Mixing
Constraining the sum of Gaussian random variables?
$begingroup$
Suppose I have $n$ Gaussian random variables $X_i$ with $i=1,2,...,n$, each with zero mean $mu=0$ and the same constant standard deviation $sigmaneq 0$. I would like to constrain the elements collectively drawn from these distributions to satisfy
$$sum_{i=1}^nX_i=0$$
How should I modify the distributions to achieve this, while maintaining zero means for each distribution separately?
In the case of $n=2$ the solution is obviously to restrict $X_2$ elements to $X_2=-X_1$ and let only $X_1$ be drawn independently. But what happens in the case $n>2$?
EDIT:
Considering the symmetry in the definition of all $X_i$ above, let us seek a solution which preserves this symmetry and keeps all $X_i$ identically distributed.
linear-algebra probability probability-distributions
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
$endgroup$
|
show 1 more comment
$begingroup$
Suppose I have $n$ Gaussian random variables $X_i$ with $i=1,2,...,n$, each with zero mean $mu=0$ and the same constant standard deviation $sigmaneq 0$. I would like to constrain the elements collectively drawn from these distributions to satisfy
$$sum_{i=1}^nX_i=0$$
How should I modify the distributions to achieve this, while maintaining zero means for each distribution separately?
In the case of $n=2$ the solution is obviously to restrict $X_2$ elements to $X_2=-X_1$ and let only $X_1$ be drawn independently. But what happens in the case $n>2$?
EDIT:
Considering the symmetry in the definition of all $X_i$ above, let us seek a solution which preserves this symmetry and keeps all $X_i$ identically distributed.
linear-algebra probability probability-distributions
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
$endgroup$
1
$begingroup$
You may "modify" $X_n$. Just take $X_n=-sum_{i=1}^{n-1}X_i$. Then $X_n$ is also normal with zero mean and some nonzero variance.
$endgroup$
– d.k.o.
Jan 30 at 23:52
$begingroup$
@d.k.o. You are right, but then $X_n$ will not be identically distributed with the rest of the $X_i$. I guess I should add this requirement to the question to make it more interesting.
$endgroup$
– Kagaratsch
Jan 30 at 23:55
2
$begingroup$
If $n$ is even, then taking $X_{n/2+i}=-X_i$ satisfies the the condition.
$endgroup$
– d.k.o.
Jan 31 at 0:05
1
$begingroup$
@d.k.o You're right again! :D But what about odd $n$...
$endgroup$
– Kagaratsch
Jan 31 at 0:10
1
$begingroup$
This question has been cross-posted and answered on SE.CV.
$endgroup$
– Ben
Jan 31 at 4:15
|
show 1 more comment
$begingroup$
Suppose I have $n$ Gaussian random variables $X_i$ with $i=1,2,...,n$, each with zero mean $mu=0$ and the same constant standard deviation $sigmaneq 0$. I would like to constrain the elements collectively drawn from these distributions to satisfy
$$sum_{i=1}^nX_i=0$$
How should I modify the distributions to achieve this, while maintaining zero means for each distribution separately?
In the case of $n=2$ the solution is obviously to restrict $X_2$ elements to $X_2=-X_1$ and let only $X_1$ be drawn independently. But what happens in the case $n>2$?
EDIT:
Considering the symmetry in the definition of all $X_i$ above, let us seek a solution which preserves this symmetry and keeps all $X_i$ identically distributed.
linear-algebra probability probability-distributions
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
$endgroup$
Suppose I have $n$ Gaussian random variables $X_i$ with $i=1,2,...,n$, each with zero mean $mu=0$ and the same constant standard deviation $sigmaneq 0$. I would like to constrain the elements collectively drawn from these distributions to satisfy
$$sum_{i=1}^nX_i=0$$
How should I modify the distributions to achieve this, while maintaining zero means for each distribution separately?
In the case of $n=2$ the solution is obviously to restrict $X_2$ elements to $X_2=-X_1$ and let only $X_1$ be drawn independently. But what happens in the case $n>2$?
EDIT:
Considering the symmetry in the definition of all $X_i$ above, let us seek a solution which preserves this symmetry and keeps all $X_i$ identically distributed.
linear-algebra probability probability-distributions
linear-algebra probability probability-distributions
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
edited Jan 30 at 23:56
Kagaratsch
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
asked Jan 30 at 23:30
KagaratschKagaratsch
1,044617
1,044617
1
$begingroup$
You may "modify" $X_n$. Just take $X_n=-sum_{i=1}^{n-1}X_i$. Then $X_n$ is also normal with zero mean and some nonzero variance.
$endgroup$
– d.k.o.
Jan 30 at 23:52
$begingroup$
@d.k.o. You are right, but then $X_n$ will not be identically distributed with the rest of the $X_i$. I guess I should add this requirement to the question to make it more interesting.
$endgroup$
– Kagaratsch
Jan 30 at 23:55
2
$begingroup$
If $n$ is even, then taking $X_{n/2+i}=-X_i$ satisfies the the condition.
$endgroup$
– d.k.o.
Jan 31 at 0:05
1
$begingroup$
@d.k.o You're right again! :D But what about odd $n$...
$endgroup$
– Kagaratsch
Jan 31 at 0:10
1
$begingroup$
This question has been cross-posted and answered on SE.CV.
$endgroup$
– Ben
Jan 31 at 4:15
|
show 1 more comment
1
$begingroup$
You may "modify" $X_n$. Just take $X_n=-sum_{i=1}^{n-1}X_i$. Then $X_n$ is also normal with zero mean and some nonzero variance.
$endgroup$
– d.k.o.
Jan 30 at 23:52
$begingroup$
@d.k.o. You are right, but then $X_n$ will not be identically distributed with the rest of the $X_i$. I guess I should add this requirement to the question to make it more interesting.
$endgroup$
– Kagaratsch
Jan 30 at 23:55
2
$begingroup$
If $n$ is even, then taking $X_{n/2+i}=-X_i$ satisfies the the condition.
$endgroup$
– d.k.o.
Jan 31 at 0:05
1
$begingroup$
@d.k.o You're right again! :D But what about odd $n$...
$endgroup$
– Kagaratsch
Jan 31 at 0:10
1
$begingroup$
This question has been cross-posted and answered on SE.CV.
$endgroup$
– Ben
Jan 31 at 4:15
1
1
$begingroup$
You may "modify" $X_n$. Just take $X_n=-sum_{i=1}^{n-1}X_i$. Then $X_n$ is also normal with zero mean and some nonzero variance.
$endgroup$
– d.k.o.
Jan 30 at 23:52
$begingroup$
You may "modify" $X_n$. Just take $X_n=-sum_{i=1}^{n-1}X_i$. Then $X_n$ is also normal with zero mean and some nonzero variance.
$endgroup$
– d.k.o.
Jan 30 at 23:52
$begingroup$
@d.k.o. You are right, but then $X_n$ will not be identically distributed with the rest of the $X_i$. I guess I should add this requirement to the question to make it more interesting.
$endgroup$
– Kagaratsch
Jan 30 at 23:55
$begingroup$
@d.k.o. You are right, but then $X_n$ will not be identically distributed with the rest of the $X_i$. I guess I should add this requirement to the question to make it more interesting.
$endgroup$
– Kagaratsch
Jan 30 at 23:55
2
2
$begingroup$
If $n$ is even, then taking $X_{n/2+i}=-X_i$ satisfies the the condition.
$endgroup$
– d.k.o.
Jan 31 at 0:05
$begingroup$
If $n$ is even, then taking $X_{n/2+i}=-X_i$ satisfies the the condition.
$endgroup$
– d.k.o.
Jan 31 at 0:05
1
1
$begingroup$
@d.k.o You're right again! :D But what about odd $n$...
$endgroup$
– Kagaratsch
Jan 31 at 0:10
$begingroup$
@d.k.o You're right again! :D But what about odd $n$...
$endgroup$
– Kagaratsch
Jan 31 at 0:10
1
1
$begingroup$
This question has been cross-posted and answered on SE.CV.
$endgroup$
– Ben
Jan 31 at 4:15
$begingroup$
This question has been cross-posted and answered on SE.CV.
$endgroup$
– Ben
Jan 31 at 4:15
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
As pointed out by d.k.o. in a comment, $X_n=-sum_{i=1}^{n-1}X_i$ represents one such solution. However, it is not homogeneous in the treatment of each $X_i$. Turns out, this can be easily fixed by considering all possible cases
$$X_j=-sum_{{{i=1},{ineq j}}}^nX_i$$
instead, and defining the new random variables $X'_j$ by superpositions of all above combinations of old ones:
$$X'_jequiv (n-1)X_j-sum_{{{i=1},{ineq j}}}^nX_i$$
This is now completely symmetric in all $X'_j$.
The explicit construction is implied as follows. To obtain a collective element $(x'_1,x'_2,...,x'_n)$ we first draw a particular collective element $(x_1,x_2,...,x_n)$ from the unrestricted distributions and then literally calculate:
$$x'_jequiv (n-1)x_j-sum_{{{i=1},{ineq j}}}^nx_i$$
for all $j$.
If needed, one could figure out the new standard deviations from this answer to a different question.
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094275%2fconstraining-the-sum-of-gaussian-random-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As pointed out by d.k.o. in a comment, $X_n=-sum_{i=1}^{n-1}X_i$ represents one such solution. However, it is not homogeneous in the treatment of each $X_i$. Turns out, this can be easily fixed by considering all possible cases
$$X_j=-sum_{{{i=1},{ineq j}}}^nX_i$$
instead, and defining the new random variables $X'_j$ by superpositions of all above combinations of old ones:
$$X'_jequiv (n-1)X_j-sum_{{{i=1},{ineq j}}}^nX_i$$
This is now completely symmetric in all $X'_j$.
The explicit construction is implied as follows. To obtain a collective element $(x'_1,x'_2,...,x'_n)$ we first draw a particular collective element $(x_1,x_2,...,x_n)$ from the unrestricted distributions and then literally calculate:
$$x'_jequiv (n-1)x_j-sum_{{{i=1},{ineq j}}}^nx_i$$
for all $j$.
If needed, one could figure out the new standard deviations from this answer to a different question.
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
$endgroup$
add a comment |
$begingroup$
As pointed out by d.k.o. in a comment, $X_n=-sum_{i=1}^{n-1}X_i$ represents one such solution. However, it is not homogeneous in the treatment of each $X_i$. Turns out, this can be easily fixed by considering all possible cases
$$X_j=-sum_{{{i=1},{ineq j}}}^nX_i$$
instead, and defining the new random variables $X'_j$ by superpositions of all above combinations of old ones:
$$X'_jequiv (n-1)X_j-sum_{{{i=1},{ineq j}}}^nX_i$$
This is now completely symmetric in all $X'_j$.
The explicit construction is implied as follows. To obtain a collective element $(x'_1,x'_2,...,x'_n)$ we first draw a particular collective element $(x_1,x_2,...,x_n)$ from the unrestricted distributions and then literally calculate:
$$x'_jequiv (n-1)x_j-sum_{{{i=1},{ineq j}}}^nx_i$$
for all $j$.
If needed, one could figure out the new standard deviations from this answer to a different question.
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
$endgroup$
add a comment |
$begingroup$
As pointed out by d.k.o. in a comment, $X_n=-sum_{i=1}^{n-1}X_i$ represents one such solution. However, it is not homogeneous in the treatment of each $X_i$. Turns out, this can be easily fixed by considering all possible cases
$$X_j=-sum_{{{i=1},{ineq j}}}^nX_i$$
instead, and defining the new random variables $X'_j$ by superpositions of all above combinations of old ones:
$$X'_jequiv (n-1)X_j-sum_{{{i=1},{ineq j}}}^nX_i$$
This is now completely symmetric in all $X'_j$.
The explicit construction is implied as follows. To obtain a collective element $(x'_1,x'_2,...,x'_n)$ we first draw a particular collective element $(x_1,x_2,...,x_n)$ from the unrestricted distributions and then literally calculate:
$$x'_jequiv (n-1)x_j-sum_{{{i=1},{ineq j}}}^nx_i$$
for all $j$.
If needed, one could figure out the new standard deviations from this answer to a different question.
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
$endgroup$
As pointed out by d.k.o. in a comment, $X_n=-sum_{i=1}^{n-1}X_i$ represents one such solution. However, it is not homogeneous in the treatment of each $X_i$. Turns out, this can be easily fixed by considering all possible cases
$$X_j=-sum_{{{i=1},{ineq j}}}^nX_i$$
instead, and defining the new random variables $X'_j$ by superpositions of all above combinations of old ones:
$$X'_jequiv (n-1)X_j-sum_{{{i=1},{ineq j}}}^nX_i$$
This is now completely symmetric in all $X'_j$.
The explicit construction is implied as follows. To obtain a collective element $(x'_1,x'_2,...,x'_n)$ we first draw a particular collective element $(x_1,x_2,...,x_n)$ from the unrestricted distributions and then literally calculate:
$$x'_jequiv (n-1)x_j-sum_{{{i=1},{ineq j}}}^nx_i$$
for all $j$.
If needed, one could figure out the new standard deviations from this answer to a different question.
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
answered Jan 31 at 0:37
KagaratschKagaratsch
1,044617
1,044617
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094275%2fconstraining-the-sum-of-gaussian-random-variables%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
You may "modify" $X_n$. Just take $X_n=-sum_{i=1}^{n-1}X_i$. Then $X_n$ is also normal with zero mean and some nonzero variance.
$endgroup$
– d.k.o.
Jan 30 at 23:52
$begingroup$
@d.k.o. You are right, but then $X_n$ will not be identically distributed with the rest of the $X_i$. I guess I should add this requirement to the question to make it more interesting.
$endgroup$
– Kagaratsch
Jan 30 at 23:55
2
$begingroup$
If $n$ is even, then taking $X_{n/2+i}=-X_i$ satisfies the the condition.
$endgroup$
– d.k.o.
Jan 31 at 0:05
1
$begingroup$
@d.k.o You're right again! :D But what about odd $n$...
$endgroup$
– Kagaratsch
Jan 31 at 0:10
1
$begingroup$
This question has been cross-posted and answered on SE.CV.
$endgroup$
– Ben
Jan 31 at 4:15