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
ANSWERED
asked 2021-05-16
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
\(Cov(\sum_{i=1}^{n}a_{i}X_{i},\sum_{j=1}^{m}b_{j}Y_{j})=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i}b_{j}Cov(X_{i}, Y_{j})\)

Answers (1)

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.
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.
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