2
$\begingroup$

Let \begin{equation} \mathbf{y}=\mathbf{X}\mathbf{\beta}+\mathbf{u} \end{equation} where

$\mathbf{y}=\begin{bmatrix}y_1 \\ \vdots \\ y_n\end{bmatrix}$,

$\mathbf{X}=\begin{bmatrix}X_{11} & \cdots & X_{1k} \\ \vdots & \ddots & \vdots \\ X_{n1} & \cdots & X_{nk}\end{bmatrix}$, or also written $\mathbf{X}=\begin{bmatrix}\mathbf{x}_1 \\ \vdots \\ \mathbf{x}_n\end{bmatrix}$ (where each $\mathbf{x}_i$ is a $k \times 1$ vector),

$\mathbf{\beta}=\begin{bmatrix}\beta_1 \\ \vdots \\ \beta_k\end{bmatrix}$, and

$\mathbf{u}=\begin{bmatrix}u_1 \\ \vdots \\ u_n\end{bmatrix}$.


Suppose that $E(\mathbf{u}|\mathbf{X})=\mathbf{0}$. It is clear this implies that $E(\mathbf{u})=\mathbf{0}$ since by the Law of Iterated Expectations, \begin{equation} E\left(\mathbf{u}\right)=E\left[E\left(\mathbf{u}\middle\vert\mathbf{X}\right)\right]=E\left(\mathbf{0}\right)=\mathbf{0}. \end{equation}

However, I want to show a similar result for the covariances, i.e. show that $E(\mathbf{u}|\mathbf{X})=\mathbf{0}$ implies $Cov(\mathbf{x},\mathbf{u})=0$ for all $\mathbf{x}$, where $\mathbf{x}$ is a column of the matrix $\mathbf{X}$. My attempt:

Let $\mathbf{x}$ denote a column of the matrix $\mathbf{X}$. Then,

\begin{align} Cov\left(\mathbf{u},\mathbf{x}\right)&=E\left[\left(\mathbf{u}-E(\mathbf{u})\right)\left(\mathbf{x}-E(\mathbf{x})\right)^{\top}\right] \\ &=E\left(\mathbf{u}\mathbf{x}^{\top}\right)-E\left(\mathbf{u}\right)E\left(\mathbf{x}^{\top}\right) \\ &=E\left(\mathbf{u}\mathbf{x}^{\top}\right) \\ &=E\left[E\left(\mathbf{u}\mathbf{x}^{\top}|\mathbf{X}\right)\right] \\ &=E\left[E\left(\mathbf{u}|\mathbf{X}\right)\mathbf{x}^{\top}\right] \\ &=E\left[\mathbf{0}\cdot\mathbf{x}^{\top}\right] \\ &=\mathbf{0}_{n\times n} \end{align}

Is this argument correct?

$\endgroup$
1
$\begingroup$

Since $E(u) = 0$, $Cov(x,u) = E(xu')=E(E(xu'|x))$. Now, $E(x_i u_j|x)=x_i$, $E(u_j|x)=0$. Hence, the matrix $E(xu'|x) = 0$, so $Cov(x,u)=0$. Hence, each column of X is uncorrelated with u.

$\endgroup$
  • $\begingroup$ Thanks for your answer! I am having a hard time interpreting the $i$'s and $j$'s in your answer. $\endgroup$ – Mathemanic Oct 7 '14 at 5:46
  • $\begingroup$ Yes, x denotes a particular column of the matrix X. Also, xi denotes the ith entry of the column x and uj means the jth entry of u (i and j vary from 1 to n). $\endgroup$ – Somabha Mukherjee Oct 7 '14 at 5:48
  • $\begingroup$ Hi Somabha, I edited my original post with an attempted proof, using the law of iterated expectations. Would you say that my attempted proof is correct? $\endgroup$ – Mathemanic Oct 7 '14 at 6:09
  • $\begingroup$ Yes, it is correct. That is exactly what I did! $\endgroup$ – Somabha Mukherjee Oct 7 '14 at 6:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.