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I am solving an exercise on multiple linear regression. Near the end I will be asked for the same data as the previous model, it is the maximum likelihood estimates.

The previous model: I have the matrix $(X'X)^{-1}$ and the matrix $X'y$ and the model is:
\begin{align} Y_i &= B_0 + B_1x_i + B_2x^2_i + e_i \\ i &= 1,...,10 \end{align}

Now I have:

\begin{align} Y_i &= g_0 + g_1x^*_i + g_2(x^*_i)^2 + e_i \\ x^*_i &= x_i -10 \\ i &= 1,...,10 \end{align}

To transform the model, can I decrease the matrix data for 10? For example the first row of the matrix $(X'X)^{-1}$ is:

$$ [4,\hspace{12mm} 0,\hspace{12mm} 0]\hspace{20mm} \\ 4-10,\ 0-10,\ 0-10,\ ... $$

The same thing can I make it also for the matrix $X'y$?


I have many doubts. I am publishing the text so it is more understandable.

Consider the regression model linear:

$$Y_i = B_0 + B_1x_i + B_2x^2_i + e_i$$

with $e_1, ...., e_n$ independent and identically distributed random variables and $x_i,\ i = 1, ..., 0$ constant fix.

The only data I have are these: https://s32.postimg.org/o20r1i5et/Immagine.png. I have to rewrite the whole thing, so I tried:

\begin{align} (X) &= \begin{bmatrix} 1 & x_1 & x^2_1\\ . & . & . \\ . & . & . \\ . & . & . \\ 1 & x_{10} & x^2_{10}\\ \end{bmatrix} \\[10pt] (X^*) &= \begin{bmatrix} 1 & x_1-10 & (x_{1}-10)^2\\ . & . & . \\ . & . & . \\ . & . & . \\ 1 & x_{10}-10 & (x_{10}-10)^2\\ \end{bmatrix} \\[10pt] X^{*'}X^* &= \begin{bmatrix} 10 & \sum_{i=1}^{10} x_i-10 & \sum_{i=1}^{10}(x_{i}-10)^2\\ \sum_{i=1}^{10} x_i-10 & \sum_{i=1}^{10} (x_{i}-10)^2 & \sum_{i=1}^{10}(x_{i}-10)^3\\ \sum_{i=1}^{10} (x_{i}-10)^2 & \sum_{i=1}^{10}(x_{i}-10)^3 & \sum_{i=1}^{10}(x_{i}-10)^4\\ \end{bmatrix} \\ \end{align}

Now do I have to calculate the inverse? Am I following the proper solution? Do I have to use the property for equivariance?

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1 Answer 1

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In general no.

You need to subtract 10 from each $x_i$ individually. In your model it is easy to see that:

$$ x_i^2-10 \neq (x_i - 10)^2 $$

So create the right variables, then do the multiplication.

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  • $\begingroup$ First of all thank you for your answer. I try to do it and I see if I have doubts. $\endgroup$
    – Valentina
    Commented Jun 20, 2016 at 11:34
  • $\begingroup$ I wrote the text and reported the calculations. I'm doing it wrong? $\endgroup$
    – Valentina
    Commented Jun 20, 2016 at 12:19

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