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Are there any distributions other than Cauchy for which the arithmetic mean of a sample follows the same distribution?
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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
asked 2 days ago
Chechy Levas
410211
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
asked 2 days ago
Chechy Levas
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.
– 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.
– whuber♦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
– 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.
– whuber♦
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.
– 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
asked 2 days ago
Chechy Levas
410211
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
asked 2 days ago
Chechy Levas
410211
asked 2 days ago
Chechy Levas
410211
edited 2 days ago
asked 2 days ago
Chechy Levas
410211
asked 2 days ago
Chechy Levas
410211
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.
– 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.
– whuber♦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
– 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.
– whuber♦
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.
– 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.
– 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.
– whuber♦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
– 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.
– whuber♦
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.
– 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.
– 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.
– 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.
– whuber♦
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.
– whuber♦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
– Chechy Levas
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
– 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.
– whuber♦
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.
– whuber♦
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.
– 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.
– 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
$$
logleft|fractnright|neqlogleft|tright|
$$
answered 2 days ago
Christoph Hanck
15.9k33870
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
– 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.
– 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 $-|ct|.$
– whuber♦
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.
– 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
$$
logleft|fractnright|neqlogleft|tright|
$$
answered 2 days ago
Christoph Hanck
15.9k33870
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
– 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.
– 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 $-|ct|.$
– whuber♦
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.
– 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
$$
logleft|fractnright|neqlogleft|tright|
$$
answered 2 days ago
Christoph Hanck
15.9k33870
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
– 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.
– 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 $-|ct|.$
– whuber♦
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.
– 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
$$
logleft|fractnright|neqlogleft|tright|
$$
answered 2 days ago
Christoph Hanck
15.9k33870
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
$$
logleft|fractnright|neqlogleft|tright|
$$
answered 2 days ago
Christoph Hanck
15.9k33870
edited 2 days ago
answered 2 days ago
Christoph Hanck
15.9k33870
answered 2 days ago
Christoph Hanck
15.9k33870
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?
– 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.
– 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 $-|ct|.$
– whuber♦
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.
– 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?
– 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.
– 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 $-|ct|.$
– whuber♦
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.
– Chechy Levas
yesterday
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
– Chechy Levas
2 days ago
So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1?
– 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.
– 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.
– 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 $-|ct|.$
– whuber♦
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 $-|ct|.$
– whuber♦
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.
– 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.
– Chechy Levas
yesterday
add a comment |Â
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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.
– 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.
– whuber♦
2 days ago
@whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations.
– 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.
– whuber♦
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.
– Chechy Levas
2 days ago