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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

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

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  • $\begingroup$ 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
    Commented Nov 27, 2017 at 22:53
  • $\begingroup$ I tried that but I got a covariance of $0$, which seems incorrect. I'll add my attempt. $\endgroup$
    – Remy
    Commented Nov 27, 2017 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 $\endgroup$ Commented Nov 27, 2017 at 22:58
  • $\begingroup$ That seems a little advanced for this kind of problem, no? I'm sure you're correct though. $\endgroup$
    – Remy
    Commented Nov 27, 2017 at 23:04
  • $\begingroup$ @vvnitram Y and Z are evidently not independent. $\endgroup$ Commented Nov 27, 2017 at 23:08

1 Answer 1

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

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  • $\begingroup$ 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
    Commented Nov 27, 2017 at 23:07
  • $\begingroup$ It is supplied that $\Var(X_2)=4$ $\endgroup$ Commented Nov 27, 2017 at 23:08
  • $\begingroup$ My mistake, that was a typo. $\endgroup$
    – Remy
    Commented Nov 27, 2017 at 23:09

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