Step 1:

It is given that \(X_{1}, ... ,X_{n} and Y_{1}, ... ,Y_{m}\) be the two sets of random variables.

Let \(a_{i} , b_{j}\) be the arbitrary constant.

It is asked to show,

\(Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}Cov(X_{i}, Y_{j})\)

Step 2

Proof :

\(Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})\)

\(=E(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})-E(\sum_{i=1}^{n}a_{i}X_{i})E(\sum_{j=1}^{m}b_{j}Y_{j})\)

\(=E\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}[E(XiYj)=E(Xi)E(Yj)]\)

\(=E\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}\ Cox(Xi, Yj)....by covariance property Cov (X , Y) = E(XY) - E(X) E(Y)\).

Hence , It is proved.

It is given that \(X_{1}, ... ,X_{n} and Y_{1}, ... ,Y_{m}\) be the two sets of random variables.

Let \(a_{i} , b_{j}\) be the arbitrary constant.

It is asked to show,

\(Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}Cov(X_{i}, Y_{j})\)

Step 2

Proof :

\(Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})\)

\(=E(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})-E(\sum_{i=1}^{n}a_{i}X_{i})E(\sum_{j=1}^{m}b_{j}Y_{j})\)

\(=E\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}[E(XiYj)=E(Xi)E(Yj)]\)

\(=E\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}\ Cox(Xi, Yj)....by covariance property Cov (X , Y) = E(XY) - E(X) E(Y)\).

Hence , It is proved.