Are there any distributions other than Cauchy for which the arithmetic mean of a sample follows the same distribution?
Clash Royale CLAN TAG#URR8PPP
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0;
up vote
10
down vote
favorite
If $X$ follows a Cauchy distribution then $Y = barX = frac1n sum_i=1^n X_i$ also follows exactly the same distribution as $X$; see this thread.
Does this property have a name?
Are there any other distributions for which this is true?
EDIT
Another way of asking this question:
let $X$ be a random variable with probability density $f(x)$.
let $Y=frac 1 nsum_i=1 ^n X_i$, where $X_i$ denotes the ith observation of $X$.
$Y$ itself can be considered as a random variable, without conditioning on any specific values of $X$.
If $X$ follows a Cauchy distribution, then the probability density function of $Y$ is $f(x)$
Are there any other kinds of (non trivial*) probability density functions for $f(x)$ that result in $Y$ having a probability density function of $f(x)$?
*The only trivial example I can think of is a Dirac delta. i.e. not a random variable.
distributions expected-value central-limit-theorem cauchy
add a comment |Â
up vote
10
down vote
favorite
If $X$ follows a Cauchy distribution then $Y = barX = frac1n sum_i=1^n X_i$ also follows exactly the same distribution as $X$; see this thread.
Does this property have a name?
Are there any other distributions for which this is true?
EDIT
Another way of asking this question:
let $X$ be a random variable with probability density $f(x)$.
let $Y=frac 1 nsum_i=1 ^n X_i$, where $X_i$ denotes the ith observation of $X$.
$Y$ itself can be considered as a random variable, without conditioning on any specific values of $X$.
If $X$ follows a Cauchy distribution, then the probability density function of $Y$ is $f(x)$
Are there any other kinds of (non trivial*) probability density functions for $f(x)$ that result in $Y$ having a probability density function of $f(x)$?
*The only trivial example I can think of is a Dirac delta. i.e. not a random variable.
distributions expected-value central-limit-theorem cauchy
This is related to a property of stable distributions. Cauchy is a stable distribution, as is Normal distribution, but the sample mean in Normal population isn't identically distributed as each $X_i$, though it is Normal.
â StubbornAtom
2 days ago
Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it.
â whuberâ¦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
â Chechy Levas
2 days ago
Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions.
â whuberâ¦
2 days ago
I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome.
â Chechy Levas
2 days ago
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
If $X$ follows a Cauchy distribution then $Y = barX = frac1n sum_i=1^n X_i$ also follows exactly the same distribution as $X$; see this thread.
Does this property have a name?
Are there any other distributions for which this is true?
EDIT
Another way of asking this question:
let $X$ be a random variable with probability density $f(x)$.
let $Y=frac 1 nsum_i=1 ^n X_i$, where $X_i$ denotes the ith observation of $X$.
$Y$ itself can be considered as a random variable, without conditioning on any specific values of $X$.
If $X$ follows a Cauchy distribution, then the probability density function of $Y$ is $f(x)$
Are there any other kinds of (non trivial*) probability density functions for $f(x)$ that result in $Y$ having a probability density function of $f(x)$?
*The only trivial example I can think of is a Dirac delta. i.e. not a random variable.
distributions expected-value central-limit-theorem cauchy
If $X$ follows a Cauchy distribution then $Y = barX = frac1n sum_i=1^n X_i$ also follows exactly the same distribution as $X$; see this thread.
Does this property have a name?
Are there any other distributions for which this is true?
EDIT
Another way of asking this question:
let $X$ be a random variable with probability density $f(x)$.
let $Y=frac 1 nsum_i=1 ^n X_i$, where $X_i$ denotes the ith observation of $X$.
$Y$ itself can be considered as a random variable, without conditioning on any specific values of $X$.
If $X$ follows a Cauchy distribution, then the probability density function of $Y$ is $f(x)$
Are there any other kinds of (non trivial*) probability density functions for $f(x)$ that result in $Y$ having a probability density function of $f(x)$?
*The only trivial example I can think of is a Dirac delta. i.e. not a random variable.
distributions expected-value central-limit-theorem cauchy
distributions expected-value central-limit-theorem cauchy
edited 2 days ago
asked 2 days ago
Chechy Levas
410211
410211
This is related to a property of stable distributions. Cauchy is a stable distribution, as is Normal distribution, but the sample mean in Normal population isn't identically distributed as each $X_i$, though it is Normal.
â StubbornAtom
2 days ago
Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it.
â whuberâ¦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
â Chechy Levas
2 days ago
Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions.
â whuberâ¦
2 days ago
I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome.
â Chechy Levas
2 days ago
add a comment |Â
This is related to a property of stable distributions. Cauchy is a stable distribution, as is Normal distribution, but the sample mean in Normal population isn't identically distributed as each $X_i$, though it is Normal.
â StubbornAtom
2 days ago
Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it.
â whuberâ¦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
â Chechy Levas
2 days ago
Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions.
â whuberâ¦
2 days ago
I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome.
â Chechy Levas
2 days ago
This is related to a property of stable distributions. Cauchy is a stable distribution, as is Normal distribution, but the sample mean in Normal population isn't identically distributed as each $X_i$, though it is Normal.
â StubbornAtom
2 days ago
This is related to a property of stable distributions. Cauchy is a stable distribution, as is Normal distribution, but the sample mean in Normal population isn't identically distributed as each $X_i$, though it is Normal.
â StubbornAtom
2 days ago
Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it.
â whuberâ¦
2 days ago
Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it.
â whuberâ¦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
â Chechy Levas
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
â Chechy Levas
2 days ago
Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions.
â whuberâ¦
2 days ago
Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions.
â whuberâ¦
2 days ago
I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome.
â Chechy Levas
2 days ago
I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome.
â Chechy Levas
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
This is not really an answer, but at least it does not seem to be easy to create such an example from a stable distribution. We would need to produce a r.v. whose characteristic function is the same as that of its average.
In general, for an iid draw, the c.f. of the average is
$$
phi_barX_n(t)=[phi_X(t/n)]^n
$$
with $phi_X$ the c.f. of a single r.v. For stable distributions with location parameter zero, we have
$$
phi_X(t)=exp-,
$$
where
$$
Phi=begincasestanleft(fracpialpha2right)&alphaneq1\-frac2pilog|t|&alpha=1endcases
$$
The Cauchy distribution corresponds to $alpha=1$, $beta=0$, so that $phi_barX_n(t)=phi_X(t)$ indeed for any scale parameter $c>0$.
In general,
$$
phi_barX_n(t)=expleft-nleft,
$$
To get $phi_barX_n(t)=phi_X(t)$, $alpha=1$ seems called for, so
begineqnarray*
phi_barX_n(t)&=&expleftright)right)right\
&=&expleftright)right)right,
endeqnarray*
but
$$
logleft|fractnright|neqlogleft|tright|
$$
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
1
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
2
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This is not really an answer, but at least it does not seem to be easy to create such an example from a stable distribution. We would need to produce a r.v. whose characteristic function is the same as that of its average.
In general, for an iid draw, the c.f. of the average is
$$
phi_barX_n(t)=[phi_X(t/n)]^n
$$
with $phi_X$ the c.f. of a single r.v. For stable distributions with location parameter zero, we have
$$
phi_X(t)=exp-,
$$
where
$$
Phi=begincasestanleft(fracpialpha2right)&alphaneq1\-frac2pilog|t|&alpha=1endcases
$$
The Cauchy distribution corresponds to $alpha=1$, $beta=0$, so that $phi_barX_n(t)=phi_X(t)$ indeed for any scale parameter $c>0$.
In general,
$$
phi_barX_n(t)=expleft-nleft,
$$
To get $phi_barX_n(t)=phi_X(t)$, $alpha=1$ seems called for, so
begineqnarray*
phi_barX_n(t)&=&expleftright)right)right\
&=&expleftright)right)right,
endeqnarray*
but
$$
logleft|fractnright|neqlogleft|tright|
$$
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
1
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
2
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
add a comment |Â
up vote
4
down vote
accepted
This is not really an answer, but at least it does not seem to be easy to create such an example from a stable distribution. We would need to produce a r.v. whose characteristic function is the same as that of its average.
In general, for an iid draw, the c.f. of the average is
$$
phi_barX_n(t)=[phi_X(t/n)]^n
$$
with $phi_X$ the c.f. of a single r.v. For stable distributions with location parameter zero, we have
$$
phi_X(t)=exp-,
$$
where
$$
Phi=begincasestanleft(fracpialpha2right)&alphaneq1\-frac2pilog|t|&alpha=1endcases
$$
The Cauchy distribution corresponds to $alpha=1$, $beta=0$, so that $phi_barX_n(t)=phi_X(t)$ indeed for any scale parameter $c>0$.
In general,
$$
phi_barX_n(t)=expleft-nleft,
$$
To get $phi_barX_n(t)=phi_X(t)$, $alpha=1$ seems called for, so
begineqnarray*
phi_barX_n(t)&=&expleftright)right)right\
&=&expleftright)right)right,
endeqnarray*
but
$$
logleft|fractnright|neqlogleft|tright|
$$
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
1
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
2
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This is not really an answer, but at least it does not seem to be easy to create such an example from a stable distribution. We would need to produce a r.v. whose characteristic function is the same as that of its average.
In general, for an iid draw, the c.f. of the average is
$$
phi_barX_n(t)=[phi_X(t/n)]^n
$$
with $phi_X$ the c.f. of a single r.v. For stable distributions with location parameter zero, we have
$$
phi_X(t)=exp-,
$$
where
$$
Phi=begincasestanleft(fracpialpha2right)&alphaneq1\-frac2pilog|t|&alpha=1endcases
$$
The Cauchy distribution corresponds to $alpha=1$, $beta=0$, so that $phi_barX_n(t)=phi_X(t)$ indeed for any scale parameter $c>0$.
In general,
$$
phi_barX_n(t)=expleft-nleft,
$$
To get $phi_barX_n(t)=phi_X(t)$, $alpha=1$ seems called for, so
begineqnarray*
phi_barX_n(t)&=&expleftright)right)right\
&=&expleftright)right)right,
endeqnarray*
but
$$
logleft|fractnright|neqlogleft|tright|
$$
This is not really an answer, but at least it does not seem to be easy to create such an example from a stable distribution. We would need to produce a r.v. whose characteristic function is the same as that of its average.
In general, for an iid draw, the c.f. of the average is
$$
phi_barX_n(t)=[phi_X(t/n)]^n
$$
with $phi_X$ the c.f. of a single r.v. For stable distributions with location parameter zero, we have
$$
phi_X(t)=exp-,
$$
where
$$
Phi=begincasestanleft(fracpialpha2right)&alphaneq1\-frac2pilog|t|&alpha=1endcases
$$
The Cauchy distribution corresponds to $alpha=1$, $beta=0$, so that $phi_barX_n(t)=phi_X(t)$ indeed for any scale parameter $c>0$.
In general,
$$
phi_barX_n(t)=expleft-nleft,
$$
To get $phi_barX_n(t)=phi_X(t)$, $alpha=1$ seems called for, so
begineqnarray*
phi_barX_n(t)&=&expleftright)right)right\
&=&expleftright)right)right,
endeqnarray*
but
$$
logleft|fractnright|neqlogleft|tright|
$$
edited 2 days ago
answered 2 days ago
Christoph Hanck
15.9k33870
15.9k33870
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
1
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
2
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
add a comment |Â
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
1
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
2
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
â Chechy Levas
2 days ago
1
1
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions.
â Christoph Hanck
2 days ago
2
2
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
You don't need to invoke the theory of stable distributions. Letting $psi=log phi$ be the cgf, your equation is $$psi(t/n) = psi(t)/n$$ for $n=1,2,3,ldots.$ Since $psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $psi$ at the origin is $-|ct|.$
â whuberâ¦
2 days ago
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
Should this be the accepted answer? Besides $alpha = 1$ the only way I can see to solve this is with $alpha = 0$, which (I think) is the Dirac delta.
â Chechy Levas
yesterday
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f366178%2fare-there-any-distributions-other-than-cauchy-for-which-the-arithmetic-mean-of-a%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
This is related to a property of stable distributions. Cauchy is a stable distribution, as is Normal distribution, but the sample mean in Normal population isn't identically distributed as each $X_i$, though it is Normal.
â StubbornAtom
2 days ago
Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it.
â whuberâ¦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
â Chechy Levas
2 days ago
Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions.
â whuberâ¦
2 days ago
I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome.
â Chechy Levas
2 days ago