For which distribution same pdf is generated for given random variable





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For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)







statistics







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$endgroup$










  • 5




    $begingroup$
    It is unclear from what you are actually asking.
    $endgroup$
    – 1123581321
    May 27 at 8:10






  • 1




    $begingroup$
    @1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
    $endgroup$
    – Henry
    May 27 at 9:03


















1












$begingroup$



For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)







statistics







share|cite|improve this question











$endgroup$










  • 5




    $begingroup$
    It is unclear from what you are actually asking.
    $endgroup$
    – 1123581321
    May 27 at 8:10






  • 1




    $begingroup$
    @1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
    $endgroup$
    – Henry
    May 27 at 9:03














1












1








1





$begingroup$



For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)







statistics







share|cite|improve this question











$endgroup$





For which of the distribution same pdf is generated for random variable X and 1/X.
Is it F(2,2)







statistics




statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 27 at 8:13







ten do

















asked May 27 at 8:08









ten doten do

415 bronze badges




415 bronze badges











  • 5




    $begingroup$
    It is unclear from what you are actually asking.
    $endgroup$
    – 1123581321
    May 27 at 8:10






  • 1




    $begingroup$
    @1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
    $endgroup$
    – Henry
    May 27 at 9:03














  • 5




    $begingroup$
    It is unclear from what you are actually asking.
    $endgroup$
    – 1123581321
    May 27 at 8:10






  • 1




    $begingroup$
    @1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
    $endgroup$
    – Henry
    May 27 at 9:03








5




5




$begingroup$
It is unclear from what you are actually asking.
$endgroup$
– 1123581321
May 27 at 8:10




$begingroup$
It is unclear from what you are actually asking.
$endgroup$
– 1123581321
May 27 at 8:10




1




1




$begingroup$
@1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
$endgroup$
– Henry
May 27 at 9:03




$begingroup$
@1123581321 Suppose $X$ had density function $f(x)$ and $Y=frac1X$ with density function $g(y)$. The question wants to consider cases where $g(y) = f(y)$
$endgroup$
– Henry
May 27 at 9:03










2 Answers
2






active

oldest

votes


















5














$begingroup$

Let $U,V$ be iid random variables with $P(U=0)=0$.



Then if $X$ is defined as $frac{U}{V}$ it will have the same distribution as $frac1{X}=frac{V}{U}$.



If moreover $X$ has a PDF then $frac1{X}$ will also have the same PDF.



Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.






share|cite|improve this answer









$endgroup$















  • $begingroup$
    It would be interesting if all solutions could be decomposed into a ratio like this
    $endgroup$
    – Henry
    May 27 at 9:23










  • $begingroup$
    @Henry Indeed. Uptil now I did not manage to prove that.
    $endgroup$
    – drhab
    May 27 at 9:30










  • $begingroup$
    @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
    $endgroup$
    – Eric Towers
    May 27 at 17:33



















5














$begingroup$

You are looking for a solution to $f(x) = dfrac{f(1/x)}{x^2}$



So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_{-1}^{1} g(x), dx$ is positive and finite



then a solution will be




  • $f(x) = dfrac{g(x)}{2k} text { when } -1 le x le 1$

  • $f(x) = dfrac{g(1/x)}{2kx^2} text { when } x lt -1 text{ or } x gt 1$


and I think all solutions will essentially be of this form



One solution is $f(x)=dfrac{1}{(1+x)^2}$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution



but there are many others, including the Cauchy density $f(x)=dfrac{1}{pi(1+x^2)}$ on $x in mathbb R$



Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac1{4x^2}$ otherwise.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    $begingroup$

    Let $U,V$ be iid random variables with $P(U=0)=0$.



    Then if $X$ is defined as $frac{U}{V}$ it will have the same distribution as $frac1{X}=frac{V}{U}$.



    If moreover $X$ has a PDF then $frac1{X}$ will also have the same PDF.



    Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.






    share|cite|improve this answer









    $endgroup$















    • $begingroup$
      It would be interesting if all solutions could be decomposed into a ratio like this
      $endgroup$
      – Henry
      May 27 at 9:23










    • $begingroup$
      @Henry Indeed. Uptil now I did not manage to prove that.
      $endgroup$
      – drhab
      May 27 at 9:30










    • $begingroup$
      @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
      $endgroup$
      – Eric Towers
      May 27 at 17:33
















    5














    $begingroup$

    Let $U,V$ be iid random variables with $P(U=0)=0$.



    Then if $X$ is defined as $frac{U}{V}$ it will have the same distribution as $frac1{X}=frac{V}{U}$.



    If moreover $X$ has a PDF then $frac1{X}$ will also have the same PDF.



    Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.






    share|cite|improve this answer









    $endgroup$















    • $begingroup$
      It would be interesting if all solutions could be decomposed into a ratio like this
      $endgroup$
      – Henry
      May 27 at 9:23










    • $begingroup$
      @Henry Indeed. Uptil now I did not manage to prove that.
      $endgroup$
      – drhab
      May 27 at 9:30










    • $begingroup$
      @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
      $endgroup$
      – Eric Towers
      May 27 at 17:33














    5














    5










    5







    $begingroup$

    Let $U,V$ be iid random variables with $P(U=0)=0$.



    Then if $X$ is defined as $frac{U}{V}$ it will have the same distribution as $frac1{X}=frac{V}{U}$.



    If moreover $X$ has a PDF then $frac1{X}$ will also have the same PDF.



    Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.






    share|cite|improve this answer









    $endgroup$



    Let $U,V$ be iid random variables with $P(U=0)=0$.



    Then if $X$ is defined as $frac{U}{V}$ it will have the same distribution as $frac1{X}=frac{V}{U}$.



    If moreover $X$ has a PDF then $frac1{X}$ will also have the same PDF.



    Special case: $U$ has chi-squared distribution. Then $X$ has $F$-distribution.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 27 at 9:00









    drhabdrhab

    114k5 gold badges51 silver badges143 bronze badges




    114k5 gold badges51 silver badges143 bronze badges















    • $begingroup$
      It would be interesting if all solutions could be decomposed into a ratio like this
      $endgroup$
      – Henry
      May 27 at 9:23










    • $begingroup$
      @Henry Indeed. Uptil now I did not manage to prove that.
      $endgroup$
      – drhab
      May 27 at 9:30










    • $begingroup$
      @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
      $endgroup$
      – Eric Towers
      May 27 at 17:33


















    • $begingroup$
      It would be interesting if all solutions could be decomposed into a ratio like this
      $endgroup$
      – Henry
      May 27 at 9:23










    • $begingroup$
      @Henry Indeed. Uptil now I did not manage to prove that.
      $endgroup$
      – drhab
      May 27 at 9:30










    • $begingroup$
      @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
      $endgroup$
      – Eric Towers
      May 27 at 17:33
















    $begingroup$
    It would be interesting if all solutions could be decomposed into a ratio like this
    $endgroup$
    – Henry
    May 27 at 9:23




    $begingroup$
    It would be interesting if all solutions could be decomposed into a ratio like this
    $endgroup$
    – Henry
    May 27 at 9:23












    $begingroup$
    @Henry Indeed. Uptil now I did not manage to prove that.
    $endgroup$
    – drhab
    May 27 at 9:30




    $begingroup$
    @Henry Indeed. Uptil now I did not manage to prove that.
    $endgroup$
    – drhab
    May 27 at 9:30












    $begingroup$
    @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
    $endgroup$
    – Eric Towers
    May 27 at 17:33




    $begingroup$
    @Rahul : Ah. Slow about an obvious. <sigh> Thanks.
    $endgroup$
    – Eric Towers
    May 27 at 17:33













    5














    $begingroup$

    You are looking for a solution to $f(x) = dfrac{f(1/x)}{x^2}$



    So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_{-1}^{1} g(x), dx$ is positive and finite



    then a solution will be




    • $f(x) = dfrac{g(x)}{2k} text { when } -1 le x le 1$

    • $f(x) = dfrac{g(1/x)}{2kx^2} text { when } x lt -1 text{ or } x gt 1$


    and I think all solutions will essentially be of this form



    One solution is $f(x)=dfrac{1}{(1+x)^2}$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution



    but there are many others, including the Cauchy density $f(x)=dfrac{1}{pi(1+x^2)}$ on $x in mathbb R$



    Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac1{4x^2}$ otherwise.






    share|cite|improve this answer











    $endgroup$




















      5














      $begingroup$

      You are looking for a solution to $f(x) = dfrac{f(1/x)}{x^2}$



      So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_{-1}^{1} g(x), dx$ is positive and finite



      then a solution will be




      • $f(x) = dfrac{g(x)}{2k} text { when } -1 le x le 1$

      • $f(x) = dfrac{g(1/x)}{2kx^2} text { when } x lt -1 text{ or } x gt 1$


      and I think all solutions will essentially be of this form



      One solution is $f(x)=dfrac{1}{(1+x)^2}$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution



      but there are many others, including the Cauchy density $f(x)=dfrac{1}{pi(1+x^2)}$ on $x in mathbb R$



      Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac1{4x^2}$ otherwise.






      share|cite|improve this answer











      $endgroup$


















        5














        5










        5







        $begingroup$

        You are looking for a solution to $f(x) = dfrac{f(1/x)}{x^2}$



        So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_{-1}^{1} g(x), dx$ is positive and finite



        then a solution will be




        • $f(x) = dfrac{g(x)}{2k} text { when } -1 le x le 1$

        • $f(x) = dfrac{g(1/x)}{2kx^2} text { when } x lt -1 text{ or } x gt 1$


        and I think all solutions will essentially be of this form



        One solution is $f(x)=dfrac{1}{(1+x)^2}$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution



        but there are many others, including the Cauchy density $f(x)=dfrac{1}{pi(1+x^2)}$ on $x in mathbb R$



        Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac1{4x^2}$ otherwise.






        share|cite|improve this answer











        $endgroup$



        You are looking for a solution to $f(x) = dfrac{f(1/x)}{x^2}$



        So take any non-negative function $g(x)$ on $[-1,1]$ where $k= intlimits_{-1}^{1} g(x), dx$ is positive and finite



        then a solution will be




        • $f(x) = dfrac{g(x)}{2k} text { when } -1 le x le 1$

        • $f(x) = dfrac{g(1/x)}{2kx^2} text { when } x lt -1 text{ or } x gt 1$


        and I think all solutions will essentially be of this form



        One solution is $f(x)=dfrac{1}{(1+x)^2}$ for $x gt 0$ and this is indeed an $F(2,2)$ distribution



        but there are many others, including the Cauchy density $f(x)=dfrac{1}{pi(1+x^2)}$ on $x in mathbb R$



        Another simple illustration, with $g(x)=1$ and so $k=2$, has $f(x)=frac14$ when $x in [-1,1]$ and $f(x)=frac1{4x^2}$ otherwise.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 27 at 18:05

























        answered May 27 at 8:59









        HenryHenry

        107k4 gold badges86 silver badges177 bronze badges




        107k4 gold badges86 silver badges177 bronze badges


































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