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
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Psi
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up vote
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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 4 hours ago
Psi
1335
New contributor
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
6
down vote
favorite
up vote
6
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 4 hours ago
Psi
1335
New contributor
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 4 hours ago
Psi
1335
New contributor
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Psi
1335
New contributor
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
asked 4 hours ago
Psi
1335
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Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
Psi
1335
asked 4 hours ago
Psi
1335
1335
New contributor
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Psi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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1 Answer
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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.
edited 1 hour ago
dsaxton
9,39811535
answered 1 hour ago
sega_sai
44538
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
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.
edited 1 hour ago
dsaxton
9,39811535
answered 1 hour ago
sega_sai
44538
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
add a comment |
up vote
6
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.
edited 1 hour ago
dsaxton
9,39811535
answered 1 hour ago
sega_sai
44538
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
add a comment |
up vote
6
down vote
accepted
up vote
6
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.
edited 1 hour ago
dsaxton
9,39811535
answered 1 hour ago
sega_sai
44538
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.
edited 1 hour ago
dsaxton
9,39811535
answered 1 hour ago
sega_sai
44538
edited 1 hour ago
dsaxton
9,39811535
edited 1 hour ago
dsaxton
9,39811535
edited 1 hour ago
dsaxton
9,39811535
9,39811535
answered 1 hour ago
sega_sai
44538
answered 1 hour ago
sega_sai
44538
answered 1 hour ago
sega_sai
44538
44538
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
add a comment |
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
Thank you so much, what an awesome reply! This was exactly what I was looking for and everything makes so much sense the way you explained it!!
– Psi
1 hour ago
add a comment |
Psi is a new contributor. Be nice, and check out our Code of Conduct.
Psi is a new contributor. Be nice, and check out our Code of Conduct.
Psi is a new contributor. Be nice, and check out our Code of Conduct.
Psi is a new contributor. Be nice, and check out our Code of Conduct.
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