Are there any stable law distributions where the sum of the random variables is distributed the same as the...
$begingroup$
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
asked Dec 5 '18 at 23:23
user321627user321627
909313
$endgroup$
add a comment |
$begingroup$
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
asked Dec 5 '18 at 23:23
user321627user321627
909313
$endgroup$
add a comment |
$begingroup$
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
asked Dec 5 '18 at 23:23
user321627user321627
909313
$endgroup$
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
asked Dec 5 '18 at 23:23
user321627user321627
909313
asked Dec 5 '18 at 23:23
user321627user321627
909313
asked Dec 5 '18 at 23:23
user321627user321627
909313
asked Dec 5 '18 at 23:23
user321627user321627
909313
asked Dec 5 '18 at 23:23
user321627user321627
909313
909313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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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.
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
$begingroup$
+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
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
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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.
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
$begingroup$
+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
|
show 3 more comments
$begingroup$
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.
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
$begingroup$
+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
|
show 3 more comments
$begingroup$
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.
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
$endgroup$
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.
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
answered Dec 5 '18 at 23:30
Kavi Rama MurthyKavi Rama Murthy
52.6k32055
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
|
show 3 more comments
$begingroup$
+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
|
show 3 more comments
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