Independence between random vector and event












1












$begingroup$



Let $U_1, U_2$ and $U_3$ be three independent uniformly $(0, 1)$ random variables.



Let $X_1,...,X_n$ be a sequence of independent uniformly $(0, 1)$ random variables.



Consider that $X_i$ and $U_j$ are independents, for all $i,j$.



Show that the event ${U_1 > U_2 > U_3}$ is independent of the $(U_{(3)},X_{(1)})$, where $U_{(3)} = max{U_1,U_2,U_3}$ and $X_{(1)} = min{X_1,...,X_n}$.




I have no idea to start. How'd be the definition of independence between events and random variables?






probability probability-theory probability-distributions independence uniform-distribution







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$endgroup$








  • 1




    $begingroup$
    As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[Xlt t]$, or $B=[Xin S]$, or most generally, for all $B$ in the sigma field $mathcal{F}(X)$ generated by $X$
    $endgroup$
    – kimchi lover
    Nov 28 '18 at 22:40








  • 1




    $begingroup$
    You also need independence between $U_k$ and$X_j$ for all indices.
    $endgroup$
    – herb steinberg
    Nov 28 '18 at 22:46
















1












$begingroup$



Let $U_1, U_2$ and $U_3$ be three independent uniformly $(0, 1)$ random variables.



Let $X_1,...,X_n$ be a sequence of independent uniformly $(0, 1)$ random variables.



Consider that $X_i$ and $U_j$ are independents, for all $i,j$.



Show that the event ${U_1 > U_2 > U_3}$ is independent of the $(U_{(3)},X_{(1)})$, where $U_{(3)} = max{U_1,U_2,U_3}$ and $X_{(1)} = min{X_1,...,X_n}$.




I have no idea to start. How'd be the definition of independence between events and random variables?






probability probability-theory probability-distributions independence uniform-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[Xlt t]$, or $B=[Xin S]$, or most generally, for all $B$ in the sigma field $mathcal{F}(X)$ generated by $X$
    $endgroup$
    – kimchi lover
    Nov 28 '18 at 22:40








  • 1




    $begingroup$
    You also need independence between $U_k$ and$X_j$ for all indices.
    $endgroup$
    – herb steinberg
    Nov 28 '18 at 22:46














1












1








1


2



$begingroup$



Let $U_1, U_2$ and $U_3$ be three independent uniformly $(0, 1)$ random variables.



Let $X_1,...,X_n$ be a sequence of independent uniformly $(0, 1)$ random variables.



Consider that $X_i$ and $U_j$ are independents, for all $i,j$.



Show that the event ${U_1 > U_2 > U_3}$ is independent of the $(U_{(3)},X_{(1)})$, where $U_{(3)} = max{U_1,U_2,U_3}$ and $X_{(1)} = min{X_1,...,X_n}$.




I have no idea to start. How'd be the definition of independence between events and random variables?






probability probability-theory probability-distributions independence uniform-distribution







share|cite|improve this question











$endgroup$





Let $U_1, U_2$ and $U_3$ be three independent uniformly $(0, 1)$ random variables.



Let $X_1,...,X_n$ be a sequence of independent uniformly $(0, 1)$ random variables.



Consider that $X_i$ and $U_j$ are independents, for all $i,j$.



Show that the event ${U_1 > U_2 > U_3}$ is independent of the $(U_{(3)},X_{(1)})$, where $U_{(3)} = max{U_1,U_2,U_3}$ and $X_{(1)} = min{X_1,...,X_n}$.




I have no idea to start. How'd be the definition of independence between events and random variables?






probability probability-theory probability-distributions independence uniform-distribution




probability probability-theory probability-distributions independence uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 10:00









Davide Giraudo

126k16150261




126k16150261










asked Nov 28 '18 at 21:18









Pedro SalgadoPedro Salgado

735




735








  • 1




    $begingroup$
    As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[Xlt t]$, or $B=[Xin S]$, or most generally, for all $B$ in the sigma field $mathcal{F}(X)$ generated by $X$
    $endgroup$
    – kimchi lover
    Nov 28 '18 at 22:40








  • 1




    $begingroup$
    You also need independence between $U_k$ and$X_j$ for all indices.
    $endgroup$
    – herb steinberg
    Nov 28 '18 at 22:46














  • 1




    $begingroup$
    As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[Xlt t]$, or $B=[Xin S]$, or most generally, for all $B$ in the sigma field $mathcal{F}(X)$ generated by $X$
    $endgroup$
    – kimchi lover
    Nov 28 '18 at 22:40








  • 1




    $begingroup$
    You also need independence between $U_k$ and$X_j$ for all indices.
    $endgroup$
    – herb steinberg
    Nov 28 '18 at 22:46








1




1




$begingroup$
As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[Xlt t]$, or $B=[Xin S]$, or most generally, for all $B$ in the sigma field $mathcal{F}(X)$ generated by $X$
$endgroup$
– kimchi lover
Nov 28 '18 at 22:40






$begingroup$
As for your last question: the rv. $X$ is independent of an event $A$ if all events $B$ defined in terms of $X$ are independent of $A$, such as $B=[Xlt t]$, or $B=[Xin S]$, or most generally, for all $B$ in the sigma field $mathcal{F}(X)$ generated by $X$
$endgroup$
– kimchi lover
Nov 28 '18 at 22:40






1




1




$begingroup$
You also need independence between $U_k$ and$X_j$ for all indices.
$endgroup$
– herb steinberg
Nov 28 '18 at 22:46




$begingroup$
You also need independence between $U_k$ and$X_j$ for all indices.
$endgroup$
– herb steinberg
Nov 28 '18 at 22:46










1 Answer
1






active

oldest

votes


















2












$begingroup$

An event and a random variable are independent if for all supported values of the random variable, the conditional expectation of the event given the variable equals the margial probability of the event.



In short you need to establish whether: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
    $endgroup$
    – Pedro Salgado
    Nov 29 '18 at 12:22








  • 1




    $begingroup$
    Use symmetry. @PedroSalgato
    $endgroup$
    – Graham Kemp
    Nov 29 '18 at 22:37










  • $begingroup$
    how can I use? @graham-kemp
    $endgroup$
    – Pedro Salgado
    Nov 30 '18 at 0:05






  • 1




    $begingroup$
    Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 0:23











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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

votes









2












$begingroup$

An event and a random variable are independent if for all supported values of the random variable, the conditional expectation of the event given the variable equals the margial probability of the event.



In short you need to establish whether: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
    $endgroup$
    – Pedro Salgado
    Nov 29 '18 at 12:22








  • 1




    $begingroup$
    Use symmetry. @PedroSalgato
    $endgroup$
    – Graham Kemp
    Nov 29 '18 at 22:37










  • $begingroup$
    how can I use? @graham-kemp
    $endgroup$
    – Pedro Salgado
    Nov 30 '18 at 0:05






  • 1




    $begingroup$
    Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 0:23
















2












$begingroup$

An event and a random variable are independent if for all supported values of the random variable, the conditional expectation of the event given the variable equals the margial probability of the event.



In short you need to establish whether: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
    $endgroup$
    – Pedro Salgado
    Nov 29 '18 at 12:22








  • 1




    $begingroup$
    Use symmetry. @PedroSalgato
    $endgroup$
    – Graham Kemp
    Nov 29 '18 at 22:37










  • $begingroup$
    how can I use? @graham-kemp
    $endgroup$
    – Pedro Salgado
    Nov 30 '18 at 0:05






  • 1




    $begingroup$
    Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 0:23














2












2








2





$begingroup$

An event and a random variable are independent if for all supported values of the random variable, the conditional expectation of the event given the variable equals the margial probability of the event.



In short you need to establish whether: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$






share|cite|improve this answer











$endgroup$



An event and a random variable are independent if for all supported values of the random variable, the conditional expectation of the event given the variable equals the margial probability of the event.



In short you need to establish whether: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 22:35

























answered Nov 29 '18 at 0:53









Graham KempGraham Kemp

85.3k43378




85.3k43378












  • $begingroup$
    $min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
    $endgroup$
    – Pedro Salgado
    Nov 29 '18 at 12:22








  • 1




    $begingroup$
    Use symmetry. @PedroSalgato
    $endgroup$
    – Graham Kemp
    Nov 29 '18 at 22:37










  • $begingroup$
    how can I use? @graham-kemp
    $endgroup$
    – Pedro Salgado
    Nov 30 '18 at 0:05






  • 1




    $begingroup$
    Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 0:23


















  • $begingroup$
    $min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
    $endgroup$
    – Pedro Salgado
    Nov 29 '18 at 12:22








  • 1




    $begingroup$
    Use symmetry. @PedroSalgato
    $endgroup$
    – Graham Kemp
    Nov 29 '18 at 22:37










  • $begingroup$
    how can I use? @graham-kemp
    $endgroup$
    – Pedro Salgado
    Nov 30 '18 at 0:05






  • 1




    $begingroup$
    Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 0:23
















$begingroup$
$min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
$endgroup$
– Pedro Salgado
Nov 29 '18 at 12:22






$begingroup$
$min{X_j)_{j=1}^n}$ is independent of ${U_1{>}U_2{>}U_3}$. How can I prove that ${U_1{>}U_2{>}U_3}$ and $max{U_i}_{i=1}^3$ are independent, or $P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3) = P({U_1{>}U_2{>}U_3})$ ?
$endgroup$
– Pedro Salgado
Nov 29 '18 at 12:22






1




1




$begingroup$
Use symmetry. @PedroSalgato
$endgroup$
– Graham Kemp
Nov 29 '18 at 22:37




$begingroup$
Use symmetry. @PedroSalgato
$endgroup$
– Graham Kemp
Nov 29 '18 at 22:37












$begingroup$
how can I use? @graham-kemp
$endgroup$
– Pedro Salgado
Nov 30 '18 at 0:05




$begingroup$
how can I use? @graham-kemp
$endgroup$
– Pedro Salgado
Nov 30 '18 at 0:05




1




1




$begingroup$
Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
$endgroup$
– Graham Kemp
Nov 30 '18 at 0:23




$begingroup$
Argue that that $mathsf P({U_1>U_2>U_3}mid u{=}max{U_i}_{i=1}^3)=mathsf P({U_3>U_1>U_2}mid u{=}max{U_i}_{i=1}^3)$ and so forth, by symmetry. @Pedro-Salgado
$endgroup$
– Graham Kemp
Nov 30 '18 at 0:23


















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