When is a sequence of jointly Gaussian r.v. exchangeable?
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A sequence of random variables $X_1, X_2,.., X_n$ is exchangeable if
begin{align*}
(X_1,X_2,ldots,X_n) stackrel{d}{=} (X_{pi_1}, X_{pi_2}, ldots, X_{pi_n}),
end{align*}
for every permuation $pi$.
Suppose we have a sequence $X_1, X_2,.., X_n$ which is jointly Gaussian with covariance $K$.
What are the conditions on $K$ for this sequence to be exchangeable?
For example, if $K=I$ then the sequence is exchangeable. I am looking for more general conditions.
probability-theory
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
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add a comment |
$begingroup$
A sequence of random variables $X_1, X_2,.., X_n$ is exchangeable if
begin{align*}
(X_1,X_2,ldots,X_n) stackrel{d}{=} (X_{pi_1}, X_{pi_2}, ldots, X_{pi_n}),
end{align*}
for every permuation $pi$.
Suppose we have a sequence $X_1, X_2,.., X_n$ which is jointly Gaussian with covariance $K$.
What are the conditions on $K$ for this sequence to be exchangeable?
For example, if $K=I$ then the sequence is exchangeable. I am looking for more general conditions.
probability-theory
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
$endgroup$
1
$begingroup$
Necessary and sufficient conditions: the variances and covariances must be constant hence $K$ is a linear combination of $I$ and the all-ones matrix.
$endgroup$
– Did
Dec 21 '18 at 14:56
$begingroup$
@Did Thanks. Do you have a reference of the top of your head?
$endgroup$
– Boby
Dec 21 '18 at 15:23
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No need to, the proof is a direct one-liner.
$endgroup$
– Did
Dec 21 '18 at 16:33
add a comment |
$begingroup$
A sequence of random variables $X_1, X_2,.., X_n$ is exchangeable if
begin{align*}
(X_1,X_2,ldots,X_n) stackrel{d}{=} (X_{pi_1}, X_{pi_2}, ldots, X_{pi_n}),
end{align*}
for every permuation $pi$.
Suppose we have a sequence $X_1, X_2,.., X_n$ which is jointly Gaussian with covariance $K$.
What are the conditions on $K$ for this sequence to be exchangeable?
For example, if $K=I$ then the sequence is exchangeable. I am looking for more general conditions.
probability-theory
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
$endgroup$
A sequence of random variables $X_1, X_2,.., X_n$ is exchangeable if
begin{align*}
(X_1,X_2,ldots,X_n) stackrel{d}{=} (X_{pi_1}, X_{pi_2}, ldots, X_{pi_n}),
end{align*}
for every permuation $pi$.
Suppose we have a sequence $X_1, X_2,.., X_n$ which is jointly Gaussian with covariance $K$.
What are the conditions on $K$ for this sequence to be exchangeable?
For example, if $K=I$ then the sequence is exchangeable. I am looking for more general conditions.
probability-theory
probability-theory
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
asked Dec 21 '18 at 14:27
BobyBoby
1,0451929
1,0451929
1
$begingroup$
Necessary and sufficient conditions: the variances and covariances must be constant hence $K$ is a linear combination of $I$ and the all-ones matrix.
$endgroup$
– Did
Dec 21 '18 at 14:56
$begingroup$
@Did Thanks. Do you have a reference of the top of your head?
$endgroup$
– Boby
Dec 21 '18 at 15:23
$begingroup$
No need to, the proof is a direct one-liner.
$endgroup$
– Did
Dec 21 '18 at 16:33
add a comment |
1
$begingroup$
Necessary and sufficient conditions: the variances and covariances must be constant hence $K$ is a linear combination of $I$ and the all-ones matrix.
$endgroup$
– Did
Dec 21 '18 at 14:56
$begingroup$
@Did Thanks. Do you have a reference of the top of your head?
$endgroup$
– Boby
Dec 21 '18 at 15:23
$begingroup$
No need to, the proof is a direct one-liner.
$endgroup$
– Did
Dec 21 '18 at 16:33
1
1
$begingroup$
Necessary and sufficient conditions: the variances and covariances must be constant hence $K$ is a linear combination of $I$ and the all-ones matrix.
$endgroup$
– Did
Dec 21 '18 at 14:56
$begingroup$
Necessary and sufficient conditions: the variances and covariances must be constant hence $K$ is a linear combination of $I$ and the all-ones matrix.
$endgroup$
– Did
Dec 21 '18 at 14:56
$begingroup$
@Did Thanks. Do you have a reference of the top of your head?
$endgroup$
– Boby
Dec 21 '18 at 15:23
$begingroup$
@Did Thanks. Do you have a reference of the top of your head?
$endgroup$
– Boby
Dec 21 '18 at 15:23
$begingroup$
No need to, the proof is a direct one-liner.
$endgroup$
– Did
Dec 21 '18 at 16:33
$begingroup$
No need to, the proof is a direct one-liner.
$endgroup$
– Did
Dec 21 '18 at 16:33
add a comment |
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$begingroup$
Necessary and sufficient conditions: the variances and covariances must be constant hence $K$ is a linear combination of $I$ and the all-ones matrix.
$endgroup$
– Did
Dec 21 '18 at 14:56
$begingroup$
@Did Thanks. Do you have a reference of the top of your head?
$endgroup$
– Boby
Dec 21 '18 at 15:23
$begingroup$
No need to, the proof is a direct one-liner.
$endgroup$
– Did
Dec 21 '18 at 16:33