Independence between random vector and event
$begingroup$
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
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
$endgroup$
add a comment |
$begingroup$
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
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
$endgroup$
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
add a comment |
$begingroup$
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
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
$endgroup$
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
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
edited Nov 29 '18 at 10:00
Davide Giraudo
126k16150261
126k16150261
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
asked Nov 28 '18 at 21:18
Pedro SalgadoPedro Salgado
735
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
$endgroup$
$begingroup$
$min{X_j)_{j=1}^n}$ 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
add a comment |
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$begingroup$
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: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
$endgroup$
$begingroup$
$min{X_j)_{j=1}^n}$ 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
add a comment |
$begingroup$
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: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
$endgroup$
$begingroup$
$min{X_j)_{j=1}^n}$ 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
add a comment |
$begingroup$
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: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
$endgroup$
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: $${forall uin(0;1)~forall xin(0;1):\quadmathsf P({U_1{>}U_2{>}U_3})=mathsf P({U_1{>}U_2{>}U_3}mid u{=}max{U_i}_{i=1}^3,x{=}min{X_j}_{j=1}^n)}$$
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
edited Nov 29 '18 at 22:35
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
answered Nov 29 '18 at 0:53
Graham KempGraham Kemp
85.3k43378
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
add a comment |
$begingroup$
$min{X_j)_{j=1}^n}$ 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
add a comment |
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$begingroup$
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