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Suppose $X$ and $Y$ are random variables with $E(X)=2, E(Y)=3 Var(X)= 4, Var(Y)=10$ and $Cov(X,Y)=-5$

  1. Find $Var (5X+2Y)$

From my book I know $$Var(5X+2Y)= Var(5X)+Var(2Y)+2Cov(5X,2Y)$$ but after that do I just substitute in values so $Var(5X)= 5\cdot 4=20$? If so, how do I figure out $Cov(5x,2y)$?

  1. Find $Cov(3X+Y,Y)$
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    $\begingroup$ No, $\text{Var}(5X) = 25\text{Var}(X)$. Why? $\endgroup$ – Ted Shifrin Jul 13 '15 at 17:15
  • $\begingroup$ I don't know. Any suggestions of websites to visit. $\endgroup$ – user219081 Jul 13 '15 at 17:50
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Hints Note that $$Var (kZ) = \mathbb{E}[Z^2] - \mathbb{E}[Z]^2 = \mathbb{E}[(kZ)^2] - \mathbb{E}[kZ]^2 = \mathbb{E}[k^2 Z^2] - (k\mathbb{E}[Z])^2$$ Can you finish this and express $Var(kZ)$ in terms of $k$ and $Var Z$?

Then do the same thing to the $Covar(X,Y)$, directly using the definition.

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  • $\begingroup$ How would you calculate $E(k^2Z^2)$? $\endgroup$ – user219081 Jul 13 '15 at 17:53
  • $\begingroup$ @AlyssaWallace don't calculate it, factor $k^2$ out of both terms and see what remains. $\endgroup$ – gt6989b Jul 13 '15 at 18:43
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Let's compute:

$$Var(5X+2Y)= Var(5X)+Var(2Y)+2Cov(5X,2Y) = 25 Var (X) + 4 Var (Y) + 20 Cov(X,Y) = 25 (4) + 4(10) + 20 (-5) = 40$$

To understand the manipulations, you need to see the properties of the expectation and the definition of covariance.

\begin{align*}\Bbb{E}[a X + bY] &= a \Bbb{E}[X] + b \Bbb{E}[Y]\\ Cov(X,Y) &= \Bbb{E}[(X - \Bbb{E}[X])(Y - \Bbb{E}[Y])]\\ Var(X) &= Cov (X,X) \end{align*}

take a look at https://en.wikipedia.org/wiki/Expected_value,

and https://en.wikipedia.org/wiki/Covariance

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