Isolation Forest Score Function Theory Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Moderator Election Q&A - Questionnaire 2019 Community Moderator Election ResultsScore probability groups with Random Forest in RRandom Forest variable Importance Z ScoreMinimum number of trees for Random Forest classifierIsolation Forest height limit absent in SkLearn implementationIsolation Forest Prediction Mechanics: Does it compare value with every tree (and the original training subset)?Multivariate outlier detection with isolation forest..How to detect most effective features?Random Forest vs. RainForestIsolation forest: how to deal with identical values?Isolation Forest Feature ImportanceIs there any implementation of Extended Isolation Forest algorithm in R/Python?
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Isolation Forest Score Function Theory
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsScore probability groups with Random Forest in RRandom Forest variable Importance Z ScoreMinimum number of trees for Random Forest classifierIsolation Forest height limit absent in SkLearn implementationIsolation Forest Prediction Mechanics: Does it compare value with every tree (and the original training subset)?Multivariate outlier detection with isolation forest..How to detect most effective features?Random Forest vs. RainForestIsolation forest: how to deal with identical values?Isolation Forest Feature ImportanceIs there any implementation of Extended Isolation Forest algorithm in R/Python?
$begingroup$
I am currently reading this paper on isolation forests. In the section about the score function, they mention the following. For context, $h(x)$ is definded as the path length of a data point traversing an iTree, and $n$ is the sample size used to grow the iTree.
The difficulty in deriving such a score from $h(x)$
is that while the maximum possible height of iTree grows
in the order of $n$, the average height grows in the order of
$log(n)$. Normalization of $h(x)$ by any of the above terms
is either not bounded or cannot be directly compared.
So herein lies my first question. What do they mean that the normalization of $h(x)$ by any of the above terms is either not bounded or cannot be directly compared? The final score function in this paper is given by
$$s(x,n)=2^-fracE(h(x))c(n)$$
where
$$c(n)=2H_n-1-2(n-1)/n$$
is the average path length of an unsuccessful search from BST Theory. Note how they are taking the expectation of the path. So in that case we are averaging all the values anyway, so why can we not use the growth of the average height of the tree? Additionally they don't even mention the analytical form of average height here, which from what I gather is $2sqrtpi n$ reference, though I have not read this reference thoroughly and may be mistaken.
Am I missing something here?
random-forest decision-trees anomaly-detection
asked 1 hour ago
Samyak ShahSamyak Shah
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am currently reading this paper on isolation forests. In the section about the score function, they mention the following. For context, $h(x)$ is definded as the path length of a data point traversing an iTree, and $n$ is the sample size used to grow the iTree.
The difficulty in deriving such a score from $h(x)$
is that while the maximum possible height of iTree grows
in the order of $n$, the average height grows in the order of
$log(n)$. Normalization of $h(x)$ by any of the above terms
is either not bounded or cannot be directly compared.
So herein lies my first question. What do they mean that the normalization of $h(x)$ by any of the above terms is either not bounded or cannot be directly compared? The final score function in this paper is given by
$$s(x,n)=2^-fracE(h(x))c(n)$$
where
$$c(n)=2H_n-1-2(n-1)/n$$
is the average path length of an unsuccessful search from BST Theory. Note how they are taking the expectation of the path. So in that case we are averaging all the values anyway, so why can we not use the growth of the average height of the tree? Additionally they don't even mention the analytical form of average height here, which from what I gather is $2sqrtpi n$ reference, though I have not read this reference thoroughly and may be mistaken.
Am I missing something here?
random-forest decision-trees anomaly-detection
asked 1 hour ago
Samyak ShahSamyak Shah
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am currently reading this paper on isolation forests. In the section about the score function, they mention the following. For context, $h(x)$ is definded as the path length of a data point traversing an iTree, and $n$ is the sample size used to grow the iTree.
The difficulty in deriving such a score from $h(x)$
is that while the maximum possible height of iTree grows
in the order of $n$, the average height grows in the order of
$log(n)$. Normalization of $h(x)$ by any of the above terms
is either not bounded or cannot be directly compared.
So herein lies my first question. What do they mean that the normalization of $h(x)$ by any of the above terms is either not bounded or cannot be directly compared? The final score function in this paper is given by
$$s(x,n)=2^-fracE(h(x))c(n)$$
where
$$c(n)=2H_n-1-2(n-1)/n$$
is the average path length of an unsuccessful search from BST Theory. Note how they are taking the expectation of the path. So in that case we are averaging all the values anyway, so why can we not use the growth of the average height of the tree? Additionally they don't even mention the analytical form of average height here, which from what I gather is $2sqrtpi n$ reference, though I have not read this reference thoroughly and may be mistaken.
Am I missing something here?
random-forest decision-trees anomaly-detection
asked 1 hour ago
Samyak ShahSamyak Shah
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am currently reading this paper on isolation forests. In the section about the score function, they mention the following. For context, $h(x)$ is definded as the path length of a data point traversing an iTree, and $n$ is the sample size used to grow the iTree.
The difficulty in deriving such a score from $h(x)$
is that while the maximum possible height of iTree grows
in the order of $n$, the average height grows in the order of
$log(n)$. Normalization of $h(x)$ by any of the above terms
is either not bounded or cannot be directly compared.
So herein lies my first question. What do they mean that the normalization of $h(x)$ by any of the above terms is either not bounded or cannot be directly compared? The final score function in this paper is given by
$$s(x,n)=2^-fracE(h(x))c(n)$$
where
$$c(n)=2H_n-1-2(n-1)/n$$
is the average path length of an unsuccessful search from BST Theory. Note how they are taking the expectation of the path. So in that case we are averaging all the values anyway, so why can we not use the growth of the average height of the tree? Additionally they don't even mention the analytical form of average height here, which from what I gather is $2sqrtpi n$ reference, though I have not read this reference thoroughly and may be mistaken.
Am I missing something here?
random-forest decision-trees anomaly-detection
random-forest decision-trees anomaly-detection
asked 1 hour ago
Samyak ShahSamyak Shah
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Samyak ShahSamyak Shah
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Samyak ShahSamyak Shah
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Samyak ShahSamyak Shah
1
asked 1 hour ago
Samyak ShahSamyak Shah
1
1
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Samyak Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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