IGNOU MST-018 Solved Assignment | June 2024 to June 2025 | Multivariate Analysis | MSCAST

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Sample Solution

IGNOU MST-018 Solved Assignment | June 2024 to June 2025 | Multivariate Analysis | MSCAST Sample Solution
  1. 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 X XXX and Y Y YYY is symmetric.
Answer:
The statement is true. Let’s provide a justification for this:

Covariance Matrix Definition

The covariance matrix of two random vectors X X XXX and Y Y YYY is defined as:
Σ = [ Cov ( X , X ) Cov ( X , Y ) Cov ( Y , X ) Cov ( Y , Y ) ] Σ = Cov ( X , X ) Cov ( X , Y ) Cov ( Y , X ) Cov ( Y , Y ) Sigma=[[“Cov”(X”,”X),”Cov”(X”,”Y)],[“Cov”(Y”,”X),”Cov”(Y”,”Y)]]\Sigma = \begin{bmatrix} \text{Cov}(X, X) & \text{Cov}(X, Y) \\ \text{Cov}(Y, X) & \text{Cov}(Y, Y) \end{bmatrix}Σ=[Cov(X,X)Cov(X,Y)Cov(Y,X)Cov(Y,Y)]
Here, Cov ( X , X ) Cov ( X , X ) “Cov”(X,X)\text{Cov}(X, X)Cov(X,X) is the variance of X X XXX, Cov ( Y , Y ) Cov ( Y , Y ) “Cov”(Y,Y)\text{Cov}(Y, Y)Cov(Y,Y) is the variance of Y Y YYY, Cov ( X , Y ) Cov ( X , Y ) “Cov”(X,Y)\text{Cov}(X, Y)Cov(X,Y) is the covariance between X X XXX and Y Y YYY, and Cov ( Y , X ) Cov ( Y , X ) “Cov”(Y,X)\text{Cov}(Y, X)Cov(Y,X) is the covariance between Y Y YYY and X X XXX.

Symmetry of Covariance

To show that the covariance matrix is symmetric, we need to prove that Cov ( X , Y ) = Cov ( Y , X ) Cov ( X , Y ) = Cov ( Y , X ) “Cov”(X,Y)=”Cov”(Y,X)\text{Cov}(X, Y) = \text{Cov}(Y, X)Cov(X,Y)=Cov(Y,X).

Definition of Covariance

The covariance between two random variables X X XXX and Y Y YYY is defined as:
Cov ( X , Y ) = E [ ( X − E [ X ] ) ( Y − E [ Y ] ) ] Cov ( X , Y ) = E [ ( X − E [ X ] ) ( Y − E [ Y ] ) ] “Cov”(X,Y)=E[(X-E[X])(Y-E[Y])]\text{Cov}(X, Y) = \mathbb{E}[(X – \mathbb{E}[X])(Y – \mathbb{E}[Y])]Cov(X,Y)=E[(X−E[X])(Y−E[Y])]
Similarly, the covariance between Y Y YYY and X X XXX is:
Cov ( Y , X ) = E [ ( Y − E [ Y ] ) ( X − E [ X ] ) ] Cov ( Y , X ) = E [ ( Y − E [ Y ] ) ( X − E [ X ] ) ] “Cov”(Y,X)=E[(Y-E[Y])(X-E[X])]\text{Cov}(Y, X) = \mathbb{E}[(Y – \mathbb{E}[Y])(X – \mathbb{E}[X])]Cov(Y,X)=E[(Y−E[Y])(X−E[X])]

Proof of Symmetry

We can see that:
Cov ( X , Y ) = E [ ( X − E [ X ] ) ( Y − E [ Y ] ) ] Cov ( X , Y ) = E [ ( X − E [ X ] ) ( Y − E [ Y ] ) ] “Cov”(X,Y)=E[(X-E[X])(Y-E[Y])]\text{Cov}(X, Y) = \mathbb{E}[(X – \mathbb{E}[X])(Y – \mathbb{E}[Y])]Cov(X,Y)=E[(X−E[X])(Y−E[Y])]
Since multiplication is commutative, we have:
E [ ( X − E [ X ] ) ( Y − E [ Y ] ) ] = E [ ( Y − E [ Y ] ) ( X − E [ X ] ) ] E [ ( X − E [ X ] ) ( Y − E [ Y ] ) ] = E [ ( Y − E [ Y ] ) ( X − E [ X ] ) ] E[(X-E[X])(Y-E[Y])]=E[(Y-E[Y])(X-E[X])]\mathbb{E}[(X – \mathbb{E}[X])(Y – \mathbb{E}[Y])] = \mathbb{E}[(Y – \mathbb{E}[Y])(X – \mathbb{E}[X])]E[(X−E[X])(Y−E[Y])]=E[(Y−E[Y])(X−E[X])]
Thus:
Cov ( X , Y ) = Cov ( Y , X ) Cov ( X , Y ) = Cov ( Y , X ) “Cov”(X,Y)=”Cov”(Y,X)\text{Cov}(X, Y) = \text{Cov}(Y, X)Cov(X,Y)=Cov(Y,X)

Conclusion

Since Cov ( X , Y ) = Cov ( Y , X ) Cov ( X , Y ) = Cov ( Y , X ) “Cov”(X,Y)=”Cov”(Y,X)\text{Cov}(X, Y) = \text{Cov}(Y, X)Cov(X,Y)=Cov(Y,X), the off-diagonal elements of the covariance matrix are equal, making the covariance matrix symmetric.

Symmetric Covariance Matrix

Therefore, the covariance matrix Σ Σ Sigma\SigmaΣ is symmetric:
Σ = [ Cov ( X , X ) Cov ( X , Y ) Cov ( Y , X ) Cov ( Y , Y ) ] = [ Var ( X ) Cov ( X , Y ) Cov ( X , Y ) Var ( Y ) ] Σ = Cov ( X , X ) Cov ( X , Y ) Cov ( Y , X ) Cov ( Y , Y ) = Var ( X ) Cov ( X , Y ) Cov ( X , Y ) Var ( Y ) Sigma=[[“Cov”(X”,”X),”Cov”(X”,”Y)],[“Cov”(Y”,”X),”Cov”(Y”,”Y)]]=[[“Var”(X),”Cov”(X”,”Y)],[“Cov”(X”,”Y),”Var”(Y)]]\Sigma = \begin{bmatrix} \text{Cov}(X, X) & \text{Cov}(X, Y) \\ \text{Cov}(Y, X) & \text{Cov}(Y, Y) \end{bmatrix} = \begin{bmatrix} \text{Var}(X) & \text{Cov}(X, Y) \\ \text{Cov}(X, Y) & \text{Var}(Y) \end{bmatrix}Σ=[Cov(X,X)Cov(X,Y)Cov(Y,X)Cov(Y,Y)]=[Var(X)Cov(X,Y)Cov(X,Y)Var(Y)]

Conclusion

The statement "The covariance matrix of random vectors X X XXX and Y Y YYY is symmetric" is true, as demonstrated by the equality Cov ( X , Y ) = Cov ( Y , X ) Cov ( X , Y ) = Cov ( Y , X ) “Cov”(X,Y)=”Cov”(Y,X)\text{Cov}(X, Y) = \text{Cov}(Y, X)Cov(X,Y)=Cov(Y,X).
(b) If X X XXX is a p p ppp-variate normal random vector, then every linear combination c ′ X c ′ X c^(‘)Xc^{\prime} Xc′X, where c p × 1 c p × 1 c_(pxx1)\mathrm{c}_{\mathrm{p} \times 1}cp×1 is a scalar vector, is also p p p\mathrm{p}p-variate normal vector.
Answer:
The statement is false. Let’s analyze and justify this.

Understanding the Statement

The statement says:
"If X X XXX is a p p ppp-variate normal random vector, then every linear combination c ′ X c ′ X c^(‘)Xc^{\prime} Xc′X, where c c ccc is a p × 1 p × 1 p xx1p \times 1p×1 scalar vector, is also a p p ppp-variate normal vector."

Breaking Down the Components

  1. p p ppp-variate Normal Random Vector: X X XXX is a random vector with a multivariate normal distribution in p p ppp-dimensional space.
  2. Linear Combination c ′ X c ′ X c^(‘)Xc^{\prime}Xc′X: c ′ X c ′ X c^(‘)Xc^{\prime}Xc′X represents a linear combination of the elements of X X XXX, where c c ccc is a vector of coefficients.

Correct Interpretation

Let’s rephrase the correct version of the statement for clarity:
  • If X X XXX is a p p ppp-variate normal random vector, then every linear combination c ′ X c ′ X c^(‘)Xc^{\prime} Xc′X, where c c ccc is a p × 1 p × 1 p xx1p \times 1p×1 vector, is also normally distributed (not necessarily a p p ppp-variate normal vector, but a univariate normal distribution).

Proof of Correct Interpretation

If X X XXX is a p p ppp-variate normal vector, then:
X ∼ N p ( μ , Σ ) X ∼ N p ( μ , Σ ) X∼N_(p)(mu,Sigma)X \sim N_p(\mu, \Sigma)X∼Np(μ,Σ)
where μ μ mu\muμ is the mean vector and Σ Σ Sigma\SigmaΣ is the covariance matrix.
Now consider a linear combination c ′ X c ′ X c^(‘)Xc^{\prime}Xc′X:
Y = c ′ X Y = c ′ X Y=c^(‘)XY = c^{\prime}XY=c′X
We need to show that Y Y YYY is normally distributed.

Mean of Y Y YYY

The mean of Y Y YYY is given by:
E [ Y ] = E [ c ′ X ] = c ′ E [ X ] = c ′ μ E [ Y ] = E [ c ′ X ] = c ′ E [ X ] = c ′ μ E[Y]=E[c^(‘)X]=c^(‘)E[X]=c^(‘)mu\mathbb{E}[Y] = \mathbb{E}[c^{\prime}X] = c^{\prime}\mathbb{E}[X] = c^{\prime}\muE[Y]=E[c′X]=c′E[X]=c′μ

Variance of Y Y YYY

The variance of Y Y YYY is given by:
Var ( Y ) = Var ( c ′ X ) = c ′ Σ c Var ( Y ) = Var ( c ′ X ) = c ′ Σ c “Var”(Y)=”Var”(c^(‘)X)=c^(‘)Sigma c\text{Var}(Y) = \text{Var}(c^{\prime}X) = c^{\prime}\Sigma cVar(Y)=Var(c′X)=c′Σc

Distribution of Y Y YYY

Since X X XXX is multivariate normal, any linear combination of its components is also normally distributed. Therefore, Y = c ′ X Y = c ′ X Y=c^(‘)XY = c^{\prime}XY=c′X is normally distributed with mean c ′ μ c ′ μ c^(‘)muc^{\prime}\muc′μ and variance c ′ Σ c c ′ Σ c c^(‘)Sigma cc^{\prime}\Sigma cc′Σc:
Y ∼ N ( c ′ μ , c ′ Σ c ) Y ∼ N ( c ′ μ , c ′ Σ c ) Y∼N(c^(‘)mu,c^(‘)Sigma c)Y \sim N(c^{\prime}\mu, c^{\prime}\Sigma c)Y∼N(c′μ,c′Σc)

Conclusion

While Y = c ′ X Y = c ′ X Y=c^(‘)XY = c^{\prime}XY=c′X is normally distributed, it is not a p p ppp-variate normal vector. It is a univariate normal random variable. Hence, the original statement is false.

Correct Statement

The correct statement should be:
  • "If X X XXX is a p p ppp-variate normal random vector, then every linear combination c ′ X c ′ X c^(‘)Xc^{\prime} Xc′X, where c c ccc is a p × 1 p × 1 p xx1p \times 1p×1 vector, is also normally distributed (a univariate normal random variable)."
This highlights the distinction between being normally distributed in a univariate sense and being a p p ppp-variate normal vector.
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