IGNOU MST-018 Solved Assignment 2024 MSCAST Program

IGNOU MST-018 Solved Assignment 2024 | MSCAST | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MST-018 Assignment Question Paper 2024

mst-018-solved-assignment-2024-qp-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

State whether the following statements are true or false and also give the reason in support of your answer:

(a) The covariance matrix of random vectors

\underline{X}

and

\underline{Y}

is symmetric.
(b) If

X

is a

p

-variate normal random vector, then every linear combination

c^{'} X

, where

c_{p \times 1}

is a scalar vector, is also

p

-variate normal vector.
(c) The trace of matrix

(\begin{array}{cc} 3 & - 2 \\ - 2 & 6 \end{array})

is 9 .
(d) If a matrix is positive definite then its inverse is also positive definite.
(e) If

X \sim N_{2} ((\begin{array}{l} 2 \\ 1 \end{array}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}))

and

Y \sim N_{2} ((\begin{matrix} - 1 \\ 3 \end{matrix}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}))

, then

X + \underset{\sim}{Y} - N_{2} ((\begin{array}{l} 1 \\ 1 \end{array}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array})) .

2 (a) Let

X = (\begin{array}{l} X_{1} \\ X_{2} \end{array})

has the following joint density function

f (x_{1}, x_{2}) = {\begin{array}{lr} 4 x_{1} x_{2} & , 0 < x_{1} < 1, 0 < x_{2} < 1, \\ 0 & , otherwise. \end{array}

Find the marginal distributions, mean vector and variance-covariance matrix. Also, comment on the independence of

X_{1}

and

X_{2}

E ({\underline{X}}^{(2)} ∣ {\underline{X}}^{(1)} = {\underline{x}}^{(1)}) and Cov ({\underline{X}}^{(2)} ∣ {\underline{X}}^{(1)} = {\underline{x}}^{(1)}) .

3 (a) Let

X

be a 3-dimensional random vector with dispersion matrix

Σ = (\begin{array}{ccc} 4 & - 2 & 0 \\ - 2 & 4 & 0 \\ 0 & 0 & 2 \end{array})

Determine the first principal component and the proportion of the total variability that it explains.

(b) Let

X \sim N_{4} (μ, Σ)

, where

\underset{\sim}{μ} = (\begin{matrix} 3 \\ - 2 \\ 1 \\ - 2 \end{matrix})

and

Σ = (\begin{array}{cccc} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & - 2 \\ 0 & 0 & - 2 & 5 \end{array})

. Check the independence of the (i)

X_{2}

and

X_{1}

(ii)

(X_{2}, X_{4})

and

(X_{1}, X_{3})

(iii)

(X_{1}, X_{2})

and

(X_{3}, X_{4})

4 (a) Consider the following data of 11 samples on 8 variables :

$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$	$y_{1}$	$y_{2}$	$y_{3}$	$y_{4}$
10	10	10	8	8.04	9.14	7.46	6.58
8	8	8	8	6.95	8.14	6.77	5.76
13	13	13	8	7.58	8.74	12.74	7.71
9	9	9	8	8.81	8.77	7.11	8.84
11	11	11	8	8.33	9.26	7.81	8.47
14	14	14	8	9.96	8.10	8.84	7.04
6	6	6	8	7.24	6.13	6.08	5.25
4	4	4	19	4.26	3.10	5.39	12.50
12	12	12	8	10.84	9.13	8.15	5.56
7	7	7	8	4.82	7.26	6.42	7.91
5	5	5	8	5.68	4.74	5.73	6.89

If the vector

\underline{x} = (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array})

and

\underline{y} = (\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{array})

, then obtain the sample covariance matrix between

x

and

y

(b) Obtain the maximum likelihood estimator of the mean vector and variance-covariance matrix of the multivariate normal distribution.

5 (a) Define the following:
(i) Covariance Matrix
(ii) Mahalanobis

D^{2}

(iii) Hotelling’s

T^{2}

(iv) Clustering
(v) Relationship between (ii) and (iii).
(b) If

X \sim N_{3} (\underset{\sim}{μ}, Σ)

with

\underset{\sim}{μ} = (\begin{array}{l} 2 \\ 1 \\ 2 \end{array})

and

Σ = (\begin{array}{ccc} 5 & 3 & 0 \\ 3 & 3 & - 2 \\ 0 & - 2 & 5 \end{array})

. Then find the joint distribution of

X_{1} + 2 X_{2}, 2 X_{1} - X_{2}

and

X_{3}

MST-018 Sample Solution 2024

mst-018-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

State whether the following statements are true or false and also give the reason in support of your answer:

(a) The covariance matrix of random vectors

X

and

Y

is symmetric.

Answer:

The statement "The covariance matrix of random vectors

X

and

Y

is symmetric" is true. This can be justified as follows:

The covariance matrix of two random vectors

X

and

Y

, denoted as

Cov (X, Y)

, is defined as:

Cov (X, Y) = E [(X - E [X]) (Y - E [Y])^{⊤}]

Where

E [\cdot]

denotes the expectation, and

⊤

denotes the transpose of a vector. The

(i, j)

-th element of this matrix is the covariance between the

i

-th element of

X

and the

j

-th element of

Y

, which is:

Cov (X_{i}, Y_{j}) = E [(X_{i} - E [X_{i}]) (Y_{j} - E [Y_{j}])]

Now, let’s consider the covariance between the

j

-th element of

Y

and the

i

-th element of

X

Cov (Y_{j}, X_{i}) = E [(Y_{j} - E [Y_{j}]) (X_{i} - E [X_{i}])]

Since expectation is a linear operator and multiplication is commutative, we have:

Cov (Y_{j}, X_{i}) = E [(X_{i} - E [X_{i}]) (Y_{j} - E [Y_{j}])] = Cov (X_{i}, Y_{j})

This implies that the

(j, i)

-th element of the covariance matrix

Cov (Y, X)

is equal to the

(i, j)

-th element of the covariance matrix

Cov (X, Y)

. Therefore, the covariance matrix

Cov (X, Y)

is symmetric.

(b) If

X

is a

p

-variate normal random vector, then every linear combination

c^{'} X

, where

c_{p \times 1}

is a scalar vector, is also

p

-variate normal vector.

Answer:

The statement "If

X

is a

p

-variate normal random vector, then every linear combination

c^{'} X

, where

c_{p \times 1}

is a scalar vector, is also

p

-variate normal vector" is false. The correct statement should be that every linear combination

c^{'} X

is a univariate normal random variable, not a

p

-variate normal vector. Here’s the justification:

X

is a

p

-variate normal random vector, then any linear combination of its elements, say

c^{'} X

, where

c

is a

p \times 1

vector, is a univariate normal random variable. This is because the linear combination of normally distributed variables is also normally distributed.

However, the result

c^{'} X

is a scalar (a single number), not a vector. Therefore, it is not correct to say that

c^{'} X

is a

p

-variate normal vector. It is a univariate (single-variable) normal random variable.

(\begin{array}{cc} 3 & - 2 \\ - 2 & 6 \end{array})

is 9 .

Answer:

The statement "The trace of matrix

(\begin{array}{cc} 3 & - 2 \\ - 2 & 6 \end{array})

is 9" is true. The trace of a matrix is defined as the sum of its diagonal elements. For the given matrix

(\begin{array}{cc} 3 & - 2 \\ - 2 & 6 \end{array})

, the diagonal elements are 3 and 6. Therefore, the trace of this matrix is:

Trace = 3 + 6 = 9

Hence, the statement is true.

(d) If a matrix is positive definite then its inverse is also positive definite.

Answer:

The statement "If a matrix is positive definite then its inverse is also positive definite" is true. Here’s the justification:

A matrix

A

is positive definite if, for any nonzero vector

x

, the quadratic form

x^{⊤} A x > 0

Now, suppose

A

is positive definite and has an inverse

A^{- 1}

. We want to show that

A^{- 1}

is also positive definite.

Let

y

be any nonzero vector. Then, there exists a nonzero vector

x

such that

y = A x

. This is because

A

is invertible and, therefore, maps nonzero vectors to nonzero vectors.

Now, consider the quadratic form involving

A^{- 1}

y^{⊤} A^{- 1} y = (A x)^{⊤} A^{- 1} (A x) = x^{⊤} (A^{⊤} A^{- 1} A) x = x^{⊤} A x > 0

The last inequality follows from the positive definiteness of

A

. Since

y

was an arbitrary nonzero vector, this shows that

A^{- 1}

is positive definite.

(e) If

X \sim N_{2} ((\begin{array}{l} 2 \\ 1 \end{array}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}))

and

Y \sim N_{2} ((\begin{matrix} - 1 \\ 3 \end{matrix}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}))

, then

\underline{X} + \underline{Y} \sim N_{2} ((\begin{array}{l} 1 \\ 1 \end{array}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array})) .

Answer:

The statement is false. The correct statement should be that the sum of the two normally distributed vectors

X

and

Y

has a mean vector that is the sum of the individual mean vectors, and a covariance matrix that is the sum of the individual covariance matrices.

Given:

$X \sim N_{2} ((\begin{array}{l} 2 \\ 1 \end{array}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}))$
$Y \sim N_{2} ((\begin{matrix} - 1 \\ 3 \end{matrix}), (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}))$

The mean vector of the sum

X + Y

is the sum of the individual mean vectors:

(\begin{array}{l} 2 \\ 1 \end{array}) + (\begin{matrix} - 1 \\ 3 \end{matrix}) = (\begin{array}{l} 2 - 1 \\ 1 + 3 \end{array}) = (\begin{array}{l} 1 \\ 4 \end{array})

The covariance matrix of the sum

X + Y

is the sum of the individual covariance matrices (assuming

X

and

Y

are independent):

(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}) + (\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}) = (\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array})

Therefore, the correct distribution of

X + Y

should be:

X + Y \sim N_{2} ((\begin{array}{l} 1 \\ 4 \end{array}), (\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}))

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IGNOU MST-018 Solved Assignment 2024 | MSCAST | IGNOU

IGNOU MST-018 Assignment Question Paper 2024

mst-018-solved-assignment-2024-qp-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

MST-018 Sample Solution 2024

mst-018-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

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IGNOU MST-018 Solved Assignment 2024 | MSCAST | IGNOU

IGNOU MST-018 Assignment Question Paper 2024

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MST-018 Sample Solution 2024

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