find a general formula for $E(X^t)$ when X has the log-normal distribution












1














Finding the answer to this integral is looking trivial to me.



I have to start with that:



$$E(X^t)=int_0^infty x^t frac
{1}{sqrt {2pi}x}e^{-ln(x)^2/2}dx$$



I am basically stuck on this integral: with $ln x=y$



$$frac{1}{sqrt {2 pi}}int_0^infty e^{yt} e^{-y^2/2}dy$$



I know that the answer is: $$e^{frac{t^2}{2}}$$






probability probability-distributions expected-value







share|cite|improve this question




















  • 2




    Let $y=ln x-t$.
    – Jimmy R.
    Nov 29 '18 at 2:29










  • After the variable change the lower bound of the integral becomes $-infty$ and not $0$.
    – Jimmy R.
    Nov 29 '18 at 2:40
















1














Finding the answer to this integral is looking trivial to me.



I have to start with that:



$$E(X^t)=int_0^infty x^t frac
{1}{sqrt {2pi}x}e^{-ln(x)^2/2}dx$$



I am basically stuck on this integral: with $ln x=y$



$$frac{1}{sqrt {2 pi}}int_0^infty e^{yt} e^{-y^2/2}dy$$



I know that the answer is: $$e^{frac{t^2}{2}}$$






probability probability-distributions expected-value







share|cite|improve this question




















  • 2




    Let $y=ln x-t$.
    – Jimmy R.
    Nov 29 '18 at 2:29










  • After the variable change the lower bound of the integral becomes $-infty$ and not $0$.
    – Jimmy R.
    Nov 29 '18 at 2:40














1












1








1







Finding the answer to this integral is looking trivial to me.



I have to start with that:



$$E(X^t)=int_0^infty x^t frac
{1}{sqrt {2pi}x}e^{-ln(x)^2/2}dx$$



I am basically stuck on this integral: with $ln x=y$



$$frac{1}{sqrt {2 pi}}int_0^infty e^{yt} e^{-y^2/2}dy$$



I know that the answer is: $$e^{frac{t^2}{2}}$$






probability probability-distributions expected-value







share|cite|improve this question















Finding the answer to this integral is looking trivial to me.



I have to start with that:



$$E(X^t)=int_0^infty x^t frac
{1}{sqrt {2pi}x}e^{-ln(x)^2/2}dx$$



I am basically stuck on this integral: with $ln x=y$



$$frac{1}{sqrt {2 pi}}int_0^infty e^{yt} e^{-y^2/2}dy$$



I know that the answer is: $$e^{frac{t^2}{2}}$$






probability probability-distributions expected-value




probability probability-distributions expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 2:28









Jimmy R.

33k42157




33k42157










asked Nov 29 '18 at 2:19









Mahamad A. Kanouté

312111




312111








  • 2




    Let $y=ln x-t$.
    – Jimmy R.
    Nov 29 '18 at 2:29










  • After the variable change the lower bound of the integral becomes $-infty$ and not $0$.
    – Jimmy R.
    Nov 29 '18 at 2:40














  • 2




    Let $y=ln x-t$.
    – Jimmy R.
    Nov 29 '18 at 2:29










  • After the variable change the lower bound of the integral becomes $-infty$ and not $0$.
    – Jimmy R.
    Nov 29 '18 at 2:40








2




2




Let $y=ln x-t$.
– Jimmy R.
Nov 29 '18 at 2:29




Let $y=ln x-t$.
– Jimmy R.
Nov 29 '18 at 2:29












After the variable change the lower bound of the integral becomes $-infty$ and not $0$.
– Jimmy R.
Nov 29 '18 at 2:40




After the variable change the lower bound of the integral becomes $-infty$ and not $0$.
– Jimmy R.
Nov 29 '18 at 2:40










2 Answers
2






active

oldest

votes


















1














Hint:$$e^{yt}e^{-y^2/2}=e^{-frac12(y^2-2yt+t^2)+frac12t^2}=e^{t^2/2}e^{-frac12(y-t)^2}$$






share|cite|improve this answer





























    1














    (I believe you've made two small mistakes - in the second integral it should be $dy$ instead of $dx$, and the integral should be taken over $-infty$ to $infty$. Be careful when doing substitutions!)



    The trick here is in completing the square:



    $$
    begin{align}
    e^{yt} e^{-y^2/2} &= e^{-frac{1}{2} (y^2 - 2yt)} \
    &=e^{-frac{1}{2} (y^2 - 2yt+ t^2)}e^{frac{1}{2}t^2}\
    &=e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}\
    end{align}
    $$
    Putting this expression back into the original integral that you were stuck on, you'll get:
    $$
    begin{align}
    int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}dy
    &=e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
    end{align}
    $$
    Observing that what you have in the integral is the PDF of the Gaussian distribution, which integrates out to 1, you will get what you want:
    $$
    begin{align}
    e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
    end{align} = e^{frac{1}{2}t^2}
    $$






    share|cite|improve this answer





















    • you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
      – Mahamad A. Kanouté
      Nov 29 '18 at 3:10











    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%2f3018068%2ffind-a-general-formula-for-ext-when-x-has-the-log-normal-distribution%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









    1














    Hint:$$e^{yt}e^{-y^2/2}=e^{-frac12(y^2-2yt+t^2)+frac12t^2}=e^{t^2/2}e^{-frac12(y-t)^2}$$






    share|cite|improve this answer


























      1














      Hint:$$e^{yt}e^{-y^2/2}=e^{-frac12(y^2-2yt+t^2)+frac12t^2}=e^{t^2/2}e^{-frac12(y-t)^2}$$






      share|cite|improve this answer
























        1












        1








        1






        Hint:$$e^{yt}e^{-y^2/2}=e^{-frac12(y^2-2yt+t^2)+frac12t^2}=e^{t^2/2}e^{-frac12(y-t)^2}$$






        share|cite|improve this answer












        Hint:$$e^{yt}e^{-y^2/2}=e^{-frac12(y^2-2yt+t^2)+frac12t^2}=e^{t^2/2}e^{-frac12(y-t)^2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 2:31









        Jimmy R.

        33k42157




        33k42157























            1














            (I believe you've made two small mistakes - in the second integral it should be $dy$ instead of $dx$, and the integral should be taken over $-infty$ to $infty$. Be careful when doing substitutions!)



            The trick here is in completing the square:



            $$
            begin{align}
            e^{yt} e^{-y^2/2} &= e^{-frac{1}{2} (y^2 - 2yt)} \
            &=e^{-frac{1}{2} (y^2 - 2yt+ t^2)}e^{frac{1}{2}t^2}\
            &=e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}\
            end{align}
            $$
            Putting this expression back into the original integral that you were stuck on, you'll get:
            $$
            begin{align}
            int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}dy
            &=e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align}
            $$
            Observing that what you have in the integral is the PDF of the Gaussian distribution, which integrates out to 1, you will get what you want:
            $$
            begin{align}
            e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align} = e^{frac{1}{2}t^2}
            $$






            share|cite|improve this answer





















            • you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
              – Mahamad A. Kanouté
              Nov 29 '18 at 3:10
















            1














            (I believe you've made two small mistakes - in the second integral it should be $dy$ instead of $dx$, and the integral should be taken over $-infty$ to $infty$. Be careful when doing substitutions!)



            The trick here is in completing the square:



            $$
            begin{align}
            e^{yt} e^{-y^2/2} &= e^{-frac{1}{2} (y^2 - 2yt)} \
            &=e^{-frac{1}{2} (y^2 - 2yt+ t^2)}e^{frac{1}{2}t^2}\
            &=e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}\
            end{align}
            $$
            Putting this expression back into the original integral that you were stuck on, you'll get:
            $$
            begin{align}
            int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}dy
            &=e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align}
            $$
            Observing that what you have in the integral is the PDF of the Gaussian distribution, which integrates out to 1, you will get what you want:
            $$
            begin{align}
            e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align} = e^{frac{1}{2}t^2}
            $$






            share|cite|improve this answer





















            • you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
              – Mahamad A. Kanouté
              Nov 29 '18 at 3:10














            1












            1








            1






            (I believe you've made two small mistakes - in the second integral it should be $dy$ instead of $dx$, and the integral should be taken over $-infty$ to $infty$. Be careful when doing substitutions!)



            The trick here is in completing the square:



            $$
            begin{align}
            e^{yt} e^{-y^2/2} &= e^{-frac{1}{2} (y^2 - 2yt)} \
            &=e^{-frac{1}{2} (y^2 - 2yt+ t^2)}e^{frac{1}{2}t^2}\
            &=e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}\
            end{align}
            $$
            Putting this expression back into the original integral that you were stuck on, you'll get:
            $$
            begin{align}
            int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}dy
            &=e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align}
            $$
            Observing that what you have in the integral is the PDF of the Gaussian distribution, which integrates out to 1, you will get what you want:
            $$
            begin{align}
            e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align} = e^{frac{1}{2}t^2}
            $$






            share|cite|improve this answer












            (I believe you've made two small mistakes - in the second integral it should be $dy$ instead of $dx$, and the integral should be taken over $-infty$ to $infty$. Be careful when doing substitutions!)



            The trick here is in completing the square:



            $$
            begin{align}
            e^{yt} e^{-y^2/2} &= e^{-frac{1}{2} (y^2 - 2yt)} \
            &=e^{-frac{1}{2} (y^2 - 2yt+ t^2)}e^{frac{1}{2}t^2}\
            &=e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}\
            end{align}
            $$
            Putting this expression back into the original integral that you were stuck on, you'll get:
            $$
            begin{align}
            int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}e^{frac{1}{2}t^2}dy
            &=e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align}
            $$
            Observing that what you have in the integral is the PDF of the Gaussian distribution, which integrates out to 1, you will get what you want:
            $$
            begin{align}
            e^{frac{1}{2}t^2}int_{-infty}^infty frac{1}{sqrt{2pi}} e^{-frac{1}{2} (y-t)^2}dy
            end{align} = e^{frac{1}{2}t^2}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 '18 at 2:38









            Sean Lee

            1578




            1578












            • you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
              – Mahamad A. Kanouté
              Nov 29 '18 at 3:10


















            • you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
              – Mahamad A. Kanouté
              Nov 29 '18 at 3:10
















            you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
            – Mahamad A. Kanouté
            Nov 29 '18 at 3:10




            you're right about the bounds. I completely forgot about it! However, my dx and dy are correctly placed
            – Mahamad A. Kanouté
            Nov 29 '18 at 3:10


















            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3018068%2ffind-a-general-formula-for-ext-when-x-has-the-log-normal-distribution%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

            Le Mesnil-Réaume

            Image
            Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more

            web3.py web3.isConnected() returns false always

            Image
            1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
            Read more