0
$\begingroup$

Consider independent random variables $X_1$, $X_2$, and $X_3$ such that $X_1$ is a random variable having mean $1$ and variance $1$, $X_2$ is a random variable having mean $2$ and variance $4$, and $X_3$ is a random variable having mean $3$ and variance $9$.

(a) Give the value of the variance of $X_1+\frac{1}{2}X_2+\frac{1}{3}X_3$.

(b) Give the value of the correlation of $Y=X_1−X_2$ and $Z=X_2+X_3$.

(a) $$\begin{align*} Var\left(X_1+\frac{1}{2}X_2+\frac{1}{3}X_3\right) &= 1^2Var(X_1)+\frac{1}{2}^2Var(X_2)+\frac{1}{3}^2Var(X_3) \\\\ &= 1^2\cdot1+\frac{1}{4}\cdot4+\frac{1}{9}\cdot9 \\\\ &=1+1+1\\\\ &= 3 \end{align*}$$

(b)

I have the following:

$E(Y)=E(X_1-X_2)=E(X_1)-E(X_2)=1-2=-1$

$Var(Y)=Var(X_1-X_2)=1^2Var(X_1)+{-1}^2Var(X_2)=Var(X_1)+Var(X_2)=1+4=5$

$E(Z)=E(X_2+X_3)=E(X_2)+E(X_3)=2+3=5$

$Var(Z)=Var(X_2+X_3)=1^2Var(X_2)+1^2Var(X_3)=Var(X_2)+Var(X_3)=4+9=13$

I also have the formula

$$\begin{align*} \rho_{YZ} &= Corr(Y,Z) \\\\ &= \frac{Cov(Y,Z)}{\sigma_Y\sigma_Z} \\\\ &= \frac{Cov(Y,Z)}{\sqrt{5}\cdot\sqrt{13}} \\\\ \end{align*}$$

Is what I have so far correct? How could I use the information provided to find $Cov(Y,Z)$?

I tried

$$\begin{align*} Cov(Y,Z) &= E(YZ)-E(Y)E(Z) \\\\ &= E((X_1-X_2)(X_2+X_3))-E(X_1-X_2)\cdot E(X_2+X_3) \\\\ &= E(X_1\cdot X_2 - X_2^2 +X_1\cdot X_3 -X_2\cdot X_3)-E(X_1-X_2)\cdot E(X_2+X_3) \\\\ &= E(X_1\cdot X_2) - E(X_2^2) +E(X_1\cdot X_3) -E(X_2\cdot X_3))-E(X_1-X_2)\cdot E(X_2+X_3) \\\\ &=(1\cdot2) - (2^2) + (1\cdot 3) - (2\cdot3) - (-1\cdot 5) \\\\ &= 0 \end{align*}$$

$\endgroup$
  • $\begingroup$ Everything is correct so far, recall that $Cov(X,Y)=E(XY)-E(X)E(Y)$. The expected value of $Y$ and $Z$ are already known so all you need to find is $E(YZ)$ $\endgroup$ – Teh Rod Nov 27 '17 at 22:53
  • $\begingroup$ I tried that but I got a covariance of $0$, which seems incorrect. I'll add my attempt. $\endgroup$ – Remy Nov 27 '17 at 22:54
  • $\begingroup$ Hint: $Y$ and $Z$ are independent. Consider the random vector $(X_1,X_2,X_3)$ and the Borel-function $g(x,y,z)=(x-y,y+z,z)$ and prove that $g$ is bijective $\endgroup$ – Martín Vacas Vignolo Nov 27 '17 at 22:58
  • $\begingroup$ That seems a little advanced for this kind of problem, no? I'm sure you're correct though. $\endgroup$ – Remy Nov 27 '17 at 23:04
  • $\begingroup$ @vvnitram Y and Z are evidently not independent. $\endgroup$ – Graham Kemp Nov 27 '17 at 23:08
1
$\begingroup$

Apply the Bilinearity of Covariance: $$\def\Cov{\mathsf{Cov}}\def\Var{\mathsf{Var}} {\Cov(Y,Z) ~{=~\Cov(X_1-X_2, X_2+X_3) \\=~ \Cov(X_1,X_2)+\Cov(X_1,X_3)-\Cov(X_2,X_2)-\Cov(X_2,X_3)\\=~0+0-\Var(X_2)-0\\ =~ -4}}$$

$\endgroup$
  • $\begingroup$ And the covariance between $3$ of them are $0$ because of independence and $Corr(X_2,X_2)=1$ since they are 100% correlated (same variable)? $\endgroup$ – Remy Nov 27 '17 at 23:07
  • $\begingroup$ It is supplied that $\Var(X_2)=4$ $\endgroup$ – Graham Kemp Nov 27 '17 at 23:08
  • $\begingroup$ My mistake, that was a typo. $\endgroup$ – Remy Nov 27 '17 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.