Conditional expectation of a random variable conditioned to another conditionally independent variable
$begingroup$
I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.
Now, assume that $Z$ can take values in ${1,...,k}$.
I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$
Now, can I say what follows?
$$
E[X | Y] = E[X]
$$
I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$
Is this correct? If not, where am I wrong?
probability conditional-expectation
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
$endgroup$
add a comment |
$begingroup$
I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.
Now, assume that $Z$ can take values in ${1,...,k}$.
I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$
Now, can I say what follows?
$$
E[X | Y] = E[X]
$$
I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$
Is this correct? If not, where am I wrong?
probability conditional-expectation
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
$endgroup$
1
$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02
$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13
$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55
add a comment |
$begingroup$
I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.
Now, assume that $Z$ can take values in ${1,...,k}$.
I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$
Now, can I say what follows?
$$
E[X | Y] = E[X]
$$
I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$
Is this correct? If not, where am I wrong?
probability conditional-expectation
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
$endgroup$
I have a two variables $X$ and $Y$ conditionally independent given a thirs variable $Z$.
Now, assume that $Z$ can take values in ${1,...,k}$.
I will have:
$$
E[XY] = sum_{i=1}^k P[Z = i]E[X|Z = i]E[Y|Z = i]
$$
Now, can I say what follows?
$$
E[X | Y] = E[X]
$$
I have a proof for this, but I am not sure:
$$
E[X | Y] = sum_{i=1}^k P[Z = i]E[X|Z = i, Y] = sum_{i=1}^k P[Z = i]E[X|Z = i] = E[X]
$$
Is this correct? If not, where am I wrong?
probability conditional-expectation
probability conditional-expectation
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
asked Dec 1 '18 at 15:56
Ulderique DemoitreUlderique Demoitre
19810
19810
1
$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02
$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13
$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55
add a comment |
1
$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02
$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13
$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55
1
1
$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02
$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02
$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13
$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13
$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55
$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55
add a comment |
1 Answer
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$begingroup$
This was trickier than it looked ... to me.
It can be put in a more general-compact form thus:
Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
$$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.
This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).
The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.
In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
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add a comment |
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$begingroup$
This was trickier than it looked ... to me.
It can be put in a more general-compact form thus:
Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
$$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.
This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).
The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.
In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
$endgroup$
add a comment |
$begingroup$
This was trickier than it looked ... to me.
It can be put in a more general-compact form thus:
Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
$$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.
This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).
The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.
In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
$endgroup$
add a comment |
$begingroup$
This was trickier than it looked ... to me.
It can be put in a more general-compact form thus:
Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
$$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.
This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).
The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.
In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
$endgroup$
This was trickier than it looked ... to me.
It can be put in a more general-compact form thus:
Let $X,Y$ be conditionally independent given $Z$; that is, $P(X ,Y mid Z ) = P(X mid Z) P(Y mid Z)$. Then
$$E[X mid Y] = E[E[X mid Y, Z]] = E[E[X mid Z]] = E[X] tag{1}$$
where in the first and third equality we have applied total expectation (in the third to $X$, in the first to $X mid Y$), and in the second we've applied the conditional independence.
This is wrong, of course (it would imply that for any Markov chain $X to Z to Y$ the extremes $X,Y$ are uncorrelated - false in general).
The wrong equality is the first one. Because $X mid Y$ is not a random variable. Actually it's... nothing.
In your question, the first equality in your "proof" is (like my first equality in $(1)$) wrong, it's a mistaken application of the law of total expectation.
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
edited Dec 7 '18 at 23:15
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
answered Dec 2 '18 at 21:19
leonbloyleonbloy
40.4k645107
40.4k645107
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1
$begingroup$
Nope, it is not correct, and to see why it is not, simply spot the step where you feel as if you were passing through some flames holding your breath (you should know which step this is).
$endgroup$
– Did
Dec 1 '18 at 16:02
$begingroup$
@Did an infallible way of assessing the soundness of proof passages :)
$endgroup$
– Easymode44
Dec 1 '18 at 17:13
$begingroup$
Related math.stackexchange.com/questions/730702/…
$endgroup$
– leonbloy
Dec 2 '18 at 20:55