Question

Suppose that \({\bf{X}}\) and \({\bf{Y}}\) are independent \({\bf{n - }}\)dimensional random vectors for which the covariance matrices \({\bf{Cov(X)}}\) and \({\bf{Cov(Y)}}\) exist . Show that \({\bf{Cov(X + Y) = Cov(X) + Cov(Y)}}\).

Short answer

By using the properties for random variables we proved that \(Cov(X + Y) = Cov(X) + Cov(Y)\).

Step-by-step solution

Define linear statistical models

A linear model describes the correlation between the dependent and independent variables as a straight line.

\(y = {a_0} + {a_1}{x_1} + {a_2}{x_2} + \tilde A,\hat A1/4 + {a_n}{x_n}\)

Models using only one predictor are simple linear regression models. Multiple predictors are used in multiple linear regression models. For many response variables, multiple regression analysis models are used.

Find the proof of the given statement

\(X\)and \(Y\) are \(n - \)dimensional random vectors and are independent.

Let,

\(\begin{array}{*{20}{c}}{X = \left[ {\begin{array}{*{20}{c}}{{X_1}}\\ \cdots \\{{X_n}}\end{array}} \right]}\\{Y = \left[ {\begin{array}{*{20}{c}}{{Y_1}}\\ \cdots \\{{Y_n}}\end{array}} \right]}\end{array}\)

For random variables we know that \(Cov(X + Y) = Cov(X) + Cov(Y)\).

\(\begin{array}{*{20}{c}}{{\mathop{\rm cov}} \left( {{X_i},{Y_j}} \right) = 0}\\{{\mathop{\rm cov}} \left( {{Y_i},{X_j}} \right) = 0}\end{array}\)

So,

\(\begin{array}{c}{\mathop{\rm cov}} \left( {{X_i} + {Y_i},{X_j} + {Y_j}} \right) = {\mathop{\rm cov}} \left( {{X_i},{X_j}} \right) + {\mathop{\rm cov}} \left( {{X_i},{Y_j}} \right) + {\mathop{\rm cov}} \left( {{Y_i},{X_j}} \right) + {\mathop{\rm cov}} \left( {{Y_i},{Y_j}} \right)\\ = {\mathop{\rm cov}} \left( {{X_i},{X_j}} \right) + 0 + 0 + {\mathop{\rm cov}} \left( {{Y_i},{Y_j}} \right)\\ = {\mathop{\rm cov}} \left( {{X_i},{X_j}} \right) + {\mathop{\rm cov}} \left( {{Y_i},{Y_j}} \right)\end{array}\)

\(Cov(X + Y) = Cov(X) + Cov(Y)\)

Hence the given statement is proved.