Ex. 14.1
Ex. 14.1
Weights for clustering. Show that weighted Euclidean distance
\[\begin{equation}
d_e^{(w)}(x_i, x_{i'}) = \frac{\sum_{l=1}^pw_l(x_{il}-x_{i'l})^2}{\sum_{l=1}^pw_l}\non
\end{equation}\]
satisfies
\[\begin{equation}
d_e^{(w)}(x_i, x_{i'}) = d_e(z_i, z_{i'}) = \sum_{l=1}^p(z_{il}-z_{i'l})^2,\non
\end{equation}\]
where
\[\begin{equation}
z_{il} = x_{il}\cdot \left(\frac{w_l}{\sum_{l=1}^pw_l}\right)^{1/2}.\non
\end{equation}\]
Thus weighted Euclidean distance based on \(x\) is equivalent to unweighted Euclidean distance based on \(z\).
Soln. 14.1
By definition of \(z_{il}\) we have
\[\begin{eqnarray}
d_e(z_i, z_{i'}) &=& \sum_{l=1}^p \left(\frac{w_l}{\sum_{l=1}^pw_l}\right)(x_{il}-x_{i'l})^2\non\\
&=&\frac{\sum_{l=1}^pw_l(x_{il}-x_{i'l})^2}{\sum_{l=1}^pw_l}\non\\
&=&d_e^{(w)}(x_i, x_{i'}).\non
\end{eqnarray}\]