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Chapter 4: Q 5E (page 207)

For all random variables\({\bf{X}}\)and\({\bf{Y}}\)and all constants\({\bf{a}}\),\({\bf{b}}\),\({\bf{c}}\), and\({\bf{d}}\), show that\({\bf{Cov}}\left( {{\bf{aX + b,cY + d}}} \right){\bf{ = acCov}}\left( {{\bf{X,Y}}} \right)\).

Short Answer

Expert verified

For all random variables\(X\)and\(Y\)all constants ,

\(Cov\left( {aX + b,cY + d} \right) = acCov\left( {X,Y} \right)\).

Step by step solution

01

Given information

X and Y are both random variables and a, b, c, dare constants.

02

Compute the covariance 

The expectation of\(aX + b\) is

\(E\left( {aX + b} \right) = a{\mu _X} + b\)

The expectation of\(cY + d\)is

\(E\left( {cY + d} \right) = c{\mu _Y} + d\)

Referring to Definition 4.6.1 for the below statement.

The covariance of\(aX + b\)and\(cY + d\)is

\(\begin{align}Cov\left( {aX + b,cY + d} \right) &= E\left( {\left( {aX + b - a\mu x - b} \right)\left( {cY + d - c\mu y - d} \right)} \right)\\ &= E\left( {ac\left( {X - \mu x} \right)\left( {Y - \mu y} \right)} \right)\\ &= acCov\left( {X,Y} \right)\end{align}\)

Hence, for all random variables\(X\)and\(Y\)all constants ,

\(Cov\left( {aX + b,cY + d} \right) = acCov\left( {X,Y} \right)\).

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