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





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    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


















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