Random variables $X,Y$ such that $E(X|Y)=E(Y|X)$ a.s.
$begingroup$
Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
$endgroup$
Let $X$ and $Y$ be random variables such that $E(|X|), E(|Y|)<infty$ and $E(X|Y)=E(Y|X)$ a.s. Then is it true that $X=Y$ a.s. ? If this is not true in general, what happens if we also assume $X,Y$ are identically distributed ?
probability-theory measure-theory random-variables conditional-expectation expected-value
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
asked Dec 20 '18 at 2:33
user521337user521337
1,0801415
1,0801415
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Not true at all.
Let $X, Y$ be indepedently, identically distributed.
$$
mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
but this does not mean $P(X = Y) = 1$.
Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.
Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.
So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).
But since $X, Y$ are independent,
$$
mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
$endgroup$
1
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
1
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Not true at all.
Let $X, Y$ be indepedently, identically distributed.
$$
mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
but this does not mean $P(X = Y) = 1$.
Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.
Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.
So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).
But since $X, Y$ are independent,
$$
mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
$endgroup$
1
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
1
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
|
show 2 more comments
$begingroup$
Not true at all.
Let $X, Y$ be indepedently, identically distributed.
$$
mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
but this does not mean $P(X = Y) = 1$.
Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.
Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.
So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).
But since $X, Y$ are independent,
$$
mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
$endgroup$
1
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
1
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
|
show 2 more comments
$begingroup$
Not true at all.
Let $X, Y$ be indepedently, identically distributed.
$$
mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
but this does not mean $P(X = Y) = 1$.
Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.
Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.
So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).
But since $X, Y$ are independent,
$$
mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
$endgroup$
Not true at all.
Let $X, Y$ be indepedently, identically distributed.
$$
mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
but this does not mean $P(X = Y) = 1$.
Suppose $X, Y overset{iid}{sim} Normal(mu, sigma^2)$.
Note that $P(Z = 0) = 0$, where $Z := X -Y$, because $Z sim N(0, 2sigma^2)$ is a continuous R.V.
So the probability measure of the event $X = Y$ is $0$, not $1$ (as in almost sure case).
But since $X, Y$ are independent,
$$
mu = mathbb{E}(X ) = mathbb{E}(X | Y ) = mathbb{E}(Y|X ) = mathbb{E}(Y )
$$
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
edited Dec 20 '18 at 3:07
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
answered Dec 20 '18 at 2:38
MoreblueMoreblue
8791216
8791216
1
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
1
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
|
show 2 more comments
1
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
1
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
1
1
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
You mean to say $E(X|Y)$ is a constant random variable !? Also I would like to point out that $E(X|X)=X$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:40
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
@user521337 Yes, since $X$ is independent of $Y$.
$endgroup$
– Moreblue
Dec 20 '18 at 2:42
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
Also, $Z$ is $N(0 , sigma^2)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:49
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
$begingroup$
@user521337 No, $Z sim N(0, 2 sigma^2)$. For the variance, check here
$endgroup$
– Moreblue
Dec 20 '18 at 2:50
1
1
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
$begingroup$
again ... I would like to point out that $E(X|X)=X$ and not $E(X)$ ...
$endgroup$
– user521337
Dec 20 '18 at 2:52
|
show 2 more comments
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