Compute $mathbb{E} left(min(X, Y) | max(X, Y) right)$ for $(X,Y)$ i.i.d. uniform on $(0,1)$












3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    How did you get $mathcal F$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:35










  • $begingroup$
    @KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:37








  • 2




    $begingroup$
    My question is why it is ${emptyset, [0,1]}$. This is not true.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:38










  • $begingroup$
    @KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:41






  • 1




    $begingroup$
    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 $frac 1 4$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:44


















3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    How did you get $mathcal F$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:35










  • $begingroup$
    @KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:37








  • 2




    $begingroup$
    My question is why it is ${emptyset, [0,1]}$. This is not true.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:38










  • $begingroup$
    @KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:41






  • 1




    $begingroup$
    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 $frac 1 4$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:44
















3












3








3


2



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 10:52









Did

247k23222457




247k23222457










asked Dec 3 '18 at 8:33









HendrraHendrra

1,161516




1,161516












  • $begingroup$
    How did you get $mathcal F$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:35










  • $begingroup$
    @KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:37








  • 2




    $begingroup$
    My question is why it is ${emptyset, [0,1]}$. This is not true.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:38










  • $begingroup$
    @KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:41






  • 1




    $begingroup$
    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 $frac 1 4$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:44




















  • $begingroup$
    How did you get $mathcal F$?
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:35










  • $begingroup$
    @KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:37








  • 2




    $begingroup$
    My question is why it is ${emptyset, [0,1]}$. This is not true.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:38










  • $begingroup$
    @KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
    $endgroup$
    – Hendrra
    Dec 3 '18 at 8:41






  • 1




    $begingroup$
    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 $frac 1 4$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 8:44


















$begingroup$
How did you get $mathcal F$?
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 8:35




$begingroup$
How did you get $mathcal F$?
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 8:35












$begingroup$
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
$endgroup$
– Hendrra
Dec 3 '18 at 8:37






$begingroup$
@KaviRamaMurthy It is the $sigma$-algebra made of $max(X, Y)$. It's just the new symbol for $sigma(max(X,Y))$
$endgroup$
– Hendrra
Dec 3 '18 at 8:37






2




2




$begingroup$
My question is why it is ${emptyset, [0,1]}$. This is not true.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 8:38




$begingroup$
My question is why it is ${emptyset, [0,1]}$. This is not true.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 8:38












$begingroup$
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
$endgroup$
– Hendrra
Dec 3 '18 at 8:41




$begingroup$
@KaviRamaMurthy I thought that $max{X, Y}$ will be $X$ or $Y$. Because both are distributed uniformly thus only two inverse images exist.
$endgroup$
– Hendrra
Dec 3 '18 at 8:41




1




1




$begingroup$
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 $frac 1 4$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 8:44






$begingroup$
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 $frac 1 4$.
$endgroup$
– Kavi Rama Murthy
Dec 3 '18 at 8:44












1 Answer
1






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    (+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
    $endgroup$
    – Did
    Dec 3 '18 at 10:51













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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    (+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
    $endgroup$
    – Did
    Dec 3 '18 at 10:51


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    (+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
    $endgroup$
    – Did
    Dec 3 '18 at 10:51
















5












5








5





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 10:53

























answered Dec 3 '18 at 10:29









Tki DenebTki Deneb

29210




29210








  • 1




    $begingroup$
    (+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
    $endgroup$
    – Did
    Dec 3 '18 at 10:51
















  • 1




    $begingroup$
    (+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
    $endgroup$
    – Did
    Dec 3 '18 at 10:51










1




1




$begingroup$
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
$endgroup$
– Did
Dec 3 '18 at 10:51






$begingroup$
(+1) The phrase "a $cap$-stable generator of $sigma(Z)$" could be replaced by the more standard "a $pi$-system generating $sigma(Z)$.
$endgroup$
– Did
Dec 3 '18 at 10:51




















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