Find the distribution of $X+Y+Z$

covariancenormal distributionprobabilityrandom variables

Suppose $(X,Y,Z)$ follow the multivariate normal distribution:
$$
(X,Y,Z)\sim\mathcal{N}\left[\begin{pmatrix}1\\3\\4\end{pmatrix},\begin{pmatrix}1 & 0.4 & 0.2\\0.4 & 1 & 0.3\\0.2 & 0.3 & 1\end{pmatrix}\right]
$$

  1. Find the distribution of $X+Y+Z$.
  2. Find $\text{Cov}(2X-3Y+Z,X+Y-Z)$.

My Attempt

  1. I would assume that it is a standard normal distribution, which is defined by its expectation and variance. Then the expectation is
    $$\mathbb{E}[X+Y+Z]=\mathbb{E}[X]+\mathbb{E}[Y]+\mathbb{E}[Z]=1+3+4=8$$
    and the variance is
    \begin{align}
    \text{Var}(X+Y+Z)&=\text{Var}(X)+\text{Var}(Y)+\text{Var}(Z)+2\text{Cov}(X,Y)+2\text{Cov}(Y,Z)+2\text{Cov}(X,Z)\\
    &= 1+1+1+2(0.4)+2(0.2)+2(0.3)=4.8
    \end{align}
    Thus, the distribution is $X+Y+Z\sim\mathcal{N}(8,4.8)$

\begin{align}
\text{Cov}(2X-3Y+Z,X+Y-Z)&=\mathbb{E}[(2X-3Y+Z)(X+Y-Z)]-\mathbb{E}[2X-3Y+Z]\mathbb{E}[X+Y-Z]\\
&=2\mathbb{E}[X^2]-3\mathbb{E}[Y^2]-\mathbb{E}[Z^2]-\mathbb{E}[XY]-\mathbb{E}[XZ]+4\mathbb{E}[YZ]…
\end{align}
Before I finish the entire calculation, how do you figure out $\mathbb{E}[X^2]$ from what we are given?

Best Answer

For 1.

A standard normal distribution has mean of $0$ and variance of $1^2$. As your normal distribution does not have those as parameters, it is not standard.

Otherwise, correct.

For 2.

You could find $\mathsf E(X^2)=\mathsf {Var}(X)+\mathsf E(X)^2$, and so forth, however, do not do that.

Just use the Bilinearity of Covariance. (Which you might notice you did in part 1.)

$$\def\Cov{\operatorname{\mathsf{Cov}}}\def\Var{\operatorname{\mathsf{Var}}}\begin{align}\Cov(aX+bY+cZ,dX+eY+fZ)&={ad\Var(X)+be\Var(Y)+cf\Var(Z)\\+(ae+bd)\Cov(X,Y)\\+(af+cd)\Cov(X,Z)\\+(bf+ce)\Cov(Y,Z)}\\[3ex]\Cov(2X−3Y+Z,X+Y−Z)&=\ldots\end{align}$$

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