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I recently came across this problem, although trivial to compute by hand - is a little challenging for me to interpret. Notably, we have three matrices:

$$\vec{c}= \begin{bmatrix} 0.5 \\ 0.5 \end{bmatrix},\hspace{0.2in} \vec{x}= \begin{bmatrix} 2 \\ 3 \end{bmatrix},\hspace{0.2in} \vec{\mu}= \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \hspace{0.2in} \mathbf{\Sigma} = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} $$

We then have the following calculation, which I have solved below.

$\vec{c}^{\,T}(\vec{x}-\vec{\mu}) (\vec{c}^{\ T}\mathbf{\Sigma}\vec{c})^{-1}$ ${}=\begin{bmatrix} \ 0.5 \end{bmatrix}$

The calculation is trivial. However, how do I interpret this solution - assuming $\vec{x}$ is a data vector, $\vec{\mu}$ is a mean vector, and $\mathbf{\Sigma}$ is a covariance matrix?

Workhorse
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1 Answers1

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Consider the random variable $$y = c^T x$$Then the mean of $y$ is $$E(y) = c^T E(x) = c^T \mu$$ and variance of $y$ is $$var(y) = E(y - c^T \mu)^2 = E(y^2) - (c^T \mu)^2 = E(y^2) + c^T \mu\mu^T c \tag{*}$$ But $$y^2 = (c^Tx)^2 = c^Tx c^Tx= c^Txx^Tc$$ So $$E(y^2) = c^TE(xx^T)c = c^T(\Sigma - \mu\mu^T)c \tag{**}$$ Replace $(**)$ is $(*)$, we get $$var(y) = c^T\Sigma c$$ Let $z$ be a standard version of $y$, i.e. $$z = \frac{y - E(y)}{var(y)} = \frac{c^Tx - c^T\mu}{c^T\Sigma c} = c^T(x - \mu)(c^T\Sigma c)^{-1} $$ So, you have computed the $z$ here.

Ahmad Bazzi
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