Inequalities in probability theory
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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 2 hours ago
Neymar
374113
add a comment |
up vote
2
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 2 hours ago
Neymar
374113
add a comment |
up vote
2
down vote
favorite
up vote
2
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 2 hours ago
Neymar
374113
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 2 hours ago
Neymar
374113
asked 2 hours ago
Neymar
374113
asked 2 hours ago
Neymar
374113
asked 2 hours ago
Neymar
374113
asked 2 hours ago
Neymar
374113
374113
add a comment |
add a comment |
2 Answers
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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 1 hour ago
Siong Thye Goh
97.9k1463116
add a comment |
up vote
1
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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 1 hour ago
Mindlack
88716
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
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 1 hour ago
Siong Thye Goh
97.9k1463116
add a comment |
up vote
2
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 1 hour ago
Siong Thye Goh
97.9k1463116
add a comment |
up vote
2
down vote
accepted
up vote
2
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 1 hour 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 1 hour ago
Siong Thye Goh
97.9k1463116
answered 1 hour ago
Siong Thye Goh
97.9k1463116
answered 1 hour ago
Siong Thye Goh
97.9k1463116
answered 1 hour ago
Siong Thye Goh
97.9k1463116
97.9k1463116
add a comment |
add a comment |
up vote
1
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 1 hour ago
Mindlack
88716
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
1
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 1 hour ago
Mindlack
88716
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
1
down vote
up vote
1
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 1 hour ago
Mindlack
88716
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 1 hour ago
Mindlack
88716
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 1 hour ago
Mindlack
88716
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 1 hour ago
Mindlack
88716
answered 1 hour ago
Mindlack
88716
88716
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 |
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