Almost sure convergence of quadratic form x'Ax
$begingroup$
Let $x_n$ be an $ntimes 1$ vector of random variables, and $A_n=(a_{ij,n})$ be an $ntimes n$ constant matrix. Suppose that $n^{-1}x_n'x_n$ converges almost surely to some limit as $nrightarrow infty$. What conditions are needed for $A_n$ in order to establish almost sure converge of $n^{-1}x_n'A_nx_n$?
probability-theory convergence probability-limit-theorems strong-convergence
asked Jan 22 at 19:46
TinaTina
789
$endgroup$
add a comment |
$begingroup$
Let $x_n$ be an $ntimes 1$ vector of random variables, and $A_n=(a_{ij,n})$ be an $ntimes n$ constant matrix. Suppose that $n^{-1}x_n'x_n$ converges almost surely to some limit as $nrightarrow infty$. What conditions are needed for $A_n$ in order to establish almost sure converge of $n^{-1}x_n'A_nx_n$?
probability-theory convergence probability-limit-theorems strong-convergence
asked Jan 22 at 19:46
TinaTina
789
$endgroup$
$begingroup$
The notation is confusing, because $A$ and $x$ aren't a particular matrix and vector, but apparently a sequence of different matrices and vectors depending on $n$. Please clarify.
$endgroup$
– Robert Israel
Jan 22 at 19:53
$begingroup$
Thanks Robert. I have edited the question.
$endgroup$
– Tina
Jan 22 at 19:59
add a comment |
$begingroup$
Let $x_n$ be an $ntimes 1$ vector of random variables, and $A_n=(a_{ij,n})$ be an $ntimes n$ constant matrix. Suppose that $n^{-1}x_n'x_n$ converges almost surely to some limit as $nrightarrow infty$. What conditions are needed for $A_n$ in order to establish almost sure converge of $n^{-1}x_n'A_nx_n$?
probability-theory convergence probability-limit-theorems strong-convergence
asked Jan 22 at 19:46
TinaTina
789
$endgroup$
Let $x_n$ be an $ntimes 1$ vector of random variables, and $A_n=(a_{ij,n})$ be an $ntimes n$ constant matrix. Suppose that $n^{-1}x_n'x_n$ converges almost surely to some limit as $nrightarrow infty$. What conditions are needed for $A_n$ in order to establish almost sure converge of $n^{-1}x_n'A_nx_n$?
probability-theory convergence probability-limit-theorems strong-convergence
probability-theory convergence probability-limit-theorems strong-convergence
asked Jan 22 at 19:46
TinaTina
789
asked Jan 22 at 19:46
TinaTina
789
edited Jan 22 at 19:57
Tina
asked Jan 22 at 19:46
TinaTina
789
asked Jan 22 at 19:46
TinaTina
789
asked Jan 22 at 19:46
TinaTina
789
789
$begingroup$
The notation is confusing, because $A$ and $x$ aren't a particular matrix and vector, but apparently a sequence of different matrices and vectors depending on $n$. Please clarify.
$endgroup$
– Robert Israel
Jan 22 at 19:53
$begingroup$
Thanks Robert. I have edited the question.
$endgroup$
– Tina
Jan 22 at 19:59
add a comment |
$begingroup$
The notation is confusing, because $A$ and $x$ aren't a particular matrix and vector, but apparently a sequence of different matrices and vectors depending on $n$. Please clarify.
$endgroup$
– Robert Israel
Jan 22 at 19:53
$begingroup$
Thanks Robert. I have edited the question.
$endgroup$
– Tina
Jan 22 at 19:59
$begingroup$
The notation is confusing, because $A$ and $x$ aren't a particular matrix and vector, but apparently a sequence of different matrices and vectors depending on $n$. Please clarify.
$endgroup$
– Robert Israel
Jan 22 at 19:53
$begingroup$
The notation is confusing, because $A$ and $x$ aren't a particular matrix and vector, but apparently a sequence of different matrices and vectors depending on $n$. Please clarify.
$endgroup$
– Robert Israel
Jan 22 at 19:53
$begingroup$
Thanks Robert. I have edited the question.
$endgroup$
– Tina
Jan 22 at 19:59
$begingroup$
Thanks Robert. I have edited the question.
$endgroup$
– Tina
Jan 22 at 19:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We may as well assume $A_n$ are symmetric, since $x' A x$ depends only on $A + A'$. A sufficient condition is that the maximum and minimum eigenvalues of $A_n$ both approach the same limit as $n to infty$. Conversely, if this is not the case we can construct (non-random) $x_n$ such that $n^{-1} x_n' x_n$ converges but $n^{-1} x_n' A_n x_n$ does not.
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
$endgroup$
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%2f3083623%2falmost-sure-convergence-of-quadratic-form-xax%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
$begingroup$
We may as well assume $A_n$ are symmetric, since $x' A x$ depends only on $A + A'$. A sufficient condition is that the maximum and minimum eigenvalues of $A_n$ both approach the same limit as $n to infty$. Conversely, if this is not the case we can construct (non-random) $x_n$ such that $n^{-1} x_n' x_n$ converges but $n^{-1} x_n' A_n x_n$ does not.
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
$endgroup$
add a comment |
$begingroup$
We may as well assume $A_n$ are symmetric, since $x' A x$ depends only on $A + A'$. A sufficient condition is that the maximum and minimum eigenvalues of $A_n$ both approach the same limit as $n to infty$. Conversely, if this is not the case we can construct (non-random) $x_n$ such that $n^{-1} x_n' x_n$ converges but $n^{-1} x_n' A_n x_n$ does not.
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
$endgroup$
add a comment |
$begingroup$
We may as well assume $A_n$ are symmetric, since $x' A x$ depends only on $A + A'$. A sufficient condition is that the maximum and minimum eigenvalues of $A_n$ both approach the same limit as $n to infty$. Conversely, if this is not the case we can construct (non-random) $x_n$ such that $n^{-1} x_n' x_n$ converges but $n^{-1} x_n' A_n x_n$ does not.
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
$endgroup$
We may as well assume $A_n$ are symmetric, since $x' A x$ depends only on $A + A'$. A sufficient condition is that the maximum and minimum eigenvalues of $A_n$ both approach the same limit as $n to infty$. Conversely, if this is not the case we can construct (non-random) $x_n$ such that $n^{-1} x_n' x_n$ converges but $n^{-1} x_n' A_n x_n$ does not.
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
answered Jan 22 at 20:17
Robert IsraelRobert Israel
325k23215469
325k23215469
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%2f3083623%2falmost-sure-convergence-of-quadratic-form-xax%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
$begingroup$
The notation is confusing, because $A$ and $x$ aren't a particular matrix and vector, but apparently a sequence of different matrices and vectors depending on $n$. Please clarify.
$endgroup$
– Robert Israel
Jan 22 at 19:53
$begingroup$
Thanks Robert. I have edited the question.
$endgroup$
– Tina
Jan 22 at 19:59