ > 0.
2πθ 2θ
The likelihood is !
n
−n/2 1 X
L(θ; x) = (2πθ) exp − (xi − θ)2 .
2θ i=1
Taking the log-likelihood:
n n 1 X 2 nθ
ℓ(θ) = − log(2π) − log(θ) − xi + nX̄n − .
2 2 2θ 2
Differentiating:
∂ℓ n 1 X 2 n
=− + 2 xi − .
∂θ 2θ 2θ 2
Setting this to zero and solving:

′ ′ 1X 2
θ2 + θ − M2,n = 0, M2,n = xi .
n
The positive solution gives the MLE: q
−1 + ′
1 + 4M2,n
θbn = .
2
Second derivative check confirms it is a maximum.

Solution of (b). The Fisher Information is

n(1 + 2θ)
I(θ) = .
2θ2

So, the asymptotic variance of θbn is

2θ2
Var(θbn ) ≈ .
n(1 + 2θ)
A variance-stabilizing transformation satisfies:
r
′ 1 + 2θ
g (θ) ∝ .
2θ2
√
Using substitution u = 1 + 2θ, we integrate to get:
√
√

1 1 + 2θ − 1
g(θ) = 1 + 2θ + ln √ .
2 1 + 2θ + 1
Applying the Delta method, the transformed estimator has asymptotic variance 1:
√
d
n g(θbn ) − g(θ) −→ N (0, 1).

Question 15. Suppose we have k independent samples of sizes n1 , . . . , nk drawn from bivariate normal
distributions: ! !!
0 1 ρi
N2 ,
0 ρi 1
for each i = 1, . . . , k. Let ri denote the sample correlation coefficient for the i-th sample. We want to test the
null hypothesis:
H0 : ρ1 = · · · = ρk = ρ.

10 May 2025 19 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Define the Fisher z-transform Zi = tanh−1 (ri ) and the weighted average:
Pk
i=1 ni Zi
Z̄ = P k
.
i=1 ni

Consider the test statistic:

k
X
T = ni (Zi − Z̄)2 .
i=1

Show that under H0 , as min(n1 , . . . , nk ) → ∞, the asymptotic distribution of T is χ2k−1 .

√ d
Proof. From previous results, we know that ni (ri − ρi ) − → N (0, (1 − ρ2i )2 ). Applying the Delta method to
−1 −1
the function g(r) = tanh (r), whose derivative is g ′ (r) = 1−r
1
2 , we define ζi = tanh (ρi ). Then:
√ d
ni (Zi − ζi ) −
→ N (0, 1).
Therefore, for large ni , Zi ∼ N (ζi , 1/ni ) approximately and the Zi are independent.
Under H0 , all ζi = ζ = tanh−1 (ρ), so:
Zi ∼ N (ζ, 1/ni ) for all i.
√
Define Yi = ni (Zi − ζ), so Yi ∼ N (0, 1). We write:
Pk √
Yi j=1 nj Yj
Zi = ζ + √ , Z̄ = ζ + Pk .
ni j=1 nj

Then: Pk √
Yi j=1 nj Yj
Zi − Z̄ = √ − Pk .
ni j=1 nj
The test statistic becomes:
k
P
k √ 2
X i=1 ni Yi
T = Yi2 − Pk .
i=1 i=1 ni
√ √
Let Y = (Y1 , . . . , Yk )T ∼ Nk (0, Ik ), and define a = √1N ( n1 , . . . , nk )T , where N = ni . Note ∥a∥2 = 1.
P

Then:
T = YT (Ik − aaT )Y.
Define Q = Ik − aaT . Then Q is symmetric and idempotent:
Q2 = Q.
Hence, T = YT QY ∼ χ2rank(Q) . Since tr(Q) = k − tr(aaT ) = k − 1,
d
→ χ2k−1 .
T −
■
Question 16. Let X1 , . . . , Xn be an i.i.d. sample from N (0, 1) distribution with sample size n = 5. Consider
the function g(x) = ex . We are interested in the distribution of the statistic
√
n(g(X n ) − g(0)) √ X n
Yn = = n(e − 1).
g ′ (0)
The standard delta method approximates the cumulative distribution function (CDF) of Yn by F1 (z) = Φ(z),
where Φ is the standard normal CDF. A two-term delta√method can√provide a refined √ approximation, let’s
call it F2 (z). The exact CDF is given by FExact (z) = Φ( n ln(1 + z/ n)) for z > − n.
Compare the accuracy of the standard delta method and the two-term delta method approximation (derived
in the solution) against the exact distribution for this specific case (n = 5, g(x) = ex , underlying N (0, 1)),
using the table provided in the solution.

10 May 2025 20 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Solution.

Derivation of the Two-Term Delta Method Approximation √ Let Tn = X n . Since Xi ∼ N (0, 1), we
√
have µ = E[Tn ] = 0 and σn2 = Var(Tn ) = 1/n. Let Zn = nTn = nX n , so Zn ∼ N (0, 1). We expand
g(Tn ) = eTn around µ = 0:
1
g(Tn ) = g(0) + g ′ (0)Tn + g ′′ (0)Tn2 + Op (n−3/2 )
2
Since g(x) = ex , g ′ (x) = ex , g ′′ (x) = ex , we have g(0) = 1, g ′ (0) = 1, g ′′ (0) = 1.
1 2
eX n = 1 + 1 · X n + · 1 · X n + Op (n−3/2 )
2
Zn Z2
= 1 + √ + n + Op (n−3/2 )
n 2n

Substitute this into the expression for Yn :

√
Yn = n(eX n − 1)
√ Z2
 
Zn
= n √ + n + Op (n−3/2 )
n 2n
2
Z
= Zn + √n + Op (n−1 )
2 n

2 2
The two-term approximation considers Yn ≈ Z + 2Z√n , where Z ∼ N (0, 1). Let h(Z) = Z + 2Z√n . We
approximate P(Yn ≤ z) by P(h(Z) ≤ z). The quadratic equation h(Z) = z, or 2√1 n Z 2 + Z − z = 0, has
solutions
p √
−1 ± 1 + 2z/ n
Z1,2 = √
1/ n
s !
√ 2z
= n −1 ± 1 + √
n

√ √
These
√ solutions are √ n ≥ 0, i.e., z ≥ − n/2. The function h(Z) has a minimum value of
√ real if 1 + 2z/
− n/2 at Z = − n. For z > − n/2, the event h(Z) ≤ z is equivalent to Z1 ≤ Z ≤ Z2 , where
√ q √
Z1 = n(−1 − 1 + 2z/ n)
√ q √
Z2 = n(−1 + 1 + 2z/ n)

Thus, the approximation is

F2 (z) = P(Z1 ≤ Z ≤ Z2 ) = Φ(Z2 ) − Φ(Z1 )

√ √
For z√≤ − n/2, the event h(Z) ≤ z is impossible (or has probability 0), so F2 (z) = 0. For n = 5, n ≈ 2.236,
so − n/2 ≈ −1.118.

Exact CDF The exact CDF is given by

√
FExact (z) = P( n(eX n − 1) ≤ z)
√
= P(eX n ≤ 1 + z/ n)
√
= P(X n ≤ ln(1 + z/ n))

10 May 2025 21 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

√
Since X n ∼ N (0, 1/n), nX n ∼ N (0, 1).
√ √ √
FExact (z) = P( nX n ≤ n ln(1 + z/ n))
√ √
= Φ( n ln(1 + z/ n))
√ √
for 1 + z/ n > 0, i.e., z > − n.

Comparison Table The following table compares the CDF values F1 (z), F2 (z), and FExact (z) for n = 5
at various values of z.

z Delta method (F1 (z)) Two term delta method (F2 (z)) Exact (FExact (z))

-3.0 0.0013 0.0000 0.0000

-2.5 0.0062 0.0000 0.0000
-2.0 0.0228 0.0000 0.0000
-1.6 0.0548 0.0000 0.0025
-1.2 0.1151 0.0276 0.0405
-0.8 0.2119 0.1481 0.1610
-0.4 0.3446 0.3251 0.3297
-0.2 0.4207 0.4169 0.4170
0.0 0.5000 0.5000 0.5000
0.2 0.5793 0.5719 0.5712
0.4 0.6554 0.6441 0.6436
0.8 0.7881 0.7554 0.7530
1.2 0.8849 0.8374 0.8317
1.6 0.9452 0.8944 0.8863
2.0 0.9772 0.9312 0.9211
2.5 0.9938 0.9584 0.9450
3.0 0.9987 0.9741 0.9591

√
Analysis of Accuracy The table provides values for the CDF of Y5 = 5(eX 5 −1) using the three methods.

• Standard Delta Method (F1 (z) = Φ(z)): This normal approximation performs adequately near
z = 0 but deteriorates in the tails. It systematically overestimates the true CDF at both extremes,
with substantial errors at z = −1.6 (F1 ≈ 0.0548 vs Exact 0.0025) and z = 1.6 (F1 ≈ 0.9452 vs Exact
0.8863).
2
• Two-Term Delta Method (F2 (z)): This refined approximation based on Yn ≈ Z + 2Z√n significantly
improves accuracy throughout. For z < −1.118, it correctly gives F2 (z) = 0, and near z = 0, its
accuracy is remarkable (at z = −0.2, F2 = 0.4169 vs Exact 0.4170). In both tails, F2 (z) consistently
outperforms F1 (z) in approximating the exact CDF.
• Exact CDF (FExact (z)): The distribution of Yn stems from the log-normal distribution and exhibits
right-skewness, explaining why the symmetric normal approximation performs poorly in the tails.

Conclusion For this small sample case (n = 5), the standard delta method provides only a rough approx-
imation, particularly in the tails where the exponential transformation introduces skewness. The two-term
delta method, by incorporating the quadratic term from the Taylor expansion, captures this asymmetry better

10 May 2025 22 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

and delivers a substantially more accurate approximation across all quantiles. This demonstrates the value
of higher-order approximations for non-linear transformations with small samples.

10 May 2025 23 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 17. Let X1 , . . . , Xn be i.i.d. random variables with distribution function F , possessing a density
f . Define the population p-th quantile as ξp = F −1 (p), where 0 < p < 1. Suppose that f (ξp ) > 0 and that f
is continuous near ξp . Let Yp,n denote the sample p-th quantile based on the sample X1 , . . . , Xn .
√
Provide Bahadur’s representation of Yp,n and use it to determine the asymptotic distribution of n(Yp,n −ξp ).

Solution.

Bahadur’s Representation Bahadur’s representation offers a linear approximation for the sample quantile
Yp,n under regularity conditions. Specifically:

p − Fn (ξp )
Yp,n = ξp + + Rn ,
f (ξp )

where Fn is the empirical distribution function:

n
1X
Fn (x) = 1(Xi ≤ x),
n i=1

and Rn is a remainder term satisfying

√ p
nRn −
→0 (i.e., Rn = op (n−1/2 )).
√
Under stronger conditions, Rn = Op (n−3/4 (log n)3/4 ), so nRn → 0 in probability.

Asymptotic Distribution From Bahadur’s representation:

√
√ n(p − Fn (ξp )) √
n(Yp,n − ξp ) = + nRn .
f (ξp )

Let Zi = 1(Xi ≤ ξp ). Then Z1 , . . . , Zn are i.i.d. Bernoulli with E[Zi ] = p and Var(Zi ) = p(1 − p). Hence,
n
1X √ d
Fn (ξp ) = Zi ⇒ n(Fn (ξp ) − p) −
→ N (0, p(1 − p))
n i=1

by the Central Limit Theorem.

Substituting into the main expression:
√ 1 √ √
n(Yp,n − ξp ) = − n(Fn (ξp ) − p) + nRn .
f (ξp )
√ d √ p
By Slutsky’s Theorem, since n(Fn (ξp ) − p) −
→ N (0, p(1 − p)) and nRn − → 0, we conclude:
√
 
d p(1 − p)
n(Yp,n − ξp ) −
→ N 0, .
[f (ξp )]2
p(1−p)
This demonstrates that Yp,n is asymptotically normal with mean ξp and variance n[f (ξp )]2 .

Question 18. Show that the Lindeberg-Feller condition:

n   
1 X 2 |Xk − µk |
E (Xk − µk ) 1 >ϵ →0 for all ϵ > 0,
s2n sn
k=1

implies both the uniform asymptotic negligibility (UAN) condition:

σk2
max →0 as n → ∞,
1≤k≤n s2
n

10 May 2025 24 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

and asymptotic normality:

P (Zn ≤ z) → Φ(z) for all z ∈ R,
where
n n
Tn − γn X X
Zn = , Tn = Xk , γn = E(Tn ), and s2n = σk2
sn
k=1 k=1
.

Proof that the condition implies UAN. The Lindeberg-Feller condition states that for all ϵ > 0,
n   
1 X 2 |Xk − µk |
E (Xk − µk ) 1 > ϵ →0 as n → ∞.
s2n sn
k=1

For any 1 ≤ k ≤ n, note that

  
2 |Xk − µk | h
2
i
E (Xk − µk ) 1 >ϵ ≤ E (Xk − µk ) = σk2 .
sn

Therefore,
n   n   
1 X 2 σk 1 X 2 |Xk − µk |
· σk 1 >ϵ ≤ 2 · E (Xk − µk ) 1 >ϵ .
s2n sn sn sn
k=1 k=1

Since the right-hand side tends to 0 as n → ∞, it follows that

σk2
max →0 as n → ∞. ■
1≤k≤n s2
n

Proof that the condition implies asymptotic normality. To show asymptotic normality, we use the character-
istic function of Zn . The characteristic function of Zn is given by
n   
Y Xk − µk
ϕZn (t) = E [exp (itZn )] = E exp it .
sn
k=1

Using a second-order Taylor expansion for exp(itx), we have

t2 σk2
    2
Xk − µk σk
E exp it =1− +o .
sn 2 s2n s2n

Taking the product over k = 1, . . . , n, we get

n 
t2 σk2
 2 
Y σk
ϕZn (t) = 1− 2
+o .
2 sn s2n
k=1

Using log(1 + x) ≈ x for small x, we have

n 
t2 σk2
 2 
X σk t2
log ϕZn (t) = − 2
+ o 2
= − + o(1).
2 sn sn 2
k=1

Exponentiating, we get  2
t
ϕZn (t) → exp − ,
2
which is the characteristic function of the standard normal distribution. By the Lévy continuity theorem,
d
Zn −
→ N (0, 1), which implies
P (Zn ≤ z) → Φ(z) for all z ∈ R. ■

10 May 2025 25 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 19. Show that under the uniform asymptotic negligibility (UAN) condition:

σk2
max →0 as n → ∞,
1≤k≤n s2
n

we have
max P (Ynk > ϵ) → 0 for any ϵ > 0,
1≤k≤n

where
n
Xk − µk X
Ynk = , s2n = σk2 , and σk2 = Var(Xk ).
sn
k=1

Proof. Let {Xk }nk=1 be independent random variables with E(Xk ) = µk and Var(Xk ) = σk2 . Define the
standardized random variables
n
Xk − µk X
Ynk = , where s2n = σk2 .
sn
k=1

For any 1 ≤ k ≤ n, by Chebyshev’s inequality, we have

σ2
 
Xk − µk Var(Xk )
P (Ynk > ϵ) = P >ϵ ≤ 2 2
= 2 k2 .
sn ϵ sn ϵ sn

Taking the maximum over 1 ≤ k ≤ n, we get

1 σk2
max P (Ynk > ϵ) ≤ max .
1≤k≤n ϵ2 1≤k≤n s2n

2
σk
Under the UAN condition, max1≤k≤n s2n → 0 as n → ∞. Therefore,

max P (Ynk > ϵ) → 0 as n → ∞. ■

1≤k≤n

10 May 2025 26 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

p
Question 20. Show that Xn −
→ X if and only if Xn are Cauchy in probability, i.e.,
∀ ϵ > 0, lim P (|Xn − Xm | > ϵ) = 0.
n,m→∞

Lemma 20.1. If a sequence of random variables {Xn } is Cauchy in probability, then there exists a subsequence
a.s.
{Xnk } and a random variable X such that Xnk −−→ X as k → ∞.

Proof of Lemma. Since {Xn } is Cauchy in probability, for each integer k ≥ 1, there exists an integer nk such
that for all n, m ≥ nk ,
P |Xn − Xm | > 2−k < 2−k .

We can choose the sequence {nk } to be strictly increasing, i.e., n1 < n2 < . . .. Consider the subsequence
{Xnk }. We have
P Xnk+1 − Xnk > 2−k < 2−k .


P∞ P∞
Let Ek = Xnk+1 − Xnk > 2−k . Then k=1 P(Ek ) < k=1 2−k = 1 < ∞. By the first Borel-Cantelli


lemma, P (lim supk→∞ Ek ) = 0. This means that for almost every ω, there exists K(ω) such that for all
k ≥ K(ω), ω ∈/ Ek , i.e., Xnk+1 (ω) − Xnk (ω) ≤ 2−k . For k ≥ K(ω) and any p ≥ 1,
k+p−1
X k+p−1
X ∞
X
Xnk+p (ω) − Xnk (ω) ≤ Xnj+1 (ω) − Xnj (ω) ≤ 2−j < 2−j = 2−k+1 .
j=k j=k j=k

This shows that for almost every ω, the sequence {Xnk (ω)} is a Cauchy sequence of real numbers. Therefore,
it converges to a limit, say X(ω). Define X(ω) = limk→∞ Xnk (ω) for ω in the set of convergence (which
has probability 1), and X(ω) = 0 otherwise. X is a measurable function, hence a random variable. This
a.s.
establishes almost sure convergence: Xnk −−→ X. ■
p
Proof of ( =⇒ ). Assume Xn −
→ X. By definition, for any ϵ > 0,
P (|Xn − X| > ϵ) → 0 as n → ∞.
For any n, m ≥ N (for sufficiently large N ), we have
P (|Xn − Xm | > ϵ) ≤ P (|Xn − X| > ϵ/2) + P (|Xm − X| > ϵ/2) .
p
Since Xn −
→ X, both terms on the right-hand side tend to 0 as n, m → ∞. Hence,
lim P (|Xn − Xm | > ϵ) = 0.
n,m→∞

Thus, Xn are Cauchy in probability. ■

Proof of ( ⇐= ). Assume Xn are Cauchy in probability. Then, for any ϵ > 0,

lim P (|Xn − Xm | > ϵ) = 0.
n,m→∞

Fix ϵ > 0. Since Xn are Cauchy in probability, there exists N such that for all n, m ≥ N ,
P (|Xn − Xm | > ϵ) < ϵ.
a.s.
By Lemma 20.1, there exists a subsequence {Xnk } and a random variable X such that Xnk −−→ X. Since
p
almost sure convergence implies convergence in probability, we have Xnk −
→ X. Now, for any ϵ > 0, we have
 ϵ  ϵ
P (|Xn − X| > ϵ) ≤ P |Xn − Xnk | > + P |Xnk − X| > .
2 2
p
→ X, we can choose K large enough so that for all k > K, P |Xnk − X| > 2ϵ <


For the second term, since Xnk −
ϵ
2.
Since Xn is Cauchy
in probability, for any fixed n > N , we can choose k large enough such that nk > N and
P |Xn − Xnk | > 2ϵ < 2ϵ .
p
Combining these bounds, we get P (|Xn − X| > ϵ) < ϵ for all n sufficiently large. Thus, Xn −
→ X. ■

10 May 2025 27 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

d
Question 21. (a) Let Fbn −
→ F , where Fbn and F are distribution functions of Xn and X, respectively. If
xn → x, xn ∈ CF , then
Fbn (xn ) → F (x) and Fbn (x− −
n ) → F (x ).

(b) If xn → x, then  
lim Fbn (xn ) − Fbn (x− −
n ) ≤ F (x) − F (x ).
n→∞

(c) If an → 0 and bn → 1, then

d
(Xn + an )bn −
→ X.

d
Proof of (a). Since Fbn −
→ F , we know that Fbn (y) → F (y) for all y ∈ CF , where CF is the set of continuity
points of F .
For the convergence of Fbn (xn ) to F (x), we use the given condition that xn ∈ CF for all n. For any ϵ > 0, by
the continuity of F at each xn , we can find δn > 0 such that |F (y) − F (xn )| < ϵ/2 whenever |y − xn | < δn .
Since xn → x, for sufficiently large n, |xn − x| < min(δn , ϵ/2). Thus:

|F (xn ) − F (x)| < ϵ/2

d
Also, since Fbn −
→ F and xn ∈ CF , for sufficiently large n:

|Fbn (xn ) − F (xn )| < ϵ/2

By the triangle inequality:

|Fbn (xn ) − F (x)| ≤ |Fbn (xn ) − F (xn )| + |F (xn ) − F (x)| < ϵ/2 + ϵ/2 = ϵ

Therefore, Fbn (xn ) → F (x). A similar argument applies for showing Fbn (x− −
n ) → F (x ). ■

Proof of (b). Let us denote ∆n = Fbn (xn ) − Fbn (x− n ), which represents the probability mass at point xn
according to the distribution Fn .
b
For any y < x, since xn → x, there exists N such that for all n ≥ N , y < xn < x + ϵ for any ϵ > 0. Since Fbn
is non-decreasing:
Fbn (y) ≤ Fbn (x−
n ) ≤ Fn (xn ) ≤ Fn (x + ϵ)
b b

Taking the limit as n → ∞ and using the weak convergence property:

F (y) ≤ lim inf Fbn (x−

n ) ≤ lim sup Fn (xn ) ≤ F (x + ϵ)
b
n→∞ n→∞

Since this holds for any y < x and any ϵ > 0:

F (x− ) ≤ lim inf Fbn (x−

n ) ≤ lim sup Fn (xn ) ≤ F (x)
b
n→∞ n→∞

Therefore:
lim sup ∆n = lim sup(Fbn (xn ) − Fbn (x− −
n )) ≤ F (x) − F (x )
n→∞ n→∞

Which establishes the required inequality. ■

Proof of (c). Let Yn = (Xn + an )bn . Since an → 0 and bn → 1, we can write

Yn = Xn bn + an bn .
d
By Slutsky’s theorem, Xn bn −
→ X because bn → 1. Also, an bn → 0 because an → 0 and bn → 1. Therefore,
the sum
d
Yn = Xn bn + an bn −→ X. ■

10 May 2025 28 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

Question 22. Let Xn − Yn = op (1) and Xn = Op (1). Show that for a continuous function g : R → R,

g(Xn ) − g(Yn ) = op (1).

Proof. Since Xn − Yn = op (1), for any ϵ > 0, we have

P (|Xn − Yn | > ϵ) → 0 as n → ∞.

Additionally, Xn = Op (1) implies that Xn is stochastically bounded, i.e.,

∀ ϵ > 0, ∃ M > 0 such that P (|Xn | > M ) < ϵ ∀ n.

From Xn − Yn = op (1) and Xn = Op (1), we can deduce that Yn = Op (1) as well. This is because Yn =
Xn − (Xn − Yn ), and both terms on the right are stochastically bounded.
Now, let g : R → R be a continuous function. We want to show that g(Xn ) − g(Yn ) = op (1).
Given any ϵ > 0, we need to show that P (|g(Xn ) − g(Yn )| > ϵ) → 0 as n → ∞.
Since Xn = Op (1) and Yn = Op (1), for any η > 0, there exists M > 0 such that:
η η
P (|Xn | > M ) < and P (|Yn | > M ) <
2 2
for all n.
Consider the compact interval [−M, M ]. Since g is continuous on R, it is uniformly continuous on [−M, M ].
This means for any ϵ > 0, there exists δ > 0 such that if x, y ∈ [−M, M ] and |x−y| < δ, then |g(x)−g(y)| < ϵ.
Now, consider the probability:
P (|g(Xn ) − g(Yn )| > ϵ)

We can decompose this as:

P (|g(Xn ) − g(Yn )| > ϵ)

≤ P (|g(Xn ) − g(Yn )| > ϵ, |Xn | ≤ M, |Yn | ≤ M ) + P (|Xn | > M or |Yn | > M )
≤ P (|g(Xn ) − g(Yn )| > ϵ, |Xn | ≤ M, |Yn | ≤ M ) + P (|Xn | > M ) + P (|Yn | > M )
< P (|g(Xn ) − g(Yn )| > ϵ, |Xn | ≤ M, |Yn | ≤ M ) + η

By uniform continuity of g on [−M, M ], if |Xn − Yn | < δ and both Xn and Yn are in [−M, M ], then
|g(Xn ) − g(Yn )| < ϵ. Therefore:

P (|g(Xn ) − g(Yn )| > ϵ, |Xn | ≤ M, |Yn | ≤ M )

≤ P (|Xn − Yn | ≥ δ, |Xn | ≤ M, |Yn | ≤ M )
≤ P (|Xn − Yn | ≥ δ)

Since Xn − Yn = op (1), we have P (|Xn − Yn | ≥ δ) → 0 as n → ∞.

Therefore, for any η > 0:
lim sup P (|g(Xn ) − g(Yn )| > ϵ) ≤ η
n→∞

Since η is arbitrary, we conclude that P (|g(Xn ) − g(Yn )| > ϵ) → 0 as n → ∞, which means

g(Xn ) − g(Yn ) = op (1) ■

10 May 2025 29 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

p
Question 23. Prove that Xn −
→ X if and only if
 
|Xn − X|
E →0 as n → ∞.
1 + |Xn − X|
p
Proof of ( =⇒ ). Assume Xn −
→ X. By definition,
 for any ϵ > 0, P (|Xn − X| > ϵ) → 0 as n → ∞. Let
Yn y
Yn = |Xn − X|. We want to show E 1+Yn → 0. Let f (y) = 1+y for y ≥ 0. This function is continuous,
p p
increasing, 0 ̸= f (y) < 1, and f (y) → 0 as y → 0. Since Xn −→ X, we have Yn −
→ 0. Because f is continuous,
p
by the Continuous Mapping Theorem, f (Yn ) − → f (0) = 0. That is,

|Xn − X| p
−
→ 0.
1 + |Xn − X|
|Xn −X|
Let Zn = 1+|Xn −X| . We have 0 ≤ Zn < 1. Since Zn is bounded by the integrable random variable 1, and
p
Zn −
→ 0, by the Dominated Convergence Theorem for convergence in probability (sometimes called Pratt’s
lemma or bounded convergence theorem), we have
 
|Xn − X|
E [Zn ] = E → 0.
1 + |Xn − X|
 
|Xn −X|
Thus, E 1+|X n −X|
→ 0. ■

Proof of ( ⇐= ). Assume  
|Xn − X|
E →0 as n → ∞.
1 + |Xn − X|
Yn
Let Yn = |Xn − X| and Zn = 1+Y n
. We are given E [Zn ] → 0. We want to show that for any ϵ > 0,
y
P (Yn > ϵ) → 0. Consider the event {Yn > ϵ}. On this event, since the function f (y) = 1+y is increasing for
y ≥ 0, we have
Yn ϵ
Zn = > .
1 + Yn 1+ϵ
ϵ
Let δ = 1+ϵ > 0. Then the event {Yn > ϵ} is a subset of the event {Zn > δ}. Therefore,

P (Yn > ϵ) ≤ P (Zn > δ) .

By Markov’s inequality applied to the non-negative random variable Zn ,

E [Zn ]
P (Zn > δ) ≤ .
δ
Combining these, we get
 
E [Zn ] 1+ϵ |Xn − X|
P (|Xn − X| > ϵ) = P (Yn > ϵ) ≤ P (Zn > δ) ≤ = E .
δ ϵ 1 + |Xn − X|

As n → ∞, the right-hand side tends to 1+ϵ

ϵ · 0 = 0, since we assumed E [Zn ] → 0. By the Sandwich Theorem,
since 0 ≤ P (|Xn − X| > ϵ), we must have

lim P (|Xn − X| > ϵ) = 0.

n→∞

p
This holds for any ϵ > 0, which is the definition of Xn −
→ X. ■
Question 24. Show that
p a.s.
(a) Xn −
→ X may not imply Xn −−→ X.

10 May 2025 30 Manas Sharma (MB2412)

Assignment — Indian Statistical Institute, Kolkata
Assignment Solutions M. Stat. — Large Sample Statistical Methods

p Lr
(b) Xn −
→ X may not imply Xn −−→ X for r ≥ 1.

Solution of (a). Consider the probability space (Ω, F, P) = ([0, 1], B ([0, 1]) , λ), where λ is the Lebesgue
measure on the Borel σ-algebra B ([0, 1]).
Define a sequence of intervals Ink for n = 1, 2, . . . and k = 1, . . . , n as Ink = k−1 k
 
n , n . Define a single sequence
of random variables Xm by ordering the indicator functions 1Ink first by n, then by k. That is, the sequence
of indicators is 1I11 , 1I21 , 1I22 , 1I31 , 1I32 , 1I33 , . . .. Let m be the index in this sequence, corresponding to a pair
(n, k). Define Xm (ω) = 1Ink (ω). Let X = 0 (the constant zero random variable).
p
We show Xm −
→ 0. For any ϵ ∈ (0, 1), if Xm = 1Ink , then

k k−1 1
P (|Xm − 0| > ϵ) = P (Xm = 1) = P (ω ∈ Ink ) = λ (Ink ) = − = .
n n n
1 p
As m → ∞, the corresponding n must also go to ∞. Thus, P (|Xm | > ϵ) = n → 0. So, Xm −
→ 0.
However, Xm does not converge almost surely to 0. For any ω ∈ [0, 1], the sequence Xm (ω) contains infinitely
many 1s. Specifically, for each n, there is exactly one k ∈ {1, . . . , n} such that ω ∈ Ink (ignoring endpoints,
which form a set of measure zero). For the index m corresponding to this (n, k), Xm (ω) = 1. Since this
occurs for every n, Xm (ω) = 1 infinitely often.
Therefore, the sequence Xm (ω) does not converge to 0 for any ω ∈ [0, 1]. The set {ω : limm→∞ Xm (ω) = 0}
a.s.
is empty (or has measure zero if endpoints are considered), so its probability is 0 ̸= 1. Hence, Xm −−̸ → 0.

Solution of (b). Consider the same probability space (Ω, F, P) = ([0, 1], B ([0, 1]) , λ).
Define the sequence of random variables Xn for n = 1, 2, . . . as:
(
n if ω ∈ 0, n1
 
Xn (ω) =
0 if ω ∈ n1 , 1


Let X = 0.
p
We show Xn −
→ 0. For any ϵ > 0, choose an integer N such that N > ϵ. Then for all n ≥ N ,
    
1 1 1
P (|Xn − 0| > ϵ) = P (Xn > ϵ) = P (Xn = n) = P ω ∈ 0, = λ 0, = .
n n n
1 p
As n → ∞, P (|Xn | > ϵ) = n → 0. So, Xn −
→ 0.
Now consider convergence in Lr for r ≥ 1. We compute the r-th moment:
Z 1
r r r
E [|Xn − 0| ] = E [|Xn | ] = |Xn (ω)| dλ(ω)
0
Z 1/n Z 1  
1 1
= nr dω + 0r dω = nr · λ 0, + 0 = nr · = nr−1 .
0 1/n n n
Since r ≥ 1, we have r − 1 ≥ 0. If r = 1, E [|Xn |] = n1−1 = n0 = 1. The limit as n → ∞ is 1 ̸= 0. If r > 1,
r
E [|Xn | ] = nr−1 . Since r − 1 > 0, limn→∞ nr−1 = ∞ = ̸ 0.
r Lr
In both cases, limn→∞ E [|Xn | ] ̸= 0. Therefore, Xn −−
̸ → 0 for any r ≥ 1.

10 May 2025 31 Manas Sharma (MB2412)