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Given the following data array :

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline J/I&1 & 2 & 3 & 4 & 5 & 6\\ \hline x & 1 & 0 & 0 & 2 & 1 & 2\\ y & 0 & 0 & 1 & 2 & 0 & 3\\ z & 0 & 1 & 2 & 1 & 0 & 2\\ \hline \end{array}$$

I can get the following values for the centered data $Y$ along with the variance $$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline &1 & 2 & 3 & 4 & 5 & 6 & v_2 & v_3\\ \hline \mbox{x }&1 & 0 & 0 & 2 & 1 & 2 & 1/\sqrt{6} & 1/\sqrt{2}\\ \mbox{y }&0 & 0 & 1 & 2 & 0 & 3 & 2/\sqrt{6} & 0\\ \mbox{z }&0 & 1 & 2 & 1 & 0 & 2 & 1/\sqrt{6} & -1/\sqrt{2}\\ \hline \end{array}$$

From there I can get the principal components value :

\begin{align} Vv_2&= \begin{pmatrix}4 & 4 & 0\\ 4 & 8 & 4\\ 0 & 4 & 4\end{pmatrix} \frac{1}{\sqrt 6}\begin{pmatrix} 1\\2\\1 \end{pmatrix}\\ &=\frac{2}{\sqrt 6} \begin{pmatrix} 1\\2\\1 \end{pmatrix} \end{align}

$PC_1$ value is therefore $2$.

\begin{align} Vv_3&= \begin{pmatrix}4 & 4 & 0\\ 4 & 8 & 4\\ 0 & 4 & 4\end{pmatrix} \frac{1}{\sqrt 2}\begin{pmatrix} 1\\0\\-1 \end{pmatrix}\\ &=\frac{2}{3\sqrt 2} \begin{pmatrix} 1\\0\\-1 \end{pmatrix} \end{align}

$PC_1$ value is therefore $\frac{2}{3}$

$$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline &1 & 2 & 3 & 4 & 5 & 6 & v_2 & v_3\\ \hline \mbox{value $PC_1$ }&&&&&&&2&\\ \mbox{Value $PC_2$ }&&&&&&&&2/3\\ \mbox{value coef INR }&&&&&&&&\\ \mbox{value coef CTR }&&&&&&&&\\ \mbox{value coef COR }&&&&&&&&\\ \hline \end{array}$$

How to get the INR (I think it's a French acronym) the contribution of an individual $x_i$ to the total inertia $I_T$:

$$INR(i)=\frac{p_id(0,y_i)^2}{I_T}$$

With $d$ being usually the euclidean distance. We can deduce from the definition that $\sum INR(i)=1$.

\begin{align} INR(1) &= \frac{1}{6}\times\frac{(-1)²+1²}{2+\frac{2}{3}} = \frac{1}{4}\\ INR(2) &= \frac{1}{6}\times\frac{(-1)²+(-1)²}{2+\frac{2}{3}} = \frac{1}{4}\\ INR(3) &= \frac{1}{6}\times\frac{1²+1²}{2+\frac{2}{3}} = \frac{1}{4}\\ INR(4) &= \frac{1}{6}\times\frac{1²+2²+1²}{2+\frac{2}{3}} = \frac{2}{3}???\\ \end{align}

It would be much more than $1$ now.

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  • $\begingroup$ I don't understand anything here, starting from the second table. What are v2 and v3? $\endgroup$ – amoeba May 2 '17 at 20:50
  • $\begingroup$ @amoeba they are the eigen vector of $Y$ because $v_1=\vec 0$ $\endgroup$ – Revolucion for Monica May 2 '17 at 20:59
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If I'm reading your question correctly, you're looking for inertia around an arbitrary point in your data cloud. The can be formulated as: $I_g-||\bar x-a||^2$. where. $I_g$ here is total inertia, $a$ is particular point in question.

Here's a source on that: with derivation included. Page 8-9. https://cedric.cnam.fr/fichiers/art_1827.pdf

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