If the random variables $X,Y$ and $Z$ have means: $$\bar{x}=2 \;\;\; \bar{y}=-3 \;\;\; \bar{z}=4$$ and variances $$\operatorname{Var}(x)=1 \;\;\; \operatorname{Var}(Y)=5 \;\;\; \operatorname{Var}(Z)=2$$ and co-variance $$\operatorname{Cov}(X,Y)= -2\;\;\; \operatorname{Cov}(X,Z)=-1\;\;\;\operatorname{Cov}(Y,Z)=1$$ Find

  1. Mean of $W=3X-Y+2Z$
  2. The variance of $W=3X-Y+2Z$ using the matrix method

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  • $\begingroup$ Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated. $\endgroup$ – xbh Jul 30 '18 at 16:25


$$E[aX+b] = aE[X]+b$$ $$\operatorname{Var}[aX+bY] = \mathbf{v}^T\mathbf{M}\mathbf{v}$$ where $\mathbf{v} = \left(\begin{matrix}a\\b\end{matrix}\right)$ and $\mathbf{M}$ is the matrix of covariance defined (in the case two variables) as $$\left(\begin{matrix}\operatorname{Var}[X]&\operatorname{Cov}[X,Y]\\\operatorname{Cov}[Y,X]&\operatorname{Var}[Y]\end{matrix}\right)$$


The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$\mathbf{v} = \left(\begin{matrix}3\\-1\\2\end{matrix}\right)\;\;\;\mathbf{M}=\left(\begin{matrix}\operatorname{Var[X]}&\operatorname{Cov[X,Y]}&\operatorname{Cov[X,Z]}\\\operatorname{Cov[Y,X]}&\operatorname{Var[Y]}&\operatorname{Cov[Y,Z]}\\\operatorname{Cov[Z,X]}&\operatorname{Cov[Z,Y]}&\operatorname{Var[Z]}\end{matrix}\right) = \left(\begin{matrix}1&-2&-1\\-2&5&1\\-1&1&2\end{matrix}\right)$$ Then we calculate the variance of $W$ as $$\operatorname{Var}[W] = \operatorname{Var}[3X-Y+2Z] = \underbrace{\left(\begin{matrix}3&-1&2\end{matrix}\right)\left(\begin{matrix}1&-2&-1\\-2&5&1\\-1&1&2\end{matrix}\right)}_{\text{first multiplication}}\left(\begin{matrix}3\\-1\\2\end{matrix}\right)$$ let's do first the underbraced multiplication $$\left(\begin{matrix}3&-1&2\end{matrix}\right)\left(\begin{matrix}1&-2&-1\\-2&5&1\\-1&1&2\end{matrix}\right) = \left(\begin{matrix}3&-9&6\end{matrix}\right)$$ and now let's do the second multiplication $$\left(\begin{matrix}3&-9&6\end{matrix}\right)\left(\begin{matrix}3\\-1\\2\end{matrix}\right)=30$$ So in the end we have that $$\operatorname{Var}[W]=30$$

  • $\begingroup$ Please note that $\mathbb E(aX+b)=a\cdot \mathbb E(X)+\color{red}b$ $\endgroup$ – callculus Jul 30 '18 at 17:08
  • $\begingroup$ @callculus Ops, my bad! Thank you $\endgroup$ – Davide Morgante Jul 30 '18 at 17:09
  • $\begingroup$ Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante $\endgroup$ – Eric kioko Jul 30 '18 at 18:30
  • $\begingroup$ No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$ $\endgroup$ – Davide Morgante Jul 30 '18 at 18:31
  • $\begingroup$ @DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method. $\endgroup$ – Eric kioko Jul 30 '18 at 18:47

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