Meaning of this notion in 0-1 loss?
$begingroup$
I am reading a paper and encountered this notion:
$$1_{{Y=1}}$$
To me it seems to be the expression as below, but I am not entirely sure and I don't think the author explictly explained it:
if Y==1:
return 1
else:
return 0
Can someone help me to clarify this notion? Much thanks for your time (:
It appears in:
https://papers.nips.cc/paper/5073-learning-with-noisy-labels.pdf
machine-learning loss-function math
edited 17 hours ago
Siong Thye Goh
1,367519
asked 19 hours ago
J. DoeJ. Doe
82
New contributor
J. Doe 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 reading a paper and encountered this notion:
$$1_{{Y=1}}$$
To me it seems to be the expression as below, but I am not entirely sure and I don't think the author explictly explained it:
if Y==1:
return 1
else:
return 0
Can someone help me to clarify this notion? Much thanks for your time (:
It appears in:
https://papers.nips.cc/paper/5073-learning-with-noisy-labels.pdf
machine-learning loss-function math
edited 17 hours ago
Siong Thye Goh
1,367519
asked 19 hours ago
J. DoeJ. Doe
82
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
19 hours ago
add a comment |
$begingroup$
I am reading a paper and encountered this notion:
$$1_{{Y=1}}$$
To me it seems to be the expression as below, but I am not entirely sure and I don't think the author explictly explained it:
if Y==1:
return 1
else:
return 0
Can someone help me to clarify this notion? Much thanks for your time (:
It appears in:
https://papers.nips.cc/paper/5073-learning-with-noisy-labels.pdf
machine-learning loss-function math
edited 17 hours ago
Siong Thye Goh
1,367519
asked 19 hours ago
J. DoeJ. Doe
82
New contributor
J. Doe 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 reading a paper and encountered this notion:
$$1_{{Y=1}}$$
To me it seems to be the expression as below, but I am not entirely sure and I don't think the author explictly explained it:
if Y==1:
return 1
else:
return 0
Can someone help me to clarify this notion? Much thanks for your time (:
It appears in:
https://papers.nips.cc/paper/5073-learning-with-noisy-labels.pdf
machine-learning loss-function math
machine-learning loss-function math
edited 17 hours ago
Siong Thye Goh
1,367519
asked 19 hours ago
J. DoeJ. Doe
82
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
Siong Thye Goh
1,367519
asked 19 hours ago
J. DoeJ. Doe
82
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
Siong Thye Goh
1,367519
edited 17 hours ago
Siong Thye Goh
1,367519
edited 17 hours ago
Siong Thye Goh
1,367519
1,367519
asked 19 hours ago
J. DoeJ. Doe
82
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 19 hours ago
J. DoeJ. Doe
82
asked 19 hours ago
J. DoeJ. Doe
82
82
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
J. Doe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
19 hours ago
add a comment |
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
19 hours ago
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
19 hours ago
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
19 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your understanding is correct.
This is known as the indicator function.
The indicator function of a subset $A$ of a set $X$ is a function
$$1_A(x)= begin{cases}1, & x in A \ 0, & x notin A end{cases}$$
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
$endgroup$
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
J. Doe 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%2f47657%2fmeaning-of-this-notion-in-0-1-loss%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your understanding is correct.
This is known as the indicator function.
The indicator function of a subset $A$ of a set $X$ is a function
$$1_A(x)= begin{cases}1, & x in A \ 0, & x notin A end{cases}$$
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
$endgroup$
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
add a comment |
$begingroup$
Your understanding is correct.
This is known as the indicator function.
The indicator function of a subset $A$ of a set $X$ is a function
$$1_A(x)= begin{cases}1, & x in A \ 0, & x notin A end{cases}$$
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
$endgroup$
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
add a comment |
$begingroup$
Your understanding is correct.
This is known as the indicator function.
The indicator function of a subset $A$ of a set $X$ is a function
$$1_A(x)= begin{cases}1, & x in A \ 0, & x notin A end{cases}$$
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
$endgroup$
Your understanding is correct.
This is known as the indicator function.
The indicator function of a subset $A$ of a set $X$ is a function
$$1_A(x)= begin{cases}1, & x in A \ 0, & x notin A end{cases}$$
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
answered 17 hours ago
Siong Thye GohSiong Thye Goh
1,367519
1,367519
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
add a comment |
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
$begingroup$
Also since both $I_{{Y=1}}$ and $I_{{y=1}}$ are present in the question. It is worth noting that $Z=I_{{Y=1}}$ is a random variable like $Z=2Y$ with a distribution, unlike $f(y)=I_{{y=1}}$ which is a function with no distribution.
$endgroup$
– Esmailian
16 hours ago
add a comment |
J. Doe is a new contributor. Be nice, and check out our Code of Conduct.
J. Doe is a new contributor. Be nice, and check out our Code of Conduct.
J. Doe is a new contributor. Be nice, and check out our Code of Conduct.
J. Doe 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%2f47657%2fmeaning-of-this-notion-in-0-1-loss%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
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
19 hours ago