linear modelling: how can I find variance for vector Y

Question

Linear models are

$$Y_1= 2\theta_1+3\theta_2 +\epsilon_1$$ $$Y_2= -2\theta_1+\theta_2 +\epsilon_2$$

and $\epsilon_1=3z_1-z_2$ and $\epsilon_2=4z_1+z_2$, where $z_1, z_2$ are two random variance such that its independent mean is 0 and variance is $\sigma^2$

I have written this model as $Y=X\beta + \epsilon$

$$Y=\left[\begin{matrix} Y_1 \\ Y_2 \end{matrix}\right]$$ $$X=\left[\begin{matrix} 2&3 \\ -2&1 \end{matrix}\right]$$ $$\beta=\left[\begin{matrix} Q_1 \\ Q_2 \end{matrix}\right]$$ $$\epsilon=\left[\begin{matrix} \epsilon_1 \\ \epsilon_2 \end{matrix}\right]$$

Then,

$$E(Y)=\left[\begin{matrix} E(Y_1) \\ E( Y_2) \end{matrix}\right]=\left[\begin{matrix} 2E(Q_1)+3E(Q_2) \\ -2E(Q_1)+E(Q_2) \end{matrix}\right]=\left[\begin{matrix} 2Q_1+3Q_2 \\ -2Q_1+Q_2 \end{matrix}\right]$$$$

$\epsilon_1=3z_1-z_2$ $(E(z_1)=0, E(z_2)=0)$

$E(\epsilon_1)=0, E(\epsilon_2)=0$

How can I find variance for vector Y?

Answer 1

When discussing the variance of a vector, we generally will construct a covariance matrix, which is a symmetric matrix that contains elements corresponding to the variance of each element in the vector and the covariance between If you have a vector $\begin{eqnarray} Y &=& \left[ \begin{array}{c} Y_1\\ Y_2\\ \end{array} \right] \end{eqnarray}$, then its covariance matrix is written as $cov(Y) = \left[ \begin{array}{cc} cov(Y_1,Y_1) & cov(Y_1,Y_2)\\ cov(Y_2,Y_1) & cov(Y_2,Y_2)\\ \end{array} \right]$.

We can simplify these expressions:

$\begin{eqnarray*} cov(Y_2,Y_1) &=& cov(Y_1,Y_2) \\ cov(Y_1,Y_1) &=& var(Y_1) \\ cov(Y_2,Y_2) &=& var(Y_2) \\ \end{eqnarray*}$

So your covariance matrix would be

$\begin{eqnarray*} cov(Y) &=& \left[ \begin{array}{cc} var(Y_1) & cov(Y_1,Y_2)\\ cov(Y_1,Y_2) & var(Y_2)\\ \end{array} \right] \\ &=& \left[ \begin{array}{cc} var(2\theta_1+3\theta_2+\varepsilon_1) & cov(Y_1,Y_2)\\ cov(Y_1,Y_2) & var(-2\theta_1+\theta_2+\varepsilon_2)\\ \end{array} \right] \\ &=& \left[ \begin{array}{cc} var(2Q_1+3Q_2+\varepsilon_1) & cov(2Q_1+3Q_2+\varepsilon_1,-2\theta_1+\theta_2+\varepsilon_2)\\ cov(2Q_1+3Q_2+\varepsilon_1,-2\theta_1+\theta_2+\varepsilon_2) & var(-2\theta_1+\theta_2+\varepsilon_2)\\ \end{array} \right] \end{eqnarray*}$

At this point, you can use the properties of covariance and variance to evaluate each element of the covariance matrix separately.