equivalence of definitions
$begingroup$
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
$endgroup$
add a comment |
$begingroup$
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
$endgroup$
add a comment |
$begingroup$
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
$endgroup$
Why are both definitions equivalent?
Random variable $X $ is called Infinite divisible if for each $n in N $ there exist iid random variables $X_1,dots X_n $ such that $X=frac1n sum X_i$ or the same without $frac1n $. The latter is the one I am familiar with.
for example if $Xsim poisson (lambda) $ then we can pick $ X_i sim poisson (lambda /n) $ but with the other definition it is not possible.
probability-theory
probability-theory
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
asked Nov 28 '18 at 16:55
Joey DoeyJoey Doey
1155
1155
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
$endgroup$
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
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%2f3017376%2fequivalence-of-definitions%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
$begingroup$
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
$endgroup$
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
$begingroup$
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
$endgroup$
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
$begingroup$
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
$endgroup$
If you can find $X_1,X_2,dots,X_n$ such that $X=frac{1}{n}sum X_i$ then we can simply choose $Y_i=X_i/n$ and we have $X=sum Y_i$.
One can do something analogous in the other direction.
So both definitions are equivalent.
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
answered Nov 28 '18 at 17:00
Jorge FernándezJorge Fernández
75.3k1191192
75.3k1191192
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
add a comment |
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
but what would be $X_i $ in my case if X is poison distributed. I am only able to find the distribution of $ X_i /n $ but not $X_i $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:45
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
or is it not possible to find the distribution of $X_i$ but only the distribution of $X_i/n $
$endgroup$
– Joey Doey
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
the $X_i$ would be $frac{1}{n}poisson(lambda/n)$
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:47
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
$begingroup$
if you know the distribution of random variable $X$ then you can find the distribution of $frac{1}{n}X$ rather easily.
$endgroup$
– Jorge Fernández
Nov 28 '18 at 17:48
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%2f3017376%2fequivalence-of-definitions%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
