deleted a line

2021-09-06 22:05:17 +02:00
parent 65052fe79a
commit 71f63a717e
7 changed files with 1191 additions and 15 deletions
@@ -2436,9 +2436,7 @@ $\frac{\sigma_j^2}{\sigma_j^2+\lambda}$. Recall that the SVD has
 eigenvalues ordered in a descending way, that is $\sigma_i \geq
 \sigma_{i+1}$.

-For small eigenvalues $\sigma_i$ it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom.
-Actually, calculating the variance of $\bm{X}\bm{v}_j$ shows that this quantity is equal to $\sigma_j^2/n$.
-With a parameter $\lambda$ we can thus shrink the role of specific parameters.  
+For small eigenvalues $\sigma_i$ it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom. More about this when we have covered the material on a statistical interpretation of various linear regression methods.


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