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a. Use the general formula for the variance of a linear combination to write an expression for\({\rm{V(aX + Y)}}\). Then let\({\rm{a = }}{{\rm{\sigma }}_{\rm{\gamma }}}{\rm{/}}{{\rm{\sigma }}_{\rm{X}}}\), and show that. (Hint: Variance is always, and\({\rm{Cov(X,Y) = }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ \times }}{{\rm{\sigma }}_{\rm{Y}}}{\rm{ \times \rho }}\).)

b. By considering\({\rm{V(aX - Y)}}\), conclude that\({\rm{\rho £ 1}}\).

c. Use the fact that \({\rm{V(W) = 0}}\)only if \({\rm{W}}\)is a constant to show that \({\rm{\rho = 1}}\)only if\({\rm{Y = aX + b}}\).

Short Answer

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Step by step solution

01

Definition

A linear combination is an expression made up of a series of terms multiplied by a constant and then added together. Linear combinations are a fundamental idea in linear algebra and related branches of mathematics.

02

Step 2: Write an expression for \({\rm{V(aX + Y)}}\)

a)

The following holds

\(\begin{array}{c}{\rm{V(aX + Y) = }}{{\rm{a}}^{\rm{2}}}{\rm{V(X) + V(Y) + 2aCov(X,Y)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ + 2a}}{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}{\rm{\rho }}\end{array}\)

(1): see the hint in the exercise.

When

\({\rm{a = }}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}\)

the following is true

\(\begin{array}{c}{\rm{0£ V(aX + Y) = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ + 2a}}{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}{\rm{\rho }}\\{\rm{ = }}{\left( {\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}} \right)^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ + 2}}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}{\rm{\rho }}\\{\rm{ = \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ + 2\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{\rho }}\\{\rm{ = 2\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{(1 + \rho )}}\end{array}\)

hence,

\[{{\text{2\sigma }}_{\text{Y}}^{\text{2}}{\text{(1 + \rho )0}}}\]

\[{{\text{\sigma }}_{\text{Y}}^{\text{2}}{\text{0}}}\]

implies that

\[{\text{1 + \rho 0}}\]

or equally

\(\rho \ge - 1\)

03

Calculating \({\rm{\rho £ 1}}\)

(b):

The following holds

\(\begin{array}{c}{\rm{V(aX - Y) = }}{{\rm{a}}^{\rm{2}}}{\rm{V(X) + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V(Y) - 2aCov(X,Y)}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ - 2a}}{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}{\rm{\rho }}\end{array}\)

(1): see the hint in the exercise.

When

\({\rm{a = }}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}\)

the following is true

\(\begin{array}{c}{\rm{0£ V(aX - Y) = }}{{\rm{a}}^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ - 2a}}{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}{\rm{\rho }}\\{\rm{ = }}{\left( {\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}} \right)^{\rm{2}}}{\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ - 2}}\frac{{{{\rm{\sigma }}_{\rm{Y}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}{{\rm{\sigma }}_{\rm{X}}}{{\rm{\sigma }}_{\rm{Y}}}{\rm{\rho }}\\{\rm{ = \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ + \sigma }}_{\rm{Y}}^{\rm{2}}{\rm{ - 2\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{\rho }}\\{\rm{ = 2\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{(1 - \rho )}}\end{array}\)

hence,

\(\begin{array}{*{20}{r}}{2\sigma _Y^2(1 - \rho ) \ge 0}\\{\sigma _Y^2 \ge 0}\end{array}\)

implies that

\(1 - \rho \ge 0\)

or equally

\(\rho \le 1.\)

04

Showing \({\rm{\rho  = 1}}\)

(c):

Assume that\({\rm{\rho = 1}}\), then from

\(\begin{array}{c}{\rm{V(aX - Y) = 2}}\\{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{(1 - \rho ) = 2}}\\{\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{(1 - 1) = 0}}\end{array}\)

for any random variables \({\rm{X}}\)and\({\rm{Y}}\). From implication

\[V(W) = 0W = b,\quad b - {\text{ constant }}\]

the following holds

\[V(aX - Y) = 0aX - Y = b,\quad b - {\text{ constant }}\]

or equally

\({\rm{Y = aX + b}}\)

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