If there are independent random variables $X_1,\ldots,X_3, D(X_i)=\sigma^2, i=1,\ldots,3$
Then find correlation and covariance between variables $X_1+2X_2+X_3$ and $2X_1-X_2+6X_3$. I have no idea how to do it. But here is my take:
$$\operatorname{Cov}(X_1+2X_2+X_3,2X_1-X_2+6X_3)$$
Here is my solution:
$$Cov(X_1,2X_1-X_2+6X_3)+Cov(2X_2,2X_1-X_2+6X_3)+Cov(X_3,2X_1-X_2+6X_3)=Cov(X1,2X_1)-Cov(X_1,X_2)+Cov(X_1,6X_3)+Cov(2X_2,2X_1)-Cov(2X_2,X_2)+Cov(2X_2,6X_3)+Cov(X_3,2X_1)-Cov(X_3,X_2)+Cov(X_3,6X_3)=2Var(X_1)-2Var(X_2)+6Var(X_3)=6\sigma^2$$
Can someone help me?
Best Answer
I show you the case with a linear combination of two variables. It a bit similar like multiplying out of ordinary brackets.
$$ \begin{align} & cov(a_1X_1+b_1X_2, a_2X_1+b_2X_2) \\\\ & \\\\ &=cov(a_1X_1, a_2X_1+b_2X_2)+cov(b_1X_2, a_2X_1+b_2X_2) \\\\ & \\\\ & =cov(a_1X_1, a_2X_1)+cov(a_1X_1,b_2X_2)+cov(b_1X_2, a_2X_1)+cov(b_1X_2, b_2X_2)\\\\ & \\\\ &=a_1a_2\cdot cov(X_1,X_1)+a_1b_2\cdot cov(X_1,X_2)+a_2b_1\cdot cov(X_1,X_2)+b_1b_2\cdot cov(X_2,X_2) \end{align} $$
Due the independence of $X_1$ and $X_2$ we have $cov(X_1,X_2)=0 $. And the covariance of $X_1$ and $X_1$ is the variance of $X_1$.
$$ \begin{align} & =a_1a_2\cdot Var(X_1)+b_1b_2\cdot var(X_2) \end{align} $$
You just have to transfer the calculation/result to the case of three random variables.
$$corr(Y_1,Y_2)=\frac{cov(Y_1,Y_2)}{\sqrt{var(Y_1)\cdot var(Y_2)}}$$
The term of the numerator is already calculated. And
$var(Y_i)=var(aX_1+bX_2+cX_3)=a^2var(X_1)+b^2var(X_2)+c^2var(X_3)$,
if the $X_i's$ are independent.