Question

# Let X_{1}....,X_{n} and Y_{1},...,Y_{m} be two sets of random variables. Let a_{i}, b_{j} be arbitrary constant. Show that 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})

Random variables
Let $$X_{1}....,X_{n} and Y_{1},...,Y_{m}$$ be two sets of random variables. Let $$a_{i}, b_{j}$$ be arbitrary constant.
Show that
$$Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}Cov(X_{i}, Y_{j})$$

2021-05-17
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.
$$Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}Cov(X_{i}, Y_{j})$$
$$Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})$$
$$=E(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})-E(\sum_{i=1}^{n}a_{i}X_{i})E(\sum_{j=1}^{m}b_{j}Y_{j})$$
$$=E\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}[E(XiYj)=E(Xi)E(Yj)]$$
$$=E\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}\ Cox(Xi, Yj)....by covariance property Cov (X , Y) = E(XY) - E(X) E(Y)$$.