What is the purpose of the constant in the probability density function The 2019 Stack Overflow Developer Survey Results Are InConfusion between probability distribution function and probability density functionProbability density function vs. probability mass functionBound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionCan a probability density function be used directly as probability function?Probability density of a function of a random variableis this function increasing or decreasing on what intervals?Homework: questions about probability distribution functions and probability density functionGaussian function constantDeriving the Covariance of Multivariate Gaussianprobability density function of a function of a random variable?
Multi tool use
What is the use of option -o in the useradd command?
How to deal with fear of taking dependencies
Inflated grade on resume at previous job, might former employer tell new employer?
If a poisoned arrow's piercing damage is reduced to 0, do you still get poisoned?
"Riffle" two strings
CiviEvent: Public link for events of a specific type
Does a dangling wire really electrocute me if I'm standing in water?
Carnot-Caratheodory metric
How come people say “Would of”?
Is bread bad for ducks?
Output the Arecibo Message
"To split hairs" vs "To be pedantic"
Monty Hall variation
Does duplicating a spell with Wish count as casting that spell?
Is there a name of the flying bionic bird?
What does "rabbited" mean/imply in this sentence?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Why is it "Tumoren" and not "Tumore"?
Is domain driven design an anti-SQL pattern?
Manuscript was "unsubmitted" because the manuscript was deposited in Arxiv Preprints
Pristine Bit Checking
Why do UK politicians seemingly ignore opinion polls on Brexit?
Why can Shazam do this?
What is the meaning of Triage in Cybersec world?
What is the purpose of the constant in the probability density function
The 2019 Stack Overflow Developer Survey Results Are InConfusion between probability distribution function and probability density functionProbability density function vs. probability mass functionBound 1D gaussian domain in the interval $[-3sigma, 3sigma]$ so it still is a probability density functionCan a probability density function be used directly as probability function?Probability density of a function of a random variableis this function increasing or decreasing on what intervals?Homework: questions about probability distribution functions and probability density functionGaussian function constantDeriving the Covariance of Multivariate Gaussianprobability density function of a function of a random variable?
$begingroup$
I have been studying the probability density function...
$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$
For now I remove the constant, and using the following proof, I prove that...
$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$
The way I interpret this is that the area under the gaussian distribution is $sqrt2 pi $. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?
probability statistics probability-distributions gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
$endgroup$
add a comment |
$begingroup$
I have been studying the probability density function...
$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$
For now I remove the constant, and using the following proof, I prove that...
$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$
The way I interpret this is that the area under the gaussian distribution is $sqrt2 pi $. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?
probability statistics probability-distributions gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
$endgroup$
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
add a comment |
$begingroup$
I have been studying the probability density function...
$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$
For now I remove the constant, and using the following proof, I prove that...
$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$
The way I interpret this is that the area under the gaussian distribution is $sqrt2 pi $. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?
probability statistics probability-distributions gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
$endgroup$
I have been studying the probability density function...
$$frac1sigma sqrt2 pie^frac(-(x - mu ))^22sigma ^2$$
For now I remove the constant, and using the following proof, I prove that...
$$int_-infty^inftye^frac-x^22 = sqrt2 pi $$
The way I interpret this is that the area under the gaussian distribution is $sqrt2 pi $. But I am still having a hard time figuring out what the constant is doing. It seems to divide by the area itself and by $sigma$ as well. Why is this done?
probability statistics probability-distributions gaussian-integral
probability statistics probability-distributions gaussian-integral
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
asked 1 hour ago
BolboaBolboa
398516
398516
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
add a comment |
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
1
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
1
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
add a comment |
$begingroup$
It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)
answered 1 hour ago
GReyesGReyes
2,39815
$endgroup$
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: "69"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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%2fmath.stackexchange.com%2fquestions%2f3181774%2fwhat-is-the-purpose-of-the-constant-in-the-probability-density-function%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
$begingroup$
If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
1
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
add a comment |
$begingroup$
If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
1
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
add a comment |
$begingroup$
If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
$endgroup$
If you consider every possible outcome of some event you should expect the probability of it happening to be $1$, not $sqrt2pi$ so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one.
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
answered 1 hour ago
CyclotomicFieldCyclotomicField
2,4681314
2,4681314
1
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
add a comment |
1
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
1
1
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
$begingroup$
(In case its not clear, which it probably is) More precisely, you want the sum of the probabilities of all possible events to equal $1$, and since an integral represents a sum over such continuous variables, that's what this is.
$endgroup$
– John Doe
1 hour ago
add a comment |
$begingroup$
It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)
answered 1 hour ago
GReyesGReyes
2,39815
$endgroup$
add a comment |
$begingroup$
It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)
answered 1 hour ago
GReyesGReyes
2,39815
$endgroup$
add a comment |
$begingroup$
It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)
answered 1 hour ago
GReyesGReyes
2,39815
$endgroup$
It is doing that, but observe that you are also stretching in the horizontal direction by the same factor (in the exponential). Say if $sigma>1$ you are decreasing your area by a factor $sigma$ but you are increasing it by the same factor because you replace $x$ by $x/sigma$ (the shift does not change the area of course)
answered 1 hour ago
GReyesGReyes
2,39815
answered 1 hour ago
GReyesGReyes
2,39815
answered 1 hour ago
GReyesGReyes
2,39815
answered 1 hour ago
GReyesGReyes
2,39815
2,39815
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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%2fmath.stackexchange.com%2fquestions%2f3181774%2fwhat-is-the-purpose-of-the-constant-in-the-probability-density-function%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
L,hsjcskoCKPaaFcPH,2OqxbKpqg7EN4leH4O75,jn,mvOtuYwED k,his lbQ8V EKjpTgv4i
1
$begingroup$
so the integral of the probability density function over the entire space is equal to one
$endgroup$
– J. W. Tanner
1 hour ago