Hfrxdjt

What is the correct way to use the pinch test for dehydration?

I am not a queen, who am I?

Is the Standard Deduction better than Itemized when both are the same amount?

If a contract sometimes uses the wrong name, is it still valid?

How can I fade player when goes inside or outside of the area?

Did Kevin spill real chili?

Antler Helmet: Can it work?

Do you forfeit tax refunds/credits if you aren't required to and don't file by April 15?

Can a non-EU citizen traveling with me come with me through the EU passport line?

How to find all the available tools in macOS terminal?

Is there a "higher Segal conjecture"?

Is it true to say that an hosting provider's DNS server is what links the entire hosting environment to ICANN?

How can players work together to take actions that are otherwise impossible?

Is 1 ppb equal to 1 μg/kg?

Do I really need recursive chmod to restrict access to a folder?

What causes the vertical darker bands in my photo?

What is a Meta algorithm?

3 doors, three guards, one stone

Why is black pepper both grey and black?

How much radiation do nuclear physics experiments expose researchers to nowadays?

When is phishing education going too far?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?

What's the purpose of writing one's academic bio in 3rd person?

Isolation Forest Score Function Theory

0

$begingroup$

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

share|improve this question

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 | 

0

$begingroup$

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

share|improve this question

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

share|improve this question

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.

$endgroup$

share|improve this question

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.

share|improve this question

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