Do error bars on probabilities have any meaning?
$begingroup$
People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
asked 10 hours ago
mahnamahnamahnamahna
311
New contributor
mahnamahna 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 |
$begingroup$
People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
asked 10 hours ago
mahnamahnamahnamahna
311
New contributor
mahnamahna 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 |
$begingroup$
People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
asked 10 hours ago
mahnamahnamahnamahna
311
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
probability
asked 10 hours ago
mahnamahnamahnamahna
311
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
mahnamahnamahnamahna
311
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
mahnamahnamahnamahna
311
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
mahnamahnamahnamahna
311
asked 10 hours ago
mahnamahnamahnamahna
311
311
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mahnamahna 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 |
2 Answers
2
active
oldest
votes
$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 10 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 9 hours ago
Xi'anXi'an
57k895357
$endgroup$
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins 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: "65"
};
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
});
}
});
mahnamahna 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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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%2fstats.stackexchange.com%2fquestions%2f393316%2fdo-error-bars-on-probabilities-have-any-meaning%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 10 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
add a comment |
$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 10 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
add a comment |
$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 10 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 10 hours ago
Tim♦Tim
57.5k9126218
answered 10 hours ago
Tim♦Tim
57.5k9126218
answered 10 hours ago
Tim♦Tim
57.5k9126218
answered 10 hours ago
Tim♦Tim
57.5k9126218
57.5k9126218
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
add a comment |
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
7 hours ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
$begingroup$
Even "known" probabilities can be shorthand for very small error bars. One could presumably show that a coin flip is perhaps 50.00001%- 49.99999% with enough trials to get small enough error bars that exclude 50.00000%. There's no physical law suggesting the odds should be precisely even for an asymmetrical coin, but the error bars are far too small for anyone to care.
$endgroup$
– Nuclear Wang
1 hour ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 9 hours ago
Xi'anXi'an
57k895357
$endgroup$
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 9 hours ago
Xi'anXi'an
57k895357
$endgroup$
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 9 hours ago
Xi'anXi'an
57k895357
$endgroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 9 hours ago
Xi'anXi'an
57k895357
answered 9 hours ago
Xi'anXi'an
57k895357
answered 9 hours ago
Xi'anXi'an
57k895357
answered 9 hours ago
Xi'anXi'an
57k895357
57k895357
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins ago
add a comment |
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins ago
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
4 hours ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins ago
$begingroup$
Joke apart, this means that the precision of the error bars is uncertain and that the evaluation of the uncertainty is itself uncertain, in an infinite regress.
$endgroup$
– Xi'an
10 mins ago
add a comment |
mahnamahna is a new contributor. Be nice, and check out our Code of Conduct.
mahnamahna is a new contributor. Be nice, and check out our Code of Conduct.
mahnamahna is a new contributor. Be nice, and check out our Code of Conduct.
mahnamahna is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Cross Validated!
- 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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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%2fstats.stackexchange.com%2fquestions%2f393316%2fdo-error-bars-on-probabilities-have-any-meaning%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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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');
var $window = $(window),
onScroll = function(e) {
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom)) {
StackExchange.using('gps', function() { StackExchange.gps.track('embedded_signup_form.view', { location: 'question_page' }); });
$window.unbind('scroll', onScroll);
}
};
$window.on('scroll', onScroll);
});
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
