Part 3

Let $ X $ be an exponential random variable with parameter $ \lambda $, so that $ f_X(x)=\lambda{exp}(-\lambda{x})u(x) $. Find the variance of $ X $. You must show all of your work.

Solution 1

Suppose

$ A=\left(\begin{array}{cc} a & b\\ c & d \end{array} \right) $.

Then the new 2-D random vector can be expressed as

$ Y=\left(\begin{array}{c}Y_1 \\ Y_2\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}aX_i+bX_j \\ cX_i+dX_j\end{array} \right) $

Therefore,

$ \begin{array}{l}Cov(Y_1,Y_2)=E[(aX_i+bX_j-E(aX_i+bX_j))(cX_i+dX_j-E(cX_i+dX_j))] \\ =E[(aX_i+bX_j-aE(X_i)-bE(X_j))(cX_i+dX_j-cE(X_i)-dE(X_j))] \\ =E[acX_i^2+adX_iX_j-acX_iE(X_i)-adX_iE(X_j)+bcX_iX_j+bdX_j^2-bcX_jE(X_i)\\ -bdX_jE(X_j)-acX_iE(X_i)-adX_jE(X_i)+acE(X_i)^2+adE(X_i)E(X_j)\\ -bcX_iE(X_j)-bdX_jE(X_j)+bcE(X_i)E(X_j)+bdE(X_i)^2]\\ =E(ac(X_i-E(X_i))^2+(ad+bc)(X_i-E(X_i)(X_j-E(X_j))+bd(X_j-E(X_j))^2]\\ =(ac)Cov(X_i,X_i)+(ad+bc)Cov(X-i,X_j)+(bd)Cov(X_j,X_j)\\ =ac\sigma^2+(ad+bc)\rho\sigma^2+bd\sigma^2 \end{array} $

Let the above formula equal to 0 and $ a=b=d=1 $, we get $ c=-1 $.

Therefore, a solution is

$ A=\left(\begin{array}{cc} 1 & 1\\ -1 & 1 \end{array} \right) $.

Solution 2

Assume

$ Y=\left(\begin{array}{c}Y_i \\ Y_j\end{array} \right)=A\left(\begin{array}{c}X_i \\ X_j\end{array} \right)=\left(\begin{array}{c}a_{11}X_i+a_{12}X_j \\ a_{21}X_i+a_{22}X_j\end{array} \right) $.

Then

$ \begin{array}{l}E(Y_iY_j)=E[(a_{11}X_i+a_{12}X_j)(a_{21}X_i+a_{22}X_j)]\\ =a_{11}a_{21}\sigma^2+a_{12}a_{22}\sigma^2+(a_{11}a_{21}+a_{22}a_{11})E(X_iX_j) \end{array} $

For $ |i-j|\geq1 $, $ E(X_i,X_j)=0 $. Therefore, $ a_{11}a_{21}+a_{12}a_{22}=0 $.

One solution can be

$ A=\left(\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array} \right) $.

Critique on Solution 2:

1. $ E(Y_iY_j)=0 $ is not the condition for the two random variables to be independent.

2. "For $ |i-j|\geq1 $, $ E(X_i,X_j)=0 $" is not supported by the given conditions.

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