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When consider the following DGP : $y=X\beta^{*}+\epsilon$ where $\beta^{*}$ is a $\tilde k\times1\ $ vector.

Define the projection matrices: $P_{X}=X(X^{T}X)^{-1}X^{T}$ and $M_{X}=I-X(X^{T}X)^{-1}$.

first problem is prove that $X^{T}P_{X}=X^{T}$ and $ X^{T}M_{X}=0$

I thought this problem is too simple to think because P is just I. So, I think that there is an intention for given problem considering second problem.

Second problem is Prove thath if $X_{(i)}$ is the $i^{th}$ column of $X$, then, $M_{X}X_{(i)}=0$

My question is

  1. What is the point of first problem?

  2. Can I solve these problem by considering $X_{(i)}=XA\ $ where A has a vector consisting of just 1 and the other columns are all zero? or Should I utilize first problem's solution?

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  • $\begingroup$ Please post separate questions separately. $\endgroup$ – Giskard Oct 14 '17 at 10:43
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    $\begingroup$ The "first problem" is indeed immediately proven by simply inserting the expression for the projection matrices. For the second problem, use the fact that $M_X$ is symmetric. $\endgroup$ – Alecos Papadopoulos Oct 14 '17 at 12:04
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This kind of projections in econometrics are usually employed for partialling out some covariates from a linear regression.

Observe that in general $P_X\neq I$. Consider $X=\begin{bmatrix}1&5\\1&0\\1&1\end{bmatrix}$. Then $X'X=\begin{bmatrix}3&6\\6&26\end{bmatrix}$ and $(X'X)^{-1}=\begin{bmatrix}26/42&-6/42\\-6/42&3/42\end{bmatrix}$ then $X(X'X)^{-1}X^T=\frac{1}{42}\begin{bmatrix}41&-4&5\\-4&26&20\\5&20&17\end{bmatrix}\neq I$.

First problem: $X^TP_X=X^TX(X'X)^{-1}X^T=(X^TX)(X'X)^{-1}X^T=I_{(\tilde k\times\tilde k)}X^T=X^T$

Similarly: $X^TM_X=X^T[I_{(\tilde k\times\tilde k)}-P_X]=X^TI_{(\tilde k\times\tilde k)}-X^TP_X=X^T-X^T=0$

The second problem is a direct consequence of $X^TM_X=0$, simply now you can post-multiply for $X_{(i)}$ since you are considering just one single column. The approach of $X_{(i)}=XA$ where $A$ is a vector $\tilde k\times 1$ of 0 apart from the $i$-th element which is 1. Then it should be straightforward to see: $0=(XA)^TM_X=M_XX_{(i)}$

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  • $\begingroup$ I have never thought that kind of solving. It was very confusing for me to utilize such a matrix. Thanks a lot! $\endgroup$ – Jeffrey Oct 14 '17 at 14:11
  • $\begingroup$ @GabMac Are there any econometrics books that present this linear regression thinking? $\endgroup$ – Laura Mar 14 at 23:10

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