Inequalities in probability theory
up vote
3
down vote
favorite
I got stuck while solving this problem. First of all, i tried to prove directly from definition, but this doesnt led anywhere. Perhaps Jensen inequality may help? But we dont have convexity of f or of g. Any ideas on how we can tackle this problem?
probability
asked 4 hours ago
Neymar
379113
add a comment |
up vote
3
down vote
favorite
I got stuck while solving this problem. First of all, i tried to prove directly from definition, but this doesnt led anywhere. Perhaps Jensen inequality may help? But we dont have convexity of f or of g. Any ideas on how we can tackle this problem?
probability
asked 4 hours ago
Neymar
379113
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I got stuck while solving this problem. First of all, i tried to prove directly from definition, but this doesnt led anywhere. Perhaps Jensen inequality may help? But we dont have convexity of f or of g. Any ideas on how we can tackle this problem?
probability
asked 4 hours ago
Neymar
379113
I got stuck while solving this problem. First of all, i tried to prove directly from definition, but this doesnt led anywhere. Perhaps Jensen inequality may help? But we dont have convexity of f or of g. Any ideas on how we can tackle this problem?
probability
probability
asked 4 hours ago
Neymar
379113
asked 4 hours ago
Neymar
379113
asked 4 hours ago
Neymar
379113
asked 4 hours ago
Neymar
379113
asked 4 hours ago
Neymar
379113
379113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let $X$ and $Y$ be iid, then we have
$$E[(f(X)-f(Y))(g(X)-g(Y))] ge 0$$
$$E[f(X)g(X)+f(Y)g(Y)-f(Y)g(X)-f(X)g(Y)] ge 0$$
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)g(X)]-E[f(X)g(Y)] ge 0$$
By independence,
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)]E[g(X)]-E[f(X)]E[g(Y)] ge 0$$
Since $X$ and $Y$ are identically distributed
$$E[f(X)g(X)]+E[f(X)g(X)]-E[f(X)]E[g(X)]-E[f(X)]E[g(X)] ge 0$$
$$2E[f(X)g(X)]-2E[f(X)]E[g(X)] ge 0$$
$$E[f(X)g(X)] ge E[f(X)]E[g(X)]$$
answered 4 hours ago
Siong Thye Goh
97.9k1463116
add a comment |
up vote
2
down vote
For every real number $x$, $(f(X)-f(x))(g(X)-g(x)) geq 0$, thus $mathbb{E}[f(X)g(X)] + f(x)g(x) geq f(x)mathbb{E}[g(X)] + g(x)mathbb{E}[f(x)$.
Then integrate wrt $dP_X$.
answered 4 hours ago
Mindlack
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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',
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%2f3043428%2finequalities-in-probability-theory%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $X$ and $Y$ be iid, then we have
$$E[(f(X)-f(Y))(g(X)-g(Y))] ge 0$$
$$E[f(X)g(X)+f(Y)g(Y)-f(Y)g(X)-f(X)g(Y)] ge 0$$
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)g(X)]-E[f(X)g(Y)] ge 0$$
By independence,
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)]E[g(X)]-E[f(X)]E[g(Y)] ge 0$$
Since $X$ and $Y$ are identically distributed
$$E[f(X)g(X)]+E[f(X)g(X)]-E[f(X)]E[g(X)]-E[f(X)]E[g(X)] ge 0$$
$$2E[f(X)g(X)]-2E[f(X)]E[g(X)] ge 0$$
$$E[f(X)g(X)] ge E[f(X)]E[g(X)]$$
answered 4 hours ago
Siong Thye Goh
97.9k1463116
add a comment |
up vote
3
down vote
accepted
Let $X$ and $Y$ be iid, then we have
$$E[(f(X)-f(Y))(g(X)-g(Y))] ge 0$$
$$E[f(X)g(X)+f(Y)g(Y)-f(Y)g(X)-f(X)g(Y)] ge 0$$
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)g(X)]-E[f(X)g(Y)] ge 0$$
By independence,
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)]E[g(X)]-E[f(X)]E[g(Y)] ge 0$$
Since $X$ and $Y$ are identically distributed
$$E[f(X)g(X)]+E[f(X)g(X)]-E[f(X)]E[g(X)]-E[f(X)]E[g(X)] ge 0$$
$$2E[f(X)g(X)]-2E[f(X)]E[g(X)] ge 0$$
$$E[f(X)g(X)] ge E[f(X)]E[g(X)]$$
answered 4 hours ago
Siong Thye Goh
97.9k1463116
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $X$ and $Y$ be iid, then we have
$$E[(f(X)-f(Y))(g(X)-g(Y))] ge 0$$
$$E[f(X)g(X)+f(Y)g(Y)-f(Y)g(X)-f(X)g(Y)] ge 0$$
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)g(X)]-E[f(X)g(Y)] ge 0$$
By independence,
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)]E[g(X)]-E[f(X)]E[g(Y)] ge 0$$
Since $X$ and $Y$ are identically distributed
$$E[f(X)g(X)]+E[f(X)g(X)]-E[f(X)]E[g(X)]-E[f(X)]E[g(X)] ge 0$$
$$2E[f(X)g(X)]-2E[f(X)]E[g(X)] ge 0$$
$$E[f(X)g(X)] ge E[f(X)]E[g(X)]$$
answered 4 hours ago
Siong Thye Goh
97.9k1463116
Let $X$ and $Y$ be iid, then we have
$$E[(f(X)-f(Y))(g(X)-g(Y))] ge 0$$
$$E[f(X)g(X)+f(Y)g(Y)-f(Y)g(X)-f(X)g(Y)] ge 0$$
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)g(X)]-E[f(X)g(Y)] ge 0$$
By independence,
$$E[f(X)g(X)]+E[f(Y)g(Y)]-E[f(Y)]E[g(X)]-E[f(X)]E[g(Y)] ge 0$$
Since $X$ and $Y$ are identically distributed
$$E[f(X)g(X)]+E[f(X)g(X)]-E[f(X)]E[g(X)]-E[f(X)]E[g(X)] ge 0$$
$$2E[f(X)g(X)]-2E[f(X)]E[g(X)] ge 0$$
$$E[f(X)g(X)] ge E[f(X)]E[g(X)]$$
answered 4 hours ago
Siong Thye Goh
97.9k1463116
answered 4 hours ago
Siong Thye Goh
97.9k1463116
answered 4 hours ago
Siong Thye Goh
97.9k1463116
answered 4 hours ago
Siong Thye Goh
97.9k1463116
97.9k1463116
add a comment |
add a comment |
up vote
2
down vote
For every real number $x$, $(f(X)-f(x))(g(X)-g(x)) geq 0$, thus $mathbb{E}[f(X)g(X)] + f(x)g(x) geq f(x)mathbb{E}[g(X)] + g(x)mathbb{E}[f(x)$.
Then integrate wrt $dP_X$.
answered 4 hours ago
Mindlack
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
For every real number $x$, $(f(X)-f(x))(g(X)-g(x)) geq 0$, thus $mathbb{E}[f(X)g(X)] + f(x)g(x) geq f(x)mathbb{E}[g(X)] + g(x)mathbb{E}[f(x)$.
Then integrate wrt $dP_X$.
answered 4 hours ago
Mindlack
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
up vote
2
down vote
For every real number $x$, $(f(X)-f(x))(g(X)-g(x)) geq 0$, thus $mathbb{E}[f(X)g(X)] + f(x)g(x) geq f(x)mathbb{E}[g(X)] + g(x)mathbb{E}[f(x)$.
Then integrate wrt $dP_X$.
answered 4 hours ago
Mindlack
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For every real number $x$, $(f(X)-f(x))(g(X)-g(x)) geq 0$, thus $mathbb{E}[f(X)g(X)] + f(x)g(x) geq f(x)mathbb{E}[g(X)] + g(x)mathbb{E}[f(x)$.
Then integrate wrt $dP_X$.
answered 4 hours ago
Mindlack
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Mindlack
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Mindlack
89716
answered 4 hours ago
Mindlack
89716
89716
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mindlack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3043428%2finequalities-in-probability-theory%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
