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1. Let

\boldsymbol{X}

N(\boldsymbol{\mu}, \boldsymbol{\Sigma})

with $\bo...

Sep 24, 2024

Solution by Steps

step 1

To find the distribution of

3 X_{1}-2 X_{2}+X_{3}

, we first note that if

X \sim N(\mu, \Sigma)

, then any linear combination of the components of

X

is also normally distributed. The mean of the linear combination can be calculated as:

\[ E[3X_1 - 2X_2 + X_3] = 3\mu_1 - 2\mu_2 + \mu_3 \]

step 2

Substituting the values of

\mu_1

\mu_2

, and

\mu_3

from

\boldsymbol{\mu} = (2, -3, 1)

, we get:

\[ E[3X_1 - 2X_2 + X_3] = 3(2) - 2(-3) + 1 = 6 + 6 + 1 = 13 \]

step 3

Next, we calculate the variance of the linear combination using the covariance matrix

\Sigma

\[ \text{Var}(3X_1 - 2X_2 + X_3) = \begin{pmatrix} 3 & -2 & 1 \end{pmatrix} \Sigma \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \]

step 4

Calculating this gives us:

\[ \text{Var}(3X_1 - 2X_2 + X_3) = \begin{pmatrix} 3 & -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 2 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \]

step 5

Performing the matrix multiplication results in:

\[ = 3(1)(3) + 3(-2)(1) + 3(1)(1) + (-2)(1)(3) + (-2)(3)(-2) + (-2)(2)(1) + (1)(1)(3) + (1)(2)(-2) + (1)(2)(1) \]

step 6

After calculating, we find:

\[ \text{Var}(3X_1 - 2X_2 + X_3) = 9 + 6 + 3 + 6 + 12 + 4 + 3 - 4 + 2 = 37 \]

step 7

Therefore, the distribution of

3X_1 - 2X_2 + X_3

is:

\[ 3X_1 - 2X_2 + X_3 \sim N(13, 37) \]

step 8

For part (b), we need to find a

2 \times 1

vector

\boldsymbol{a}

such that

X_2

and

\boldsymbol{a}^{T}\binom{X_1}{X_3}

are independent. This can be achieved if the covariance between

X_2

and the linear combination is zero

step 9

The covariance can be expressed as:

\[ \text{Cov}(X_2, \boldsymbol{a}^{T}\binom{X_1}{X_3}) = \boldsymbol{a}^{T} \begin{pmatrix} \text{Cov}(X_2, X_1) \\ \text{Cov}(X_2, X_3) \end{pmatrix} \]

step 10

Setting this equal to zero gives us the equations to solve for

\boldsymbol{a}

. From the covariance matrix

\Sigma

, we find:

\[ \text{Cov}(X_2, X_1) = 1, \quad \text{Cov}(X_2, X_3) = 2 \]

step 11

Thus, we need to solve:

\[ a_1(1) + a_2(2) = 0 \]

step 12

A possible solution is to set

a_1 = 2

and

a_2 = -1

, giving us:

\[ \boldsymbol{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} \]

Answer

(a)

3X_1 - 2X_2 + X_3 \sim N(13, 37)

; (b)

\boldsymbol{a} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}

Key Concept

Linear combinations of normally distributed variables are also normally distributed. Independence can be achieved through zero covariance.

Explanation

The mean and variance of the linear combination were calculated using properties of the normal distribution, and the independence condition was derived from the covariance structure.