Compute $mathbb{E} left(min(X, Y) | max(X, Y) right)$ for $(X,Y)$ i.i.d. uniform on $(0,1)$
Let $X, Y$ be independent random variables with uniform distribution on the interval $[0, 1]$. My task is to find
$$mathbb{E} big(min(X, Y) | max(X, Y) big).$$
I think it can be done in the following way
$$mathbb{E} big(min(X, Y) | max(X, Y) big) = mathbb{E}big(min(X, Y)|sigma(max(X,Y) big) = mathbb{E}big(min(X, Y)|mathcal{F} big).$$
Of course $mathcal{F} = { emptyset, [0,1] }$ thus $min(X, Y)$ is $mathcal{F}$-measurable. That leads us to
$$mathbb{E}big(min(X, Y)|mathcal{F} big) = min(X, Y).$$
Is my reasoning correct? That won't stand for other distributions will it?
probability-theory conditional-expectation
edited Dec 3 '18 at 10:52
Did
246k23220454
asked Dec 3 '18 at 8:33
Hendrra
1,079416
|
show 5 more comments
Let $X, Y$ be independent random variables with uniform distribution on the interval $[0, 1]$. My task is to find
$$mathbb{E} big(min(X, Y) | max(X, Y) big).$$
I think it can be done in the following way
$$mathbb{E} big(min(X, Y) | max(X, Y) big) = mathbb{E}big(min(X, Y)|sigma(max(X,Y) big) = mathbb{E}big(min(X, Y)|mathcal{F} big).$$
Of course $mathcal{F} = { emptyset, [0,1] }$ thus $min(X, Y)$ is $mathcal{F}$-measurable. That leads us to
$$mathbb{E}big(min(X, Y)|mathcal{F} big) = min(X, Y).$$
Is my reasoning correct? That won't stand for other distributions will it?
probability-theory conditional-expectation
edited Dec 3 '18 at 10:52
Did
246k23220454
asked Dec 3 '18 at 8:33
Hendrra
1,079416
How did you get $mathcal F$?
– Kavi Rama Murthy
Dec 3 '18 at 8:35
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
– Hendrra
Dec 3 '18 at 8:37
2
My question is why it is ${emptyset, [0,1]}$. This is not true.
– Kavi Rama Murthy
Dec 3 '18 at 8:38
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
– Hendrra
Dec 3 '18 at 8:41
1
A set like ${max (X,Y) < frac 1 2} ={X< frac 1 2 text {and} Y< frac 1 2 }$ belongs to $mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $frac 1 4$.
– Kavi Rama Murthy
Dec 3 '18 at 8:44
|
show 5 more comments
Let $X, Y$ be independent random variables with uniform distribution on the interval $[0, 1]$. My task is to find
$$mathbb{E} big(min(X, Y) | max(X, Y) big).$$
I think it can be done in the following way
$$mathbb{E} big(min(X, Y) | max(X, Y) big) = mathbb{E}big(min(X, Y)|sigma(max(X,Y) big) = mathbb{E}big(min(X, Y)|mathcal{F} big).$$
Of course $mathcal{F} = { emptyset, [0,1] }$ thus $min(X, Y)$ is $mathcal{F}$-measurable. That leads us to
$$mathbb{E}big(min(X, Y)|mathcal{F} big) = min(X, Y).$$
Is my reasoning correct? That won't stand for other distributions will it?
probability-theory conditional-expectation
edited Dec 3 '18 at 10:52
Did
246k23220454
asked Dec 3 '18 at 8:33
Hendrra
1,079416
Let $X, Y$ be independent random variables with uniform distribution on the interval $[0, 1]$. My task is to find
$$mathbb{E} big(min(X, Y) | max(X, Y) big).$$
I think it can be done in the following way
$$mathbb{E} big(min(X, Y) | max(X, Y) big) = mathbb{E}big(min(X, Y)|sigma(max(X,Y) big) = mathbb{E}big(min(X, Y)|mathcal{F} big).$$
Of course $mathcal{F} = { emptyset, [0,1] }$ thus $min(X, Y)$ is $mathcal{F}$-measurable. That leads us to
$$mathbb{E}big(min(X, Y)|mathcal{F} big) = min(X, Y).$$
Is my reasoning correct? That won't stand for other distributions will it?
probability-theory conditional-expectation
probability-theory conditional-expectation
edited Dec 3 '18 at 10:52
Did
246k23220454
asked Dec 3 '18 at 8:33
Hendrra
1,079416
edited Dec 3 '18 at 10:52
Did
246k23220454
asked Dec 3 '18 at 8:33
Hendrra
1,079416
edited Dec 3 '18 at 10:52
Did
246k23220454
edited Dec 3 '18 at 10:52
Did
246k23220454
edited Dec 3 '18 at 10:52
Did
246k23220454
246k23220454
asked Dec 3 '18 at 8:33
Hendrra
1,079416
asked Dec 3 '18 at 8:33
Hendrra
1,079416
asked Dec 3 '18 at 8:33
Hendrra
1,079416
1,079416
How did you get $mathcal F$?
– Kavi Rama Murthy
Dec 3 '18 at 8:35
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
– Hendrra
Dec 3 '18 at 8:37
2
My question is why it is ${emptyset, [0,1]}$. This is not true.
– Kavi Rama Murthy
Dec 3 '18 at 8:38
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
– Hendrra
Dec 3 '18 at 8:41
1
A set like ${max (X,Y) < frac 1 2} ={X< frac 1 2 text {and} Y< frac 1 2 }$ belongs to $mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $frac 1 4$.
– Kavi Rama Murthy
Dec 3 '18 at 8:44
|
show 5 more comments
How did you get $mathcal F$?
– Kavi Rama Murthy
Dec 3 '18 at 8:35
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
– Hendrra
Dec 3 '18 at 8:37
2
My question is why it is ${emptyset, [0,1]}$. This is not true.
– Kavi Rama Murthy
Dec 3 '18 at 8:38
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
– Hendrra
Dec 3 '18 at 8:41
1
A set like ${max (X,Y) < frac 1 2} ={X< frac 1 2 text {and} Y< frac 1 2 }$ belongs to $mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $frac 1 4$.
– Kavi Rama Murthy
Dec 3 '18 at 8:44
How did you get $mathcal F$?
– Kavi Rama Murthy
Dec 3 '18 at 8:35
How did you get $mathcal F$?
– Kavi Rama Murthy
Dec 3 '18 at 8:35
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
– Hendrra
Dec 3 '18 at 8:37
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
– Hendrra
Dec 3 '18 at 8:37
2
2
My question is why it is ${emptyset, [0,1]}$. This is not true.
– Kavi Rama Murthy
Dec 3 '18 at 8:38
My question is why it is ${emptyset, [0,1]}$. This is not true.
– Kavi Rama Murthy
Dec 3 '18 at 8:38
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
– Hendrra
Dec 3 '18 at 8:41
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
– Hendrra
Dec 3 '18 at 8:41
1
1
A set like ${max (X,Y) < frac 1 2} ={X< frac 1 2 text {and} Y< frac 1 2 }$ belongs to $mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $frac 1 4$.
– Kavi Rama Murthy
Dec 3 '18 at 8:44
A set like ${max (X,Y) < frac 1 2} ={X< frac 1 2 text {and} Y< frac 1 2 }$ belongs to $mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $frac 1 4$.
– Kavi Rama Murthy
Dec 3 '18 at 8:44
|
show 5 more comments
1 Answer
1
active
oldest
votes
drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.
To this end, let $Z = max(X,Y)$. We want to show
$$
E[min(X,Y)|sigma(Z)] = frac{Z}{2} quad text{a.s.}
$$
The right hand side is obviously $sigma(Z)$-measurable. So all we need to show is
$$
E[min(X,Y)mathbb{1}_F] = Eleft[frac{Z}{2}mathbb{1}_Fright]
$$
for all $F in sigma(Z)$.
We will show this property for $F = {Z leq a }$ with $a in mathbb{R}$. This is enough because these sets form a $pi$-system (meaning it's $cap$-stable) generating $sigma(Z)$ and we can use a dynkin system argument to get the property for all $F in sigma(Z)$. Let me know if I should go into more detail here.
Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2zmathbb{1}_{[0,1]}(z)$. So we get for the right hand side
$$
Eleft[frac{Z}{2}mathbb{1}_Fright] = frac12 int_{-infty}^azdP^Z(z) = int_{-infty}^a z^2 mathbb{1}_{[0,1]}(z)dz\
=
begin{cases}
0 quad &a < 0\
frac13 a^3 quad &a in [0,1]\
frac13 quad &a > 1
end{cases}.
$$
For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = {Z leq a } = {X leq a } cap {Y leq a }$,
$$
E[min(X,Y)mathbb{1}_F] = 2E[min(X,Y)mathbb{1}_Fmathbb{1}_{{X leq Y}}] = 2E[X mathbb{1}_{{X leq a}}mathbb{1}_{{Y leq a}}mathbb{1}_{{X leq Y}}]\
= 2int_0^1int_0^1 x mathbb{1}_{{x leq a}} mathbb{1}_{{y leq a}}mathbb{1}_{{x leq y}} dy dx\
= 2 int_0^1 x mathbb{1}_{{x leq a}} int_x^1 mathbb{1}_{{y leq a}} dy dx.
$$
Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.
answered Dec 3 '18 at 10:29
Tki Deneb
26710
1
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
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%2f3023781%2fcompute-mathbbe-left-minx-y-maxx-y-right-for-x-y-i-i-d-un%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
drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.
To this end, let $Z = max(X,Y)$. We want to show
$$
E[min(X,Y)|sigma(Z)] = frac{Z}{2} quad text{a.s.}
$$
The right hand side is obviously $sigma(Z)$-measurable. So all we need to show is
$$
E[min(X,Y)mathbb{1}_F] = Eleft[frac{Z}{2}mathbb{1}_Fright]
$$
for all $F in sigma(Z)$.
We will show this property for $F = {Z leq a }$ with $a in mathbb{R}$. This is enough because these sets form a $pi$-system (meaning it's $cap$-stable) generating $sigma(Z)$ and we can use a dynkin system argument to get the property for all $F in sigma(Z)$. Let me know if I should go into more detail here.
Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2zmathbb{1}_{[0,1]}(z)$. So we get for the right hand side
$$
Eleft[frac{Z}{2}mathbb{1}_Fright] = frac12 int_{-infty}^azdP^Z(z) = int_{-infty}^a z^2 mathbb{1}_{[0,1]}(z)dz\
=
begin{cases}
0 quad &a < 0\
frac13 a^3 quad &a in [0,1]\
frac13 quad &a > 1
end{cases}.
$$
For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = {Z leq a } = {X leq a } cap {Y leq a }$,
$$
E[min(X,Y)mathbb{1}_F] = 2E[min(X,Y)mathbb{1}_Fmathbb{1}_{{X leq Y}}] = 2E[X mathbb{1}_{{X leq a}}mathbb{1}_{{Y leq a}}mathbb{1}_{{X leq Y}}]\
= 2int_0^1int_0^1 x mathbb{1}_{{x leq a}} mathbb{1}_{{y leq a}}mathbb{1}_{{x leq y}} dy dx\
= 2 int_0^1 x mathbb{1}_{{x leq a}} int_x^1 mathbb{1}_{{y leq a}} dy dx.
$$
Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.
answered Dec 3 '18 at 10:29
Tki Deneb
26710
1
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
add a comment |
drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.
To this end, let $Z = max(X,Y)$. We want to show
$$
E[min(X,Y)|sigma(Z)] = frac{Z}{2} quad text{a.s.}
$$
The right hand side is obviously $sigma(Z)$-measurable. So all we need to show is
$$
E[min(X,Y)mathbb{1}_F] = Eleft[frac{Z}{2}mathbb{1}_Fright]
$$
for all $F in sigma(Z)$.
We will show this property for $F = {Z leq a }$ with $a in mathbb{R}$. This is enough because these sets form a $pi$-system (meaning it's $cap$-stable) generating $sigma(Z)$ and we can use a dynkin system argument to get the property for all $F in sigma(Z)$. Let me know if I should go into more detail here.
Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2zmathbb{1}_{[0,1]}(z)$. So we get for the right hand side
$$
Eleft[frac{Z}{2}mathbb{1}_Fright] = frac12 int_{-infty}^azdP^Z(z) = int_{-infty}^a z^2 mathbb{1}_{[0,1]}(z)dz\
=
begin{cases}
0 quad &a < 0\
frac13 a^3 quad &a in [0,1]\
frac13 quad &a > 1
end{cases}.
$$
For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = {Z leq a } = {X leq a } cap {Y leq a }$,
$$
E[min(X,Y)mathbb{1}_F] = 2E[min(X,Y)mathbb{1}_Fmathbb{1}_{{X leq Y}}] = 2E[X mathbb{1}_{{X leq a}}mathbb{1}_{{Y leq a}}mathbb{1}_{{X leq Y}}]\
= 2int_0^1int_0^1 x mathbb{1}_{{x leq a}} mathbb{1}_{{y leq a}}mathbb{1}_{{x leq y}} dy dx\
= 2 int_0^1 x mathbb{1}_{{x leq a}} int_x^1 mathbb{1}_{{y leq a}} dy dx.
$$
Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.
answered Dec 3 '18 at 10:29
Tki Deneb
26710
1
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
add a comment |
drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.
To this end, let $Z = max(X,Y)$. We want to show
$$
E[min(X,Y)|sigma(Z)] = frac{Z}{2} quad text{a.s.}
$$
The right hand side is obviously $sigma(Z)$-measurable. So all we need to show is
$$
E[min(X,Y)mathbb{1}_F] = Eleft[frac{Z}{2}mathbb{1}_Fright]
$$
for all $F in sigma(Z)$.
We will show this property for $F = {Z leq a }$ with $a in mathbb{R}$. This is enough because these sets form a $pi$-system (meaning it's $cap$-stable) generating $sigma(Z)$ and we can use a dynkin system argument to get the property for all $F in sigma(Z)$. Let me know if I should go into more detail here.
Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2zmathbb{1}_{[0,1]}(z)$. So we get for the right hand side
$$
Eleft[frac{Z}{2}mathbb{1}_Fright] = frac12 int_{-infty}^azdP^Z(z) = int_{-infty}^a z^2 mathbb{1}_{[0,1]}(z)dz\
=
begin{cases}
0 quad &a < 0\
frac13 a^3 quad &a in [0,1]\
frac13 quad &a > 1
end{cases}.
$$
For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = {Z leq a } = {X leq a } cap {Y leq a }$,
$$
E[min(X,Y)mathbb{1}_F] = 2E[min(X,Y)mathbb{1}_Fmathbb{1}_{{X leq Y}}] = 2E[X mathbb{1}_{{X leq a}}mathbb{1}_{{Y leq a}}mathbb{1}_{{X leq Y}}]\
= 2int_0^1int_0^1 x mathbb{1}_{{x leq a}} mathbb{1}_{{y leq a}}mathbb{1}_{{x leq y}} dy dx\
= 2 int_0^1 x mathbb{1}_{{x leq a}} int_x^1 mathbb{1}_{{y leq a}} dy dx.
$$
Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.
answered Dec 3 '18 at 10:29
Tki Deneb
26710
drhab has already commented how to informally derive the solution. I will formally prove that this is in fact the correct solution.
To this end, let $Z = max(X,Y)$. We want to show
$$
E[min(X,Y)|sigma(Z)] = frac{Z}{2} quad text{a.s.}
$$
The right hand side is obviously $sigma(Z)$-measurable. So all we need to show is
$$
E[min(X,Y)mathbb{1}_F] = Eleft[frac{Z}{2}mathbb{1}_Fright]
$$
for all $F in sigma(Z)$.
We will show this property for $F = {Z leq a }$ with $a in mathbb{R}$. This is enough because these sets form a $pi$-system (meaning it's $cap$-stable) generating $sigma(Z)$ and we can use a dynkin system argument to get the property for all $F in sigma(Z)$. Let me know if I should go into more detail here.
Because $X$ and $Y$ are iid $U([0,1])$, $Z$ has the Lebesgue density $f_Z(z) = 2zmathbb{1}_{[0,1]}(z)$. So we get for the right hand side
$$
Eleft[frac{Z}{2}mathbb{1}_Fright] = frac12 int_{-infty}^azdP^Z(z) = int_{-infty}^a z^2 mathbb{1}_{[0,1]}(z)dz\
=
begin{cases}
0 quad &a < 0\
frac13 a^3 quad &a in [0,1]\
frac13 quad &a > 1
end{cases}.
$$
For the left hand side, we get, using the symmetry of $X$ and $Y$ and $F = {Z leq a } = {X leq a } cap {Y leq a }$,
$$
E[min(X,Y)mathbb{1}_F] = 2E[min(X,Y)mathbb{1}_Fmathbb{1}_{{X leq Y}}] = 2E[X mathbb{1}_{{X leq a}}mathbb{1}_{{Y leq a}}mathbb{1}_{{X leq Y}}]\
= 2int_0^1int_0^1 x mathbb{1}_{{x leq a}} mathbb{1}_{{y leq a}}mathbb{1}_{{x leq y}} dy dx\
= 2 int_0^1 x mathbb{1}_{{x leq a}} int_x^1 mathbb{1}_{{y leq a}} dy dx.
$$
Now you can distinguish the same $3$ cases as above and notice that you arrive at the same values.
answered Dec 3 '18 at 10:29
Tki Deneb
26710
edited Dec 3 '18 at 10:53
answered Dec 3 '18 at 10:29
Tki Deneb
26710
answered Dec 3 '18 at 10:29
Tki Deneb
26710
answered Dec 3 '18 at 10:29
Tki Deneb
26710
26710
1
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
add a comment |
1
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
1
1
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
– Did
Dec 3 '18 at 10:51
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%2f3023781%2fcompute-mathbbe-left-minx-y-maxx-y-right-for-x-y-i-i-d-un%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
How did you get $mathcal F$?
– Kavi Rama Murthy
Dec 3 '18 at 8:35
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
– Hendrra
Dec 3 '18 at 8:37
2
My question is why it is ${emptyset, [0,1]}$. This is not true.
– Kavi Rama Murthy
Dec 3 '18 at 8:38
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
– Hendrra
Dec 3 '18 at 8:41
1
A set like ${max (X,Y) < frac 1 2} ={X< frac 1 2 text {and} Y< frac 1 2 }$ belongs to $mathcal F$. This set is neither empty nor $[0,1]$ because its probability is $frac 1 4$.
– Kavi Rama Murthy
Dec 3 '18 at 8:44