Currently I am preparing for my exam. Therefore, we have a sample exam with the answers included. Unfortunately, even with the answers it not all makes sense to me. This is the question with the answer provided:

Let $\bf{x}$ be a random vector with expectation $E(\textbf{x})=\boldsymbol{\mu}$ and covariance matrix $Var(\textbf{x})=\bf{\Sigma}$. Let $\textbf{x}=(x_1, \boldsymbol{x_2'})$, where $x_1$ is a scalar and $\boldsymbol{\mu}$ and $\boldsymbol{\Sigma}$ are partitioned conformably. Let $y=x_1-\bf{b'x_2}$$=(1, -\bf{b}')x$ for some conformable nonrandom vector $\textbf{b}$.

The question: Derive $\textbf{b}$ that minimizes the variance of y. Show the covariances between y and the elements of $\bf{x_2}$ are zero for this choice of $\textbf{b}$.

I know that $\boldsymbol{\Sigma}= \begin{bmatrix} \sigma^2 & \boldsymbol{\sigma_{12}} \\ \boldsymbol{\sigma_{21}} & \boldsymbol{\Sigma_{22}} \end{bmatrix}$ (where $\boldsymbol{\sigma_{21}}=\boldsymbol{\sigma_{12}'}$) and that $\boldsymbol{\mu}=(\mu_1, \boldsymbol{\mu_2})'$. Now the variance of y is given as $Var(y)=(1, -\mathbf{b'})\begin{pmatrix} \sigma^2 & \boldsymbol{\sigma_{12}} \\ \boldsymbol{\sigma_{21}} & \boldsymbol{\Sigma_{22}}\end{pmatrix}\begin{pmatrix}1 \\ -\mathbf{b}\end{pmatrix}$. Solving and taking the derivative with respect to $\mathbf{b}$ yields that it is minimized by $\mathbf{b}=\boldsymbol{\Sigma_{22}}^{-1}\boldsymbol{\sigma_{21}}$. Then it is said that the covariances between y and the elements of $\mathbf{x_2}$ are given by:

$E\{y(\mathbf{x_2}-\boldsymbol{\mu_2})\}=(1, -\mathbf{b'})\begin{pmatrix} \sigma^2 & \boldsymbol{\sigma_{12}} \\ \boldsymbol{\sigma_{21}} & \boldsymbol{\Sigma_{22}}\end{pmatrix}\begin{pmatrix}\mathbf{0} \\ \mathbf{I}\end{pmatrix}=\mathbf{0}$

Now I need help here, what is happening? I know that the covariance is given as: $cov(y, \mathbf{x_2})=E\{[y-E(y)][\mathbf{x_2}-E(\mathbf{x_2})]\}$ and $E(\mathbf{x_2})=\boldsymbol{\mu_2}$, but why is $E(y)$ disappearing? And what happens after that first equality sign when it has been set equal to zero?

Hopefully I have clarified the problem enough. Thanks in advance.


1 Answer 1


Hint (if I understand your question correctly): another formula for covariance is

$$ \text{Cov}(y,\mathbf{x_2}) = E[y\mathbf{x_2}] - E[y](E[\mathbf{x_2}])' $$


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