1
$\begingroup$

I tried to do the F test for NLMs by myself and run into a dead-end. I have a linear model with normal distributed error $$Y = X\beta + e$$ I know that the test statistic $$F = \frac{n-r}{q}\frac{\|Q_{L_H} Y\|^2_2 - \|Q_{L} Y\|^2_2}{\|Q_L Y\|_2^2} \sim F_{q,n-r}$$ where $n$ is the length of the given data, $r$ is the dimension from the model matrix $X \in \mathbb{R}^{n \times p}$, $q$ is the dimension from the hypothesis matrix $H \in \mathbb{R}^{q \times p}$ and $L_H$ is a $r-q$ dimensional linear subspace from $L := \lbrace X\beta | \beta \in \mathbb{R}^p\rbrace$. Suppose the matrix $H$ is, in my special case, $H = (0\ \ \ \ \ 1\ \ \ \ \ 0)$ because I'd like to proof the hypothesis $$H_0: H\beta = \beta_1 = 0 \qquad vs. \qquad H_1: \beta_1 \neq 0$$ (Later I want to try this for more complex matrices $H$)

I also know that $Q_L$ is the orthogonal projection into $L^\perp$. This matrix can be easy calculated by $$Q_L = E - P_L = E - X(X^TX)^{-1}X^T$$ ($P_L$ is the orthogonal projection into $L$)

My problem now is that I don't know how to compute $Q_{L_H}$?

My first try was to compute it the same way like $P_L$ and $Q_L$, just with the matrix $H$ instead of $X$. But that gives me a $3 \times 3$ matrix because my $\beta = (\beta_0\ \ \ \ \ \beta_1 \ \ \ \ \ \beta_2)^T$ and with that $p = 3$.

Maybe somebody can give me a hint or know how this works.

$\endgroup$

1 Answer 1

0
$\begingroup$

For each who want to know the answer: You have to compute $P_{L_H}$ analogous to $P_L$, just with the designmatrix $X_0$, that results under $H_0$. In my case that means if $\beta_1$ is null, the designmatrix for that is $$X_0 = X \beta_{H_0} = X \left(\begin{array}{cc} 1 \\ 0 \\ 1 \end{array}\right)$$ where $X$ is the designmatrix. Then compute $$P_{L_H} = X_0 (X_0^T X)^{-1}X_0^T$$ and with that $$Q_{L_H} = I_n - P_{L_H}.$$ And thats it.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.