Why is PCA sensitive to outliers?
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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
asked Nov 26 at 1:59
Psi
20817
add a comment |
up vote
21
down vote
favorite
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
asked Nov 26 at 1:59
Psi
20817
4
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.
– mathreadler
Nov 26 at 11:48
This answer tells you everything you need. Just picture an outlier and read attentively.
– Stephan Kolassa
Nov 26 at 20:15
add a comment |
up vote
21
down vote
favorite
up vote
21
down vote
favorite
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
asked Nov 26 at 1:59
Psi
20817
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
asked Nov 26 at 1:59
Psi
20817
asked Nov 26 at 1:59
Psi
20817
edited Nov 26 at 2:07
asked Nov 26 at 1:59
Psi
20817
asked Nov 26 at 1:59
Psi
20817
asked Nov 26 at 1:59
Psi
20817
20817
4
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.
– mathreadler
Nov 26 at 11:48
This answer tells you everything you need. Just picture an outlier and read attentively.
– Stephan Kolassa
Nov 26 at 20:15
add a comment |
4
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.
– mathreadler
Nov 26 at 11:48
This answer tells you everything you need. Just picture an outlier and read attentively.
– Stephan Kolassa
Nov 26 at 20:15
4
4
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.
– mathreadler
Nov 26 at 11:48
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.
– mathreadler
Nov 26 at 11:48
This answer tells you everything you need. Just picture an outlier and read attentively.
– Stephan Kolassa
Nov 26 at 20:15
This answer tells you everything you need. Just picture an outlier and read attentively.
– Stephan Kolassa
Nov 26 at 20:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
30
down vote
accepted
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 at 4:40
sega_sai
750610
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
30
down vote
accepted
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 at 4:40
sega_sai
750610
add a comment |
up vote
30
down vote
accepted
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 at 4:40
sega_sai
750610
add a comment |
up vote
30
down vote
accepted
up vote
30
down vote
accepted
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 at 4:40
sega_sai
750610
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 at 4:40
sega_sai
750610
edited Nov 26 at 18:51
answered Nov 26 at 4:40
sega_sai
750610
answered Nov 26 at 4:40
sega_sai
750610
answered Nov 26 at 4:40
sega_sai
750610
750610
add a comment |
add a comment |
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4
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.
– mathreadler
Nov 26 at 11:48
This answer tells you everything you need. Just picture an outlier and read attentively.
– Stephan Kolassa
Nov 26 at 20:15