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?

  • $\begingroup$ Hint: covariance is linear in both its arguments. Moreover it takes value is zero if the arguments are independent. $\endgroup$ – drhab Dec 29 '19 at 20:49
  • $\begingroup$ More specific: Covariance is bilinear. $\endgroup$ – callculus Dec 29 '19 at 20:57
  • 1
    $\begingroup$ @callculus Maybe some of you have another hint, I'm really not getting it. $\endgroup$ – user Dec 29 '19 at 21:31
  • $\begingroup$ $=2Cov(X_1,X_1)-Cov(X_1,X_2)+6Cov(X_1,X_3)+\cdots+6Cov(X_3,X_3)=...$ $\endgroup$ – Jean Marie Dec 29 '19 at 22:51
  • $\begingroup$ @user If have any further question, feel free to ask. $\endgroup$ – callculus Dec 29 '19 at 23:56

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


if the $X_i's$ are independent.

  • $\begingroup$ I added my result, wanted to check if I am correct. $\endgroup$ – user Jan 2 at 19:47
  • $\begingroup$ @user It looks perfect. $\endgroup$ – callculus Jan 2 at 20:01
  • $\begingroup$ #callculus Ok, and what about correlation, how to find that? $\endgroup$ – user Jan 2 at 20:18
  • $\begingroup$ @user I´ve made an edit. $\endgroup$ – callculus Jan 2 at 23:48

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