Mixing
Conditional Probability Mean
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I have a continuous random variable Y with its mean conditioned on another random (discrete) variable X as follows.
m0 = E(Y|X=0) = 12
m1 = 3
m2 = 7
and I want to find mY|X<2 = E(Y|X<2). Is this intuitive. Because supposedly it is not and my immediate reaction is to say it is simply the E(Y|X = 0) + E(Y|X = 1)
Thanks in advance
EDIT: I do have some information on the variable X but it seemed to not be helpful to me.
p0 = P(X = 0) = 0.2
p1 = 0.7
p2 = 0.1
probability probability-theory probability-distributions
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
$endgroup$
add a comment |
$begingroup$
I have a continuous random variable Y with its mean conditioned on another random (discrete) variable X as follows.
m0 = E(Y|X=0) = 12
m1 = 3
m2 = 7
and I want to find mY|X<2 = E(Y|X<2). Is this intuitive. Because supposedly it is not and my immediate reaction is to say it is simply the E(Y|X = 0) + E(Y|X = 1)
Thanks in advance
EDIT: I do have some information on the variable X but it seemed to not be helpful to me.
p0 = P(X = 0) = 0.2
p1 = 0.7
p2 = 0.1
probability probability-theory probability-distributions
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
$endgroup$
$begingroup$
I'm not sure you can find $E[Y mid X<2]$ only using $E[Y mid X=x]$ for $x=0,1,2$. I think you need more information about the distribution of $X$.
$endgroup$
– angryavian
Jan 13 at 21:23
$begingroup$
I added some information but do you think it's relevant?
$endgroup$
– A.Code.1
Jan 13 at 21:27
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@A.Code.1 Is $E(X)=3$, or what is meant by $m_1$ and $m_2$?
$endgroup$
– callculus
Jan 13 at 21:33
$begingroup$
m1 = E(Y|X = 1) and m2 = E(Y|X =2)
$endgroup$
– A.Code.1
Jan 13 at 21:37
$begingroup$
@angryavian I have to admit I´m not sure with my answer. I should delete it. You should write an answer.
$endgroup$
– callculus
Jan 13 at 21:55
add a comment |
$begingroup$
I have a continuous random variable Y with its mean conditioned on another random (discrete) variable X as follows.
m0 = E(Y|X=0) = 12
m1 = 3
m2 = 7
and I want to find mY|X<2 = E(Y|X<2). Is this intuitive. Because supposedly it is not and my immediate reaction is to say it is simply the E(Y|X = 0) + E(Y|X = 1)
Thanks in advance
EDIT: I do have some information on the variable X but it seemed to not be helpful to me.
p0 = P(X = 0) = 0.2
p1 = 0.7
p2 = 0.1
probability probability-theory probability-distributions
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
$endgroup$
I have a continuous random variable Y with its mean conditioned on another random (discrete) variable X as follows.
m0 = E(Y|X=0) = 12
m1 = 3
m2 = 7
and I want to find mY|X<2 = E(Y|X<2). Is this intuitive. Because supposedly it is not and my immediate reaction is to say it is simply the E(Y|X = 0) + E(Y|X = 1)
Thanks in advance
EDIT: I do have some information on the variable X but it seemed to not be helpful to me.
p0 = P(X = 0) = 0.2
p1 = 0.7
p2 = 0.1
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
edited Jan 13 at 21:26
A.Code.1
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
asked Jan 13 at 21:17
A.Code.1A.Code.1
32
32
$begingroup$
I'm not sure you can find $E[Y mid X<2]$ only using $E[Y mid X=x]$ for $x=0,1,2$. I think you need more information about the distribution of $X$.
$endgroup$
– angryavian
Jan 13 at 21:23
$begingroup$
I added some information but do you think it's relevant?
$endgroup$
– A.Code.1
Jan 13 at 21:27
$begingroup$
@A.Code.1 Is $E(X)=3$, or what is meant by $m_1$ and $m_2$?
$endgroup$
– callculus
Jan 13 at 21:33
$begingroup$
m1 = E(Y|X = 1) and m2 = E(Y|X =2)
$endgroup$
– A.Code.1
Jan 13 at 21:37
$begingroup$
@angryavian I have to admit I´m not sure with my answer. I should delete it. You should write an answer.
$endgroup$
– callculus
Jan 13 at 21:55
add a comment |
$begingroup$
I'm not sure you can find $E[Y mid X<2]$ only using $E[Y mid X=x]$ for $x=0,1,2$. I think you need more information about the distribution of $X$.
$endgroup$
– angryavian
Jan 13 at 21:23
$begingroup$
I added some information but do you think it's relevant?
$endgroup$
– A.Code.1
Jan 13 at 21:27
$begingroup$
@A.Code.1 Is $E(X)=3$, or what is meant by $m_1$ and $m_2$?
$endgroup$
– callculus
Jan 13 at 21:33
$begingroup$
m1 = E(Y|X = 1) and m2 = E(Y|X =2)
$endgroup$
– A.Code.1
Jan 13 at 21:37
$begingroup$
@angryavian I have to admit I´m not sure with my answer. I should delete it. You should write an answer.
$endgroup$
– callculus
Jan 13 at 21:55
$begingroup$
I'm not sure you can find $E[Y mid X<2]$ only using $E[Y mid X=x]$ for $x=0,1,2$. I think you need more information about the distribution of $X$.
$endgroup$
– angryavian
Jan 13 at 21:23
$begingroup$
I'm not sure you can find $E[Y mid X<2]$ only using $E[Y mid X=x]$ for $x=0,1,2$. I think you need more information about the distribution of $X$.
$endgroup$
– angryavian
Jan 13 at 21:23
$begingroup$
I added some information but do you think it's relevant?
$endgroup$
– A.Code.1
Jan 13 at 21:27
$begingroup$
I added some information but do you think it's relevant?
$endgroup$
– A.Code.1
Jan 13 at 21:27
$begingroup$
@A.Code.1 Is $E(X)=3$, or what is meant by $m_1$ and $m_2$?
$endgroup$
– callculus
Jan 13 at 21:33
$begingroup$
@A.Code.1 Is $E(X)=3$, or what is meant by $m_1$ and $m_2$?
$endgroup$
– callculus
Jan 13 at 21:33
$begingroup$
m1 = E(Y|X = 1) and m2 = E(Y|X =2)
$endgroup$
– A.Code.1
Jan 13 at 21:37
$begingroup$
m1 = E(Y|X = 1) and m2 = E(Y|X =2)
$endgroup$
– A.Code.1
Jan 13 at 21:37
$begingroup$
@angryavian I have to admit I´m not sure with my answer. I should delete it. You should write an answer.
$endgroup$
– callculus
Jan 13 at 21:55
$begingroup$
@angryavian I have to admit I´m not sure with my answer. I should delete it. You should write an answer.
$endgroup$
– callculus
Jan 13 at 21:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The conditional expectation of $Y$ given an event $A$ is
$$E[Y mid A] = frac{E[Y cdot 1_A]}{P(A)},$$
where $1_A$ is an indicator random variable that equals $1$ when the event $A$ occurs, and equals zero otherwise.
In your case you have $A = {X < 2}$.
The denominator is $P(A) = P(X=0) + P(X=1)$.
By the law of total expectation, the numerator is
$$E[Y cdot 1_A] = sum_{x=0}^2 E[Y cdot 1_A mid X=x] P(X=x) = E[Y mid X=0] P(X=0) + E[Y mid X=1] P(X=1).$$
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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active
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$begingroup$
The conditional expectation of $Y$ given an event $A$ is
$$E[Y mid A] = frac{E[Y cdot 1_A]}{P(A)},$$
where $1_A$ is an indicator random variable that equals $1$ when the event $A$ occurs, and equals zero otherwise.
In your case you have $A = {X < 2}$.
The denominator is $P(A) = P(X=0) + P(X=1)$.
By the law of total expectation, the numerator is
$$E[Y cdot 1_A] = sum_{x=0}^2 E[Y cdot 1_A mid X=x] P(X=x) = E[Y mid X=0] P(X=0) + E[Y mid X=1] P(X=1).$$
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
add a comment |
$begingroup$
The conditional expectation of $Y$ given an event $A$ is
$$E[Y mid A] = frac{E[Y cdot 1_A]}{P(A)},$$
where $1_A$ is an indicator random variable that equals $1$ when the event $A$ occurs, and equals zero otherwise.
In your case you have $A = {X < 2}$.
The denominator is $P(A) = P(X=0) + P(X=1)$.
By the law of total expectation, the numerator is
$$E[Y cdot 1_A] = sum_{x=0}^2 E[Y cdot 1_A mid X=x] P(X=x) = E[Y mid X=0] P(X=0) + E[Y mid X=1] P(X=1).$$
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
add a comment |
$begingroup$
The conditional expectation of $Y$ given an event $A$ is
$$E[Y mid A] = frac{E[Y cdot 1_A]}{P(A)},$$
where $1_A$ is an indicator random variable that equals $1$ when the event $A$ occurs, and equals zero otherwise.
In your case you have $A = {X < 2}$.
The denominator is $P(A) = P(X=0) + P(X=1)$.
By the law of total expectation, the numerator is
$$E[Y cdot 1_A] = sum_{x=0}^2 E[Y cdot 1_A mid X=x] P(X=x) = E[Y mid X=0] P(X=0) + E[Y mid X=1] P(X=1).$$
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
$endgroup$
The conditional expectation of $Y$ given an event $A$ is
$$E[Y mid A] = frac{E[Y cdot 1_A]}{P(A)},$$
where $1_A$ is an indicator random variable that equals $1$ when the event $A$ occurs, and equals zero otherwise.
In your case you have $A = {X < 2}$.
The denominator is $P(A) = P(X=0) + P(X=1)$.
By the law of total expectation, the numerator is
$$E[Y cdot 1_A] = sum_{x=0}^2 E[Y cdot 1_A mid X=x] P(X=x) = E[Y mid X=0] P(X=0) + E[Y mid X=1] P(X=1).$$
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
answered Jan 13 at 22:28
angryavianangryavian
41.1k23380
41.1k23380
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
add a comment |
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
$begingroup$
Nice answer, angryavian.
$endgroup$
– callculus
Jan 13 at 22:52
add a comment |
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$begingroup$
I'm not sure you can find $E[Y mid X<2]$ only using $E[Y mid X=x]$ for $x=0,1,2$. I think you need more information about the distribution of $X$.
$endgroup$
– angryavian
Jan 13 at 21:23
$begingroup$
I added some information but do you think it's relevant?
$endgroup$
– A.Code.1
Jan 13 at 21:27
$begingroup$
@A.Code.1 Is $E(X)=3$, or what is meant by $m_1$ and $m_2$?
$endgroup$
– callculus
Jan 13 at 21:33
$begingroup$
m1 = E(Y|X = 1) and m2 = E(Y|X =2)
$endgroup$
– A.Code.1
Jan 13 at 21:37
$begingroup$
@angryavian I have to admit I´m not sure with my answer. I should delete it. You should write an answer.
$endgroup$
– callculus
Jan 13 at 21:55