Prove that if $mathbb{E}[g(Y)]leq mathbb{E}[g(X)]$ for every nondecreasing function $g$ then $F_Xle F_Y$...
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Let $X,Y$ random variables. Every coninuous and not-decreasing monotone function $g:mathbb{R}to[0,1]$ fulfills $mathbb{E}[g(Y)]leq mathbb{E}[g(X)]$. Prove that $F_X(t)leq F_Y(t)$.
I don't have any initial idea to solve it.
probability random-variables expected-value
edited Jan 12 at 22:21
Did
247k23223460
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
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closed as off-topic by RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B Jan 15 at 23:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $X,Y$ random variables. Every coninuous and not-decreasing monotone function $g:mathbb{R}to[0,1]$ fulfills $mathbb{E}[g(Y)]leq mathbb{E}[g(X)]$. Prove that $F_X(t)leq F_Y(t)$.
I don't have any initial idea to solve it.
probability random-variables expected-value
edited Jan 12 at 22:21
Did
247k23223460
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
$endgroup$
closed as off-topic by RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B Jan 15 at 23:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What do you know about Distribution Functions?
$endgroup$
– Mark Viola
Jan 12 at 21:08
$begingroup$
Are you familiar with the Riemann Stieltjes Integral?
$endgroup$
– Mark Viola
Jan 12 at 21:17
add a comment |
$begingroup$
Let $X,Y$ random variables. Every coninuous and not-decreasing monotone function $g:mathbb{R}to[0,1]$ fulfills $mathbb{E}[g(Y)]leq mathbb{E}[g(X)]$. Prove that $F_X(t)leq F_Y(t)$.
I don't have any initial idea to solve it.
probability random-variables expected-value
edited Jan 12 at 22:21
Did
247k23223460
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
$endgroup$
Let $X,Y$ random variables. Every coninuous and not-decreasing monotone function $g:mathbb{R}to[0,1]$ fulfills $mathbb{E}[g(Y)]leq mathbb{E}[g(X)]$. Prove that $F_X(t)leq F_Y(t)$.
I don't have any initial idea to solve it.
probability random-variables expected-value
probability random-variables expected-value
edited Jan 12 at 22:21
Did
247k23223460
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
edited Jan 12 at 22:21
Did
247k23223460
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
edited Jan 12 at 22:21
Did
247k23223460
edited Jan 12 at 22:21
Did
247k23223460
edited Jan 12 at 22:21
Did
247k23223460
247k23223460
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
asked Jan 12 at 20:55
J. DoeJ. Doe
14110
14110
closed as off-topic by RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B Jan 15 at 23:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B Jan 15 at 23:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Riccardo.Alestra, José Carlos Santos, amWhy, John B
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What do you know about Distribution Functions?
$endgroup$
– Mark Viola
Jan 12 at 21:08
$begingroup$
Are you familiar with the Riemann Stieltjes Integral?
$endgroup$
– Mark Viola
Jan 12 at 21:17
add a comment |
$begingroup$
What do you know about Distribution Functions?
$endgroup$
– Mark Viola
Jan 12 at 21:08
$begingroup$
Are you familiar with the Riemann Stieltjes Integral?
$endgroup$
– Mark Viola
Jan 12 at 21:17
$begingroup$
What do you know about Distribution Functions?
$endgroup$
– Mark Viola
Jan 12 at 21:08
$begingroup$
What do you know about Distribution Functions?
$endgroup$
– Mark Viola
Jan 12 at 21:08
$begingroup$
Are you familiar with the Riemann Stieltjes Integral?
$endgroup$
– Mark Viola
Jan 12 at 21:17
$begingroup$
Are you familiar with the Riemann Stieltjes Integral?
$endgroup$
– Mark Viola
Jan 12 at 21:17
add a comment |
2 Answers
2
active
oldest
votes
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Hint: For a fixed $t$ and $epsilon>0$, define $g_epsilon(x)$ as
$$g_epsilon(x)=begin{cases}0,quad xle t\frac{x-t}{epsilon},quad t<xle t+epsilon\1,quad t+epsilon<xend{cases}.
$$ We can see that $g_epsilon uparrow 1_{(t,infty)}(x)$ as $epsilonto 0$. Now, by the assumption it holds that
$$
E[g_epsilon(Y)]leq E[g_epsilon(X)],quadforall epsilon>0.
$$ From this fact, deduce that
$$
P(t<Y)=E[1_{{t<Y}}]le E[1_{{t<X}}]=P(t<X)
$$ for all $tinmathbb{R}$. (e.g. by monotone convergence theorem)
answered Jan 12 at 21:17
SongSong
10.8k628
$endgroup$
add a comment |
$begingroup$
HINT:
Use Integration by Parts of the Riemann-Stieltjes Integrals
$$ int_{-infty}^infty g(t),dF_X(t)$$
and
$$ int_{-infty}^infty g(t),dF_Y(t)$$
Then use the fact that $g$ is increasing and $F_X$ and $F_Y$ are non-negative.
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: For a fixed $t$ and $epsilon>0$, define $g_epsilon(x)$ as
$$g_epsilon(x)=begin{cases}0,quad xle t\frac{x-t}{epsilon},quad t<xle t+epsilon\1,quad t+epsilon<xend{cases}.
$$ We can see that $g_epsilon uparrow 1_{(t,infty)}(x)$ as $epsilonto 0$. Now, by the assumption it holds that
$$
E[g_epsilon(Y)]leq E[g_epsilon(X)],quadforall epsilon>0.
$$ From this fact, deduce that
$$
P(t<Y)=E[1_{{t<Y}}]le E[1_{{t<X}}]=P(t<X)
$$ for all $tinmathbb{R}$. (e.g. by monotone convergence theorem)
answered Jan 12 at 21:17
SongSong
10.8k628
$endgroup$
add a comment |
$begingroup$
Hint: For a fixed $t$ and $epsilon>0$, define $g_epsilon(x)$ as
$$g_epsilon(x)=begin{cases}0,quad xle t\frac{x-t}{epsilon},quad t<xle t+epsilon\1,quad t+epsilon<xend{cases}.
$$ We can see that $g_epsilon uparrow 1_{(t,infty)}(x)$ as $epsilonto 0$. Now, by the assumption it holds that
$$
E[g_epsilon(Y)]leq E[g_epsilon(X)],quadforall epsilon>0.
$$ From this fact, deduce that
$$
P(t<Y)=E[1_{{t<Y}}]le E[1_{{t<X}}]=P(t<X)
$$ for all $tinmathbb{R}$. (e.g. by monotone convergence theorem)
answered Jan 12 at 21:17
SongSong
10.8k628
$endgroup$
add a comment |
$begingroup$
Hint: For a fixed $t$ and $epsilon>0$, define $g_epsilon(x)$ as
$$g_epsilon(x)=begin{cases}0,quad xle t\frac{x-t}{epsilon},quad t<xle t+epsilon\1,quad t+epsilon<xend{cases}.
$$ We can see that $g_epsilon uparrow 1_{(t,infty)}(x)$ as $epsilonto 0$. Now, by the assumption it holds that
$$
E[g_epsilon(Y)]leq E[g_epsilon(X)],quadforall epsilon>0.
$$ From this fact, deduce that
$$
P(t<Y)=E[1_{{t<Y}}]le E[1_{{t<X}}]=P(t<X)
$$ for all $tinmathbb{R}$. (e.g. by monotone convergence theorem)
answered Jan 12 at 21:17
SongSong
10.8k628
$endgroup$
Hint: For a fixed $t$ and $epsilon>0$, define $g_epsilon(x)$ as
$$g_epsilon(x)=begin{cases}0,quad xle t\frac{x-t}{epsilon},quad t<xle t+epsilon\1,quad t+epsilon<xend{cases}.
$$ We can see that $g_epsilon uparrow 1_{(t,infty)}(x)$ as $epsilonto 0$. Now, by the assumption it holds that
$$
E[g_epsilon(Y)]leq E[g_epsilon(X)],quadforall epsilon>0.
$$ From this fact, deduce that
$$
P(t<Y)=E[1_{{t<Y}}]le E[1_{{t<X}}]=P(t<X)
$$ for all $tinmathbb{R}$. (e.g. by monotone convergence theorem)
answered Jan 12 at 21:17
SongSong
10.8k628
edited Jan 12 at 21:23
answered Jan 12 at 21:17
SongSong
10.8k628
answered Jan 12 at 21:17
SongSong
10.8k628
answered Jan 12 at 21:17
SongSong
10.8k628
10.8k628
add a comment |
add a comment |
$begingroup$
HINT:
Use Integration by Parts of the Riemann-Stieltjes Integrals
$$ int_{-infty}^infty g(t),dF_X(t)$$
and
$$ int_{-infty}^infty g(t),dF_Y(t)$$
Then use the fact that $g$ is increasing and $F_X$ and $F_Y$ are non-negative.
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
add a comment |
$begingroup$
HINT:
Use Integration by Parts of the Riemann-Stieltjes Integrals
$$ int_{-infty}^infty g(t),dF_X(t)$$
and
$$ int_{-infty}^infty g(t),dF_Y(t)$$
Then use the fact that $g$ is increasing and $F_X$ and $F_Y$ are non-negative.
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
add a comment |
$begingroup$
HINT:
Use Integration by Parts of the Riemann-Stieltjes Integrals
$$ int_{-infty}^infty g(t),dF_X(t)$$
and
$$ int_{-infty}^infty g(t),dF_Y(t)$$
Then use the fact that $g$ is increasing and $F_X$ and $F_Y$ are non-negative.
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
$endgroup$
HINT:
Use Integration by Parts of the Riemann-Stieltjes Integrals
$$ int_{-infty}^infty g(t),dF_X(t)$$
and
$$ int_{-infty}^infty g(t),dF_Y(t)$$
Then use the fact that $g$ is increasing and $F_X$ and $F_Y$ are non-negative.
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
edited Jan 12 at 22:29
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
answered Jan 12 at 21:22
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
add a comment |
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
Your $F_x$ and $F_y$ should read $F_X$ and $F_Y$. (The error is obvious but it appears repeatedly on the site, and I have no idea why.)
$endgroup$
– Did
Jan 12 at 22:22
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
$begingroup$
@did Didier, thank you. I've corrected the notation. This shows the risks that arise when working from a cell phone.
$endgroup$
– Mark Viola
Jan 12 at 22:31
add a comment |
$begingroup$
What do you know about Distribution Functions?
$endgroup$
– Mark Viola
Jan 12 at 21:08
$begingroup$
Are you familiar with the Riemann Stieltjes Integral?
$endgroup$
– Mark Viola
Jan 12 at 21:17