Isolation Forest Score Function Theory
$begingroup$
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
$$s(x,n)=2^{-frac{E(h(x))}{c(n)}}$$
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 $2sqrt{pi n}$ 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 5 mins 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
$$s(x,n)=2^{-frac{E(h(x))}{c(n)}}$$
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 $2sqrt{pi n}$ 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 5 mins 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
$$s(x,n)=2^{-frac{E(h(x))}{c(n)}}$$
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 $2sqrt{pi n}$ 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 5 mins 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
$$s(x,n)=2^{-frac{E(h(x))}{c(n)}}$$
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 $2sqrt{pi n}$ 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 5 mins 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 5 mins 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 5 mins 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 5 mins ago
Samyak ShahSamyak Shah
1
asked 5 mins 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.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "557"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Samyak Shah is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f49362%2fisolation-forest-score-function-theory%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Samyak Shah is a new contributor. Be nice, and check out our Code of Conduct.
Samyak Shah is a new contributor. Be nice, and check out our Code of Conduct.
Samyak Shah is a new contributor. Be nice, and check out our Code of Conduct.
Samyak Shah is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Data Science Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f49362%2fisolation-forest-score-function-theory%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
