Example of E(E(X|F)|G) neq E(E(X|G)|F) [duplicate]
up vote
0
down vote
favorite
This question already has an answer here:
An example s.t. $E(E(X|mathcal F_1)|mathcal F_2)neq E(E(X|mathcal F_2)|mathcal F_1)$
1 answer
Can you find an example where
E(E(X|F)|G) $neq$ E(E(X|G)|F) (F and G is $sigma$-field in probability theory)
probability probability-theory conditional-expectation conditional-probability
asked Nov 14 at 15:41
wts
11
marked as duplicate by JMoravitz, Davide Giraudo probability-theory
Users with the probability-theory badge can single-handedly close probability-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 14 at 17:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
0
down vote
favorite
This question already has an answer here:
An example s.t. $E(E(X|mathcal F_1)|mathcal F_2)neq E(E(X|mathcal F_2)|mathcal F_1)$
1 answer
Can you find an example where
E(E(X|F)|G) $neq$ E(E(X|G)|F) (F and G is $sigma$-field in probability theory)
probability probability-theory conditional-expectation conditional-probability
asked Nov 14 at 15:41
wts
11
marked as duplicate by JMoravitz, Davide Giraudo probability-theory
Users with the probability-theory badge can single-handedly close probability-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 14 at 17:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Do you have any thoughts on the problem? What have you tried?
– saz
Nov 14 at 15:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question already has an answer here:
An example s.t. $E(E(X|mathcal F_1)|mathcal F_2)neq E(E(X|mathcal F_2)|mathcal F_1)$
1 answer
Can you find an example where
E(E(X|F)|G) $neq$ E(E(X|G)|F) (F and G is $sigma$-field in probability theory)
probability probability-theory conditional-expectation conditional-probability
asked Nov 14 at 15:41
wts
11
This question already has an answer here:
An example s.t. $E(E(X|mathcal F_1)|mathcal F_2)neq E(E(X|mathcal F_2)|mathcal F_1)$
1 answer
Can you find an example where
E(E(X|F)|G) $neq$ E(E(X|G)|F) (F and G is $sigma$-field in probability theory)
This question already has an answer here:
An example s.t. $E(E(X|mathcal F_1)|mathcal F_2)neq E(E(X|mathcal F_2)|mathcal F_1)$
1 answer
probability probability-theory conditional-expectation conditional-probability
probability probability-theory conditional-expectation conditional-probability
asked Nov 14 at 15:41
wts
11
asked Nov 14 at 15:41
wts
11
asked Nov 14 at 15:41
wts
11
asked Nov 14 at 15:41
wts
11
asked Nov 14 at 15:41
wts
11
11
marked as duplicate by JMoravitz, Davide Giraudo probability-theory
Users with the probability-theory badge can single-handedly close probability-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 14 at 17:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by JMoravitz, Davide Giraudo probability-theory
Users with the probability-theory badge can single-handedly close probability-theory questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Nov 14 at 17:19
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Do you have any thoughts on the problem? What have you tried?
– saz
Nov 14 at 15:58
add a comment |
2
Do you have any thoughts on the problem? What have you tried?
– saz
Nov 14 at 15:58
2
2
Do you have any thoughts on the problem? What have you tried?
– saz
Nov 14 at 15:58
Do you have any thoughts on the problem? What have you tried?
– saz
Nov 14 at 15:58
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Try an example with $F$ and $G$ the $sigma$-algebras generated by random variables $Y,Z$
respectively. $mathbb E[ldots|F]$ is a function of $Y$ and $mathbb E[ldots|G]$ is a function of $Z$. Except in "trivial" cases, these won't be equal.
answered Nov 14 at 16:10
Robert Israel
313k23206452
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Try an example with $F$ and $G$ the $sigma$-algebras generated by random variables $Y,Z$
respectively. $mathbb E[ldots|F]$ is a function of $Y$ and $mathbb E[ldots|G]$ is a function of $Z$. Except in "trivial" cases, these won't be equal.
answered Nov 14 at 16:10
Robert Israel
313k23206452
add a comment |
up vote
0
down vote
Try an example with $F$ and $G$ the $sigma$-algebras generated by random variables $Y,Z$
respectively. $mathbb E[ldots|F]$ is a function of $Y$ and $mathbb E[ldots|G]$ is a function of $Z$. Except in "trivial" cases, these won't be equal.
answered Nov 14 at 16:10
Robert Israel
313k23206452
add a comment |
up vote
0
down vote
up vote
0
down vote
Try an example with $F$ and $G$ the $sigma$-algebras generated by random variables $Y,Z$
respectively. $mathbb E[ldots|F]$ is a function of $Y$ and $mathbb E[ldots|G]$ is a function of $Z$. Except in "trivial" cases, these won't be equal.
answered Nov 14 at 16:10
Robert Israel
313k23206452
Try an example with $F$ and $G$ the $sigma$-algebras generated by random variables $Y,Z$
respectively. $mathbb E[ldots|F]$ is a function of $Y$ and $mathbb E[ldots|G]$ is a function of $Z$. Except in "trivial" cases, these won't be equal.
answered Nov 14 at 16:10
Robert Israel
313k23206452
edited Nov 14 at 17:16
answered Nov 14 at 16:10
Robert Israel
313k23206452
answered Nov 14 at 16:10
Robert Israel
313k23206452
answered Nov 14 at 16:10
Robert Israel
313k23206452
313k23206452
add a comment |
add a comment |
2
Do you have any thoughts on the problem? What have you tried?
– saz
Nov 14 at 15:58