Meaning of this notion in 0-1 loss?Prove Reccurrent Neural Network can exhibit oscillatory behaviorPerformance and architecture of neural network for increased dimensionsReinforcement learning: understanding this derivation of n-step Tree Backup algorithmUsing ML to create unique descriptors?L2 loss vs. mean squared lossAn unbiased simulator for policy simulation in reinforcement learningIs it possible to use NEAT networks for solving video games?Why is my loss function for DQN converging too quickly?Why Gradient methods work in finding the parameters in Neural Networks?Splitting image dataset with few subjects but many data
Store Credit Card Information in Password Manager?
Did arcade monitors have same pixel aspect ratio as TV sets?
Biological Blimps: Propulsion
Need a math help for the Cagan's model in macroeconomics
Argument list too long when zipping large list of certain files in a folder
On a tidally locked planet, would time be quantized?
Closed-form expression for certain product
When were female captains banned from Starfleet?
What linear sensor for a keybaord?
Reply 'no position' while the job posting is still there
Is there a single word describing earning money through any means?
Calculating Wattage for Resistor in High Frequency Application?
MTG Artifact and Enchantment Rulings
Offered money to buy a house, seller is asking for more to cover gap between their listing and mortgage owed
Drawing ramified coverings with tikz
Is it improper etiquette to ask your opponent what his/her rating is before the game?
Did US corporations pay demonstrators in the German demonstrations against article 13?
Can somebody explain the brexit thing in one or two child-proof sentences?
How should I respond when I lied about my education and the company finds out through background check?
Is it better practice to read straight from sheet music rather than memorize it?
The screen of my macbook suddenly broken down how can I do to recover
Open a doc from terminal, but not by its name
Could Intel SGX be dangerous under Linux?
How is flyblackbird.com operating under Part 91K?
Meaning of this notion in 0-1 loss?
Prove Reccurrent Neural Network can exhibit oscillatory behaviorPerformance and architecture of neural network for increased dimensionsReinforcement learning: understanding this derivation of n-step Tree Backup algorithmUsing ML to create unique descriptors?L2 loss vs. mean squared lossAn unbiased simulator for policy simulation in reinforcement learningIs it possible to use NEAT networks for solving video games?Why is my loss function for DQN converging too quickly?Why Gradient methods work in finding the parameters in Neural Networks?Splitting image dataset with few subjects but many data
$begingroup$
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 Mar 20 at 11:57
Siong Thye Goh
1,387519
asked Mar 20 at 9:49
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 Mar 20 at 11:57
Siong Thye Goh
1,387519
asked Mar 20 at 9:49
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
Mar 20 at 10:19
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 Mar 20 at 11:57
Siong Thye Goh
1,387519
asked Mar 20 at 9:49
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 Mar 20 at 11:57
Siong Thye Goh
1,387519
asked Mar 20 at 9:49
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 Mar 20 at 11:57
Siong Thye Goh
1,387519
asked Mar 20 at 9:49
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 Mar 20 at 11:57
Siong Thye Goh
1,387519
edited Mar 20 at 11:57
Siong Thye Goh
1,387519
edited Mar 20 at 11:57
Siong Thye Goh
1,387519
1,387519
asked Mar 20 at 9:49
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 Mar 20 at 9:49
J. DoeJ. Doe
82
asked Mar 20 at 9:49
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
Mar 20 at 10:19
add a comment |
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
Mar 20 at 10:19
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
Mar 20 at 10:19
$begingroup$
Your assumption is right.
$endgroup$
– Alireza Zolanvari
Mar 20 at 10:19
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
$$1_A(x)= begincases1, & x in A \ 0, & x notin A endcases$$
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
$endgroup$
$begingroup$
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
Mar 20 at 13:04
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
$$1_A(x)= begincases1, & x in A \ 0, & x notin A endcases$$
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
$endgroup$
$begingroup$
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
Mar 20 at 13:04
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
$$1_A(x)= begincases1, & x in A \ 0, & x notin A endcases$$
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
$endgroup$
$begingroup$
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
Mar 20 at 13:04
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
$$1_A(x)= begincases1, & x in A \ 0, & x notin A endcases$$
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
$endgroup$
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
$$1_A(x)= begincases1, & x in A \ 0, & x notin A endcases$$
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
answered Mar 20 at 11:56
Siong Thye GohSiong Thye Goh
1,387519
1,387519
$begingroup$
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
Mar 20 at 13:04
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
Mar 20 at 13:04
$begingroup$
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
Mar 20 at 13:04
$begingroup$
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
Mar 20 at 13:04
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
Mar 20 at 10:19