Are there any stable law distributions where the sum of the random variables is distributed the same as the...












0












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I am currently trying to find a distribution where the sum of iid random variables from the distribution is equivalent to the distribution of each of the random variables itself. More concretely, I am looking for a distribution $F$ where if $X_1, ldots, X_n overset{iid}sim F$, then



$$
sum_{i=1}^n X_i sim F
$$



for any $n$.



The closest thing I know of is the Cauchy but the Cauchy works with the average, and the Cauchy does not admit defined moments. I have tried to work with a stable law distribution and play around with terms, but have not been able to find a specification that works. Could someone guide me to a distribution which would work as above? Thanks.






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    0












    $begingroup$


    I am currently trying to find a distribution where the sum of iid random variables from the distribution is equivalent to the distribution of each of the random variables itself. More concretely, I am looking for a distribution $F$ where if $X_1, ldots, X_n overset{iid}sim F$, then



    $$
    sum_{i=1}^n X_i sim F
    $$



    for any $n$.



    The closest thing I know of is the Cauchy but the Cauchy works with the average, and the Cauchy does not admit defined moments. I have tried to work with a stable law distribution and play around with terms, but have not been able to find a specification that works. Could someone guide me to a distribution which would work as above? Thanks.






    probability statistics







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am currently trying to find a distribution where the sum of iid random variables from the distribution is equivalent to the distribution of each of the random variables itself. More concretely, I am looking for a distribution $F$ where if $X_1, ldots, X_n overset{iid}sim F$, then



      $$
      sum_{i=1}^n X_i sim F
      $$



      for any $n$.



      The closest thing I know of is the Cauchy but the Cauchy works with the average, and the Cauchy does not admit defined moments. I have tried to work with a stable law distribution and play around with terms, but have not been able to find a specification that works. Could someone guide me to a distribution which would work as above? Thanks.






      probability statistics







      share|cite|improve this question









      $endgroup$




      I am currently trying to find a distribution where the sum of iid random variables from the distribution is equivalent to the distribution of each of the random variables itself. More concretely, I am looking for a distribution $F$ where if $X_1, ldots, X_n overset{iid}sim F$, then



      $$
      sum_{i=1}^n X_i sim F
      $$



      for any $n$.



      The closest thing I know of is the Cauchy but the Cauchy works with the average, and the Cauchy does not admit defined moments. I have tried to work with a stable law distribution and play around with terms, but have not been able to find a specification that works. Could someone guide me to a distribution which would work as above? Thanks.






      probability statistics




      probability statistics






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      asked Dec 5 '18 at 23:23









      user321627user321627

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          1 Answer
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          $begingroup$

          There is no such distribution except the trivial one where $X_i=0$. If $phi$ is the characteristic function then we get $phi^{n}equiv phi$ for all $n$. If $|phi (t)| <1$ this gives (by letting $n to infty$) $phi (t)=0$. But $phi (0)=1$ and $phi$ is continuous so we must have $|phi (t)| =1$ for $|t|$ sufficiently small. It is easy to see from this $X_1=0$ almost surely.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
            $endgroup$
            – Henry
            Dec 5 '18 at 23:44












          • $begingroup$
            @Henry You are right. Thanks for the comment.
            $endgroup$
            – Kavi Rama Murthy
            Dec 5 '18 at 23:49










          • $begingroup$
            @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
            $endgroup$
            – user321627
            Dec 6 '18 at 0:00










          • $begingroup$
            @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 6 '18 at 0:04












          • $begingroup$
            Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
            $endgroup$
            – user321627
            Dec 8 '18 at 16:55











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          1 Answer
          1






          active

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          1 Answer
          1






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          oldest

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          active

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          2












          $begingroup$

          There is no such distribution except the trivial one where $X_i=0$. If $phi$ is the characteristic function then we get $phi^{n}equiv phi$ for all $n$. If $|phi (t)| <1$ this gives (by letting $n to infty$) $phi (t)=0$. But $phi (0)=1$ and $phi$ is continuous so we must have $|phi (t)| =1$ for $|t|$ sufficiently small. It is easy to see from this $X_1=0$ almost surely.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
            $endgroup$
            – Henry
            Dec 5 '18 at 23:44












          • $begingroup$
            @Henry You are right. Thanks for the comment.
            $endgroup$
            – Kavi Rama Murthy
            Dec 5 '18 at 23:49










          • $begingroup$
            @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
            $endgroup$
            – user321627
            Dec 6 '18 at 0:00










          • $begingroup$
            @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 6 '18 at 0:04












          • $begingroup$
            Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
            $endgroup$
            – user321627
            Dec 8 '18 at 16:55
















          2












          $begingroup$

          There is no such distribution except the trivial one where $X_i=0$. If $phi$ is the characteristic function then we get $phi^{n}equiv phi$ for all $n$. If $|phi (t)| <1$ this gives (by letting $n to infty$) $phi (t)=0$. But $phi (0)=1$ and $phi$ is continuous so we must have $|phi (t)| =1$ for $|t|$ sufficiently small. It is easy to see from this $X_1=0$ almost surely.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
            $endgroup$
            – Henry
            Dec 5 '18 at 23:44












          • $begingroup$
            @Henry You are right. Thanks for the comment.
            $endgroup$
            – Kavi Rama Murthy
            Dec 5 '18 at 23:49










          • $begingroup$
            @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
            $endgroup$
            – user321627
            Dec 6 '18 at 0:00










          • $begingroup$
            @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 6 '18 at 0:04












          • $begingroup$
            Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
            $endgroup$
            – user321627
            Dec 8 '18 at 16:55














          2












          2








          2





          $begingroup$

          There is no such distribution except the trivial one where $X_i=0$. If $phi$ is the characteristic function then we get $phi^{n}equiv phi$ for all $n$. If $|phi (t)| <1$ this gives (by letting $n to infty$) $phi (t)=0$. But $phi (0)=1$ and $phi$ is continuous so we must have $|phi (t)| =1$ for $|t|$ sufficiently small. It is easy to see from this $X_1=0$ almost surely.






          share|cite|improve this answer









          $endgroup$



          There is no such distribution except the trivial one where $X_i=0$. If $phi$ is the characteristic function then we get $phi^{n}equiv phi$ for all $n$. If $|phi (t)| <1$ this gives (by letting $n to infty$) $phi (t)=0$. But $phi (0)=1$ and $phi$ is continuous so we must have $|phi (t)| =1$ for $|t|$ sufficiently small. It is easy to see from this $X_1=0$ almost surely.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 23:30









          Kavi Rama MurthyKavi Rama Murthy

          52.6k32055




          52.6k32055












          • $begingroup$
            +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
            $endgroup$
            – Henry
            Dec 5 '18 at 23:44












          • $begingroup$
            @Henry You are right. Thanks for the comment.
            $endgroup$
            – Kavi Rama Murthy
            Dec 5 '18 at 23:49










          • $begingroup$
            @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
            $endgroup$
            – user321627
            Dec 6 '18 at 0:00










          • $begingroup$
            @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 6 '18 at 0:04












          • $begingroup$
            Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
            $endgroup$
            – user321627
            Dec 8 '18 at 16:55


















          • $begingroup$
            +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
            $endgroup$
            – Henry
            Dec 5 '18 at 23:44












          • $begingroup$
            @Henry You are right. Thanks for the comment.
            $endgroup$
            – Kavi Rama Murthy
            Dec 5 '18 at 23:49










          • $begingroup$
            @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
            $endgroup$
            – user321627
            Dec 6 '18 at 0:00










          • $begingroup$
            @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
            $endgroup$
            – Kavi Rama Murthy
            Dec 6 '18 at 0:04












          • $begingroup$
            Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
            $endgroup$
            – user321627
            Dec 8 '18 at 16:55
















          $begingroup$
          +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
          $endgroup$
          – Henry
          Dec 5 '18 at 23:44






          $begingroup$
          +1. Perhaps an alternative approach is to say $phi^2(t)=phi(t)$ here which implies $phi(t)=0$ or $1$, and then $phi (0)=1$ and $phi$ being continuous in general implies $phi (t)=1$ for all $t$ here and thus $X_1=0$ almost surely
          $endgroup$
          – Henry
          Dec 5 '18 at 23:44














          $begingroup$
          @Henry You are right. Thanks for the comment.
          $endgroup$
          – Kavi Rama Murthy
          Dec 5 '18 at 23:49




          $begingroup$
          @Henry You are right. Thanks for the comment.
          $endgroup$
          – Kavi Rama Murthy
          Dec 5 '18 at 23:49












          $begingroup$
          @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
          $endgroup$
          – user321627
          Dec 6 '18 at 0:00




          $begingroup$
          @KaviRamaMurthy Thanks a lot, this is great, a most wonderful argument! On a related side, would you be aware of any distributions where the average or another functional transformation would leave us with marginally identically distributed random variables? I know the Cauchy is one but it has no defined moments. Any suggestions would be appreciated!
          $endgroup$
          – user321627
          Dec 6 '18 at 0:00












          $begingroup$
          @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
          $endgroup$
          – Kavi Rama Murthy
          Dec 6 '18 at 0:04






          $begingroup$
          @user321627 In the case of Cauchy distribution $sum_{i=1}^{n}X_i$ has same distribution as $nX_1$. As similar thing happens with any symmetric stable distribution ($nX_1$ gets replaced by $n^{1/alpha}X_1$ for some $alpha in (0,2])$. In the non -symmetric stable case you get $n^{1/alpha}X_1+c_n$ for some constants $c_n$.
          $endgroup$
          – Kavi Rama Murthy
          Dec 6 '18 at 0:04














          $begingroup$
          Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
          $endgroup$
          – user321627
          Dec 8 '18 at 16:55




          $begingroup$
          Thanks for your post, may I ask if you have any resources on how I can derive these constants for the non-symmetric stable case? I have referenced wikipedia but am getting difficulty in obtaining the constants.
          $endgroup$
          – user321627
          Dec 8 '18 at 16:55


















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