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?

How are circuits which use complex ICs normally simulated?

What tool would a Roman-age civilization have to grind silver and other metals into dust?

On the insanity of kings as an argument against monarchy

How to make payment on the internet without leaving a money trail?

Time travel alters history but people keep saying nothing's changed

What is the use of option -o in the useradd command?

What effect does the “loading” weapon property have in practical terms?

Inline version of a function returns different value than non-inline version

Are USB sockets on wall outlets live all the time, even when the switch is off?

Does light intensity oscillate really fast since it is a wave?

Is "plugging out" electronic devices an American expression?

What is the steepest angle that a canal can be traversable without locks?

How long do I have to send payment?

Should I use my personal or workplace e-mail when registering to external websites for work purpose?

Inversion Puzzle

How come people say “Would of”?

What is the best strategy for white in this position?

Is it possible for the two major parties in the UK to form a coalition with each other instead of a much smaller party?

Is bread bad for ducks?

Why could you hear an Amstrad CPC working?

What does "rabbited" mean/imply in this sentence?

Limit the amount of RAM Mathematica may access?

Is there a name of the flying bionic bird?

Why can Shazam do this?



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?










1












$begingroup$


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?










share|cite|improve this question









$endgroup$







  • 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
















1












$begingroup$


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?










share|cite|improve this question









$endgroup$







  • 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














1












1








1





$begingroup$


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?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours 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













  • 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











2 Answers
2






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 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



















2












$begingroup$

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)






share|cite|improve this answer









$endgroup$













    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
    );



    );













    draft saved

    draft discarded


















    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









    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 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
















    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 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














    2












    2








    2





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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













    • 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












    2












    $begingroup$

    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)






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      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)






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        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)






        share|cite|improve this answer









        $endgroup$



        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)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        GReyesGReyes

        2,39815




        2,39815



























            draft saved

            draft discarded
















































            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.




            draft saved


            draft discarded














            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





















































            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







            Popular posts from this blog

            Oświęcim Innehåll Historia | Källor | Externa länkar | Navigeringsmeny50°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.2213950°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.221393089658Nordisk familjebok, AuschwitzInsidan tro och existensJewish Community i OświęcimAuschwitz Jewish Center: MuseumAuschwitz Jewish Center

            Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

            Typsetting diagram chases (with TikZ?) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to define the default vertical distance between nodes?Draw edge on arcNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to place nodes in an absolute coordinate system in tikzCommutative diagram with curve connecting between nodesTikz with standalone: pinning tikz coordinates to page cmDrawing a Decision Diagram with Tikz and layout manager