Why is PCA sensitive to outliers?
$begingroup$
There are many posts on this SE that discuss robust approaches to principal component analysis (PCA), but I cannot find a single good explanation of why PCA is sensitive to outliers in the first place.
machine-learning pca outliers
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
asked Nov 26 '18 at 1:59
PsiPsi
20828
$endgroup$
add a comment |
$begingroup$
There are many posts on this SE that discuss robust approaches to principal component analysis (PCA), but I cannot find a single good explanation of why PCA is sensitive to outliers in the first place.
machine-learning pca outliers
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
asked Nov 26 '18 at 1:59
PsiPsi
20828
$endgroup$
4
$begingroup$
Because L2 norm contribution is very high for outliers. Then when minimizing L2 norm (which is what PCA tries to do), those points will pull harder to fit than points closer to middle will.
$endgroup$
– mathreadler
Nov 26 '18 at 11:48
$begingroup$
This answer tells you everything you need. Just picture an outlier and read attentively.
$endgroup$
– Stephan Kolassa
Nov 26 '18 at 20:15
add a comment |
$begingroup$
There are many posts on this SE that discuss robust approaches to principal component analysis (PCA), but I cannot find a single good explanation of why PCA is sensitive to outliers in the first place.
machine-learning pca outliers
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
asked Nov 26 '18 at 1:59
PsiPsi
20828
$endgroup$
There are many posts on this SE that discuss robust approaches to principal component analysis (PCA), but I cannot find a single good explanation of why PCA is sensitive to outliers in the first place.
machine-learning pca outliers
machine-learning pca outliers
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
asked Nov 26 '18 at 1:59
PsiPsi
20828
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
asked Nov 26 '18 at 1:59
PsiPsi
20828
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
edited Dec 30 '18 at 19:39
Peter Mortensen
20328
20328
asked Nov 26 '18 at 1:59
PsiPsi
20828
asked Nov 26 '18 at 1:59
PsiPsi
20828
asked Nov 26 '18 at 1:59
PsiPsi
20828
20828
4
$begingroup$
Because L2 norm contribution is very high for outliers. Then when minimizing L2 norm (which is what PCA tries to do), those points will pull harder to fit than points closer to middle will.
$endgroup$
– mathreadler
Nov 26 '18 at 11:48
$begingroup$
This answer tells you everything you need. Just picture an outlier and read attentively.
$endgroup$
– Stephan Kolassa
Nov 26 '18 at 20:15
add a comment |
4
$begingroup$
Because L2 norm contribution is very high for outliers. Then when minimizing L2 norm (which is what PCA tries to do), those points will pull harder to fit than points closer to middle will.
$endgroup$
– mathreadler
Nov 26 '18 at 11:48
$begingroup$
This answer tells you everything you need. Just picture an outlier and read attentively.
$endgroup$
– Stephan Kolassa
Nov 26 '18 at 20:15
4
4
$begingroup$
Because L2 norm contribution is very high for outliers. Then when minimizing L2 norm (which is what PCA tries to do), those points will pull harder to fit than points closer to middle will.
$endgroup$
– mathreadler
Nov 26 '18 at 11:48
$begingroup$
Because L2 norm contribution is very high for outliers. Then when minimizing L2 norm (which is what PCA tries to do), those points will pull harder to fit than points closer to middle will.
$endgroup$
– mathreadler
Nov 26 '18 at 11:48
$begingroup$
This answer tells you everything you need. Just picture an outlier and read attentively.
$endgroup$
– Stephan Kolassa
Nov 26 '18 at 20:15
$begingroup$
This answer tells you everything you need. Just picture an outlier and read attentively.
$endgroup$
– Stephan Kolassa
Nov 26 '18 at 20:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One of the reasons is that PCA can be thought as low-rank decomposition of the data that minimizes the sum of $L_2$ norms of the residuals of the decomposition. I.e. if $Y$ is your data ($m$ vectors of $n$ dimensions), and $X$ is the PCA basis ($k$ vectors of $n$ dimensions), then the decomposition will strictly minimize
$$lVert Y-XA rVert^2_F = sum_{j=1}^{m} lVert Y_j - X A_{j.} rVert^2 $$
Here $A$ is the matrix of coefficients of PCA decomposition and $lVert
cdot rVert_F$ is a Frobenius norm of the matrix
Because the PCA minimizes the $L_2$ norms (i.e. quadratic norms) it has the same issues a least-squares or fitting a Gaussian by being sensitive to outliers. Because of the squaring of deviations from the outliers, they will dominate the total norm and therefore will drive the PCA components.
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
$endgroup$
add a comment |
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1 Answer
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$begingroup$
One of the reasons is that PCA can be thought as low-rank decomposition of the data that minimizes the sum of $L_2$ norms of the residuals of the decomposition. I.e. if $Y$ is your data ($m$ vectors of $n$ dimensions), and $X$ is the PCA basis ($k$ vectors of $n$ dimensions), then the decomposition will strictly minimize
$$lVert Y-XA rVert^2_F = sum_{j=1}^{m} lVert Y_j - X A_{j.} rVert^2 $$
Here $A$ is the matrix of coefficients of PCA decomposition and $lVert
cdot rVert_F$ is a Frobenius norm of the matrix
Because the PCA minimizes the $L_2$ norms (i.e. quadratic norms) it has the same issues a least-squares or fitting a Gaussian by being sensitive to outliers. Because of the squaring of deviations from the outliers, they will dominate the total norm and therefore will drive the PCA components.
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
$endgroup$
add a comment |
$begingroup$
One of the reasons is that PCA can be thought as low-rank decomposition of the data that minimizes the sum of $L_2$ norms of the residuals of the decomposition. I.e. if $Y$ is your data ($m$ vectors of $n$ dimensions), and $X$ is the PCA basis ($k$ vectors of $n$ dimensions), then the decomposition will strictly minimize
$$lVert Y-XA rVert^2_F = sum_{j=1}^{m} lVert Y_j - X A_{j.} rVert^2 $$
Here $A$ is the matrix of coefficients of PCA decomposition and $lVert
cdot rVert_F$ is a Frobenius norm of the matrix
Because the PCA minimizes the $L_2$ norms (i.e. quadratic norms) it has the same issues a least-squares or fitting a Gaussian by being sensitive to outliers. Because of the squaring of deviations from the outliers, they will dominate the total norm and therefore will drive the PCA components.
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
$endgroup$
add a comment |
$begingroup$
One of the reasons is that PCA can be thought as low-rank decomposition of the data that minimizes the sum of $L_2$ norms of the residuals of the decomposition. I.e. if $Y$ is your data ($m$ vectors of $n$ dimensions), and $X$ is the PCA basis ($k$ vectors of $n$ dimensions), then the decomposition will strictly minimize
$$lVert Y-XA rVert^2_F = sum_{j=1}^{m} lVert Y_j - X A_{j.} rVert^2 $$
Here $A$ is the matrix of coefficients of PCA decomposition and $lVert
cdot rVert_F$ is a Frobenius norm of the matrix
Because the PCA minimizes the $L_2$ norms (i.e. quadratic norms) it has the same issues a least-squares or fitting a Gaussian by being sensitive to outliers. Because of the squaring of deviations from the outliers, they will dominate the total norm and therefore will drive the PCA components.
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
$endgroup$
One of the reasons is that PCA can be thought as low-rank decomposition of the data that minimizes the sum of $L_2$ norms of the residuals of the decomposition. I.e. if $Y$ is your data ($m$ vectors of $n$ dimensions), and $X$ is the PCA basis ($k$ vectors of $n$ dimensions), then the decomposition will strictly minimize
$$lVert Y-XA rVert^2_F = sum_{j=1}^{m} lVert Y_j - X A_{j.} rVert^2 $$
Here $A$ is the matrix of coefficients of PCA decomposition and $lVert
cdot rVert_F$ is a Frobenius norm of the matrix
Because the PCA minimizes the $L_2$ norms (i.e. quadratic norms) it has the same issues a least-squares or fitting a Gaussian by being sensitive to outliers. Because of the squaring of deviations from the outliers, they will dominate the total norm and therefore will drive the PCA components.
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
edited Nov 26 '18 at 18:51
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
answered Nov 26 '18 at 4:40
sega_saisega_sai
830710
830710
add a comment |
add a comment |
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$begingroup$
Because L2 norm contribution is very high for outliers. Then when minimizing L2 norm (which is what PCA tries to do), those points will pull harder to fit than points closer to middle will.
$endgroup$
– mathreadler
Nov 26 '18 at 11:48
$begingroup$
This answer tells you everything you need. Just picture an outlier and read attentively.
$endgroup$
– Stephan Kolassa
Nov 26 '18 at 20:15