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Let $Y = X \beta + \varepsilon$ with $Y \in \mathbb{R} \in \mathbb{R}^n$, $X \in \mathbb{R}^{n \times p}$, $\operatorname{rank}(X) = p$, $\beta \in \mathbb{R}^p$, $\varepsilon \sim \mathcal{N}_n(0, \sigma^2 I_n)$.

What is the distribution of $\hat{\beta}$? $$\hat{\beta} = {(X^TX)}^{-1} X^T Y$$

My thoughts

Except for $\varepsilon$, everything is constant. So $\hat{\beta}$ is an affine function of $\varepsilon$. Hence $\hat{\beta}$ is also multivariate normal distributed.

\begin{align} \mathbb{E}(\hat{\beta}) &= \mathbb{E}({(X^TX)}^{-1} X^T Y)\\ &= {(X^TX)}^{-1} X^T \mathbb{E}(Y)\\ &= {(X^TX)}^{-1} X^T \mathbb{E}(X \beta + \varepsilon)\\ &= {(X^TX)}^{-1} X^T X \beta + \mathbb{E}(\varepsilon)\\ &= {(X^TX)}^{-1} X^T X \beta\\ \end{align}

First sub-question: Is ${(X^TX)}^{-1} = X^{-1} (X^T)^{-1}$, even if $X$ is not square? (And which inverse would that be? I've just read that there are multiple candidates.)

edit: I've just seen ${(X^TX)}^{-1} X^T X \beta = {(X^TX)}^{-1} (X^T X) \beta$, so I don't need this to continue.

\begin{align} \mathbb{E}(\hat{\beta}) &= X^{-1} {X^T}^{-1} X^T X \beta\\ &=\beta \end{align}

\begin{align} C(\hat{\beta}) &= C({(X^TX)}^{-1} X^T Y)\\ &={(X^TX)}^{-1} X^T C(Y) { \left ({(X^TX)}^{-1} X^T \right)}^T\\ &={(X^TX)}^{-1} X^T \sigma^2 I_n { \left ({(X^TX)}^{-1} X^T \right)}^T\\ &=\sigma^2 {(X^TX)}^{-1} X^T { \left ({(X^TX)}^{-1} X^T \right)}^T\\ &=\sigma^2 {(X^TX)}^{-1} X^T X {(X^T X)^{-1}}^T \\ &=\sigma^2 {(X^T X)^{-1}}^T \\ \end{align}

Sub-question 2: According to my notes, $\hat{\beta} \sim \mathcal{N}_n \left ({\beta, \sigma^2 (X^T X)^{-1}} \right )$. Is that the same?

I have read that $X^T X$ is always symmetric, but is ${(X^T X)}^{-1}$ also symmetric?

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    $\begingroup$ Please be careful that the $(X^TX)^{-1}$ can not be split in $X^{-1}(X^T)^{-1}$; this is not permitted. In some sense, you can think of $(X^TX)^{-1}X^T$ as an inverse of your non-square linear system (and in fact that matrix is called Moore-Penrose pseudo-inverse). $\endgroup$ – Ilis Feb 24 '17 at 13:54
  • $\begingroup$ @Ilis I thought so. This is why I continued with "if this is a valid transformation". I've just noticed that I was blind at this step and I don't need it. I'm not sure if I should remove this entirely. I think it is interesting (and also your comment is interesting). $\endgroup$ – Martin Thoma Feb 24 '17 at 13:59
  • $\begingroup$ regarding sub-question 2: yes both of them are symmetric. If you look at what the entry $(i,j)$ of $X^T X$ is, you can easily understand that is equal to $(j,i)$ entry. For the inverse, consider that $(A^T)^{-1}=(A^{-1})^T$. $\endgroup$ – Ilis Feb 24 '17 at 14:08

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