How to prove that $E(Y|D=1)=E(DY)/E(D)$
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1
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How to prove that
$$E(Y|D=1)=E(DY)/E(D)$$
and
$$E(Y|X,D=1)=E(DY|X)/E(D|X),$$
where $D$ is a binary variable and takes value of 0 and 1.
probability probability-theory statistics probability-distributions conditional-expectation
asked Nov 19 at 17:05
J.Mike
313110
add a comment |
up vote
1
down vote
favorite
How to prove that
$$E(Y|D=1)=E(DY)/E(D)$$
and
$$E(Y|X,D=1)=E(DY|X)/E(D|X),$$
where $D$ is a binary variable and takes value of 0 and 1.
probability probability-theory statistics probability-distributions conditional-expectation
asked Nov 19 at 17:05
J.Mike
313110
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
Nov 19 at 17:07
1
This is not true, in general. Say, $Y=D$, $E(D|D=1)=1$, and the r.h.s. $E(D^2)/E(D)$ need not to be $1$.
– NCh
Nov 20 at 1:56
Thank you for your comment. Sorry, I forgot to give the definition of $D$. $D$ is a binary variable which takes value of 0 and 1. So $D^2=D$, then $E(D^2)/E(D)=E(D)/E(D)=1$
– J.Mike
Nov 20 at 4:26
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How to prove that
$$E(Y|D=1)=E(DY)/E(D)$$
and
$$E(Y|X,D=1)=E(DY|X)/E(D|X),$$
where $D$ is a binary variable and takes value of 0 and 1.
probability probability-theory statistics probability-distributions conditional-expectation
asked Nov 19 at 17:05
J.Mike
313110
How to prove that
$$E(Y|D=1)=E(DY)/E(D)$$
and
$$E(Y|X,D=1)=E(DY|X)/E(D|X),$$
where $D$ is a binary variable and takes value of 0 and 1.
probability probability-theory statistics probability-distributions conditional-expectation
probability probability-theory statistics probability-distributions conditional-expectation
asked Nov 19 at 17:05
J.Mike
313110
asked Nov 19 at 17:05
J.Mike
313110
edited Nov 20 at 4:28
asked Nov 19 at 17:05
J.Mike
313110
asked Nov 19 at 17:05
J.Mike
313110
asked Nov 19 at 17:05
J.Mike
313110
313110
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
Nov 19 at 17:07
1
This is not true, in general. Say, $Y=D$, $E(D|D=1)=1$, and the r.h.s. $E(D^2)/E(D)$ need not to be $1$.
– NCh
Nov 20 at 1:56
Thank you for your comment. Sorry, I forgot to give the definition of $D$. $D$ is a binary variable which takes value of 0 and 1. So $D^2=D$, then $E(D^2)/E(D)=E(D)/E(D)=1$
– J.Mike
Nov 20 at 4:26
add a comment |
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
Nov 19 at 17:07
1
This is not true, in general. Say, $Y=D$, $E(D|D=1)=1$, and the r.h.s. $E(D^2)/E(D)$ need not to be $1$.
– NCh
Nov 20 at 1:56
Thank you for your comment. Sorry, I forgot to give the definition of $D$. $D$ is a binary variable which takes value of 0 and 1. So $D^2=D$, then $E(D^2)/E(D)=E(D)/E(D)=1$
– J.Mike
Nov 20 at 4:26
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
Nov 19 at 17:07
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
Nov 19 at 17:07
1
1
This is not true, in general. Say, $Y=D$, $E(D|D=1)=1$, and the r.h.s. $E(D^2)/E(D)$ need not to be $1$.
– NCh
Nov 20 at 1:56
This is not true, in general. Say, $Y=D$, $E(D|D=1)=1$, and the r.h.s. $E(D^2)/E(D)$ need not to be $1$.
– NCh
Nov 20 at 1:56
Thank you for your comment. Sorry, I forgot to give the definition of $D$. $D$ is a binary variable which takes value of 0 and 1. So $D^2=D$, then $E(D^2)/E(D)=E(D)/E(D)=1$
– J.Mike
Nov 20 at 4:26
Thank you for your comment. Sorry, I forgot to give the definition of $D$. $D$ is a binary variable which takes value of 0 and 1. So $D^2=D$, then $E(D^2)/E(D)=E(D)/E(D)=1$
– J.Mike
Nov 20 at 4:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
First observe that $$mathbb E[D]=1cdot P(D=1)+0cdot P(D=0)=P(D=1)tag{1}$$
So, begin{align}mathbb E[YD]&=mathbb E[Ycdot 1mid D=1]P(D=1)+mathbb E[Ycdot 0mid D=0]P(D=0)\&overset{(1)}=mathbb E[Ymid D=1]mathbb E[D]+0end{align} and the result follows by rearranging terms and assuming that $P(D=1)>0$. Can you do the same for the $X$ part?
answered Nov 20 at 4:36
Jimmy R.
33k42157
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
add a comment |
up vote
1
down vote
The unconditional case is shown as above. For the conditional case, first observe that
$$mathbb E[D|X]=1cdot P(D=1|X)+0cdot P(D=0|X)=P(D=1|X)tag{1}$$
So,
begin{align}mathbb E[YD|X]&=mathbb E[YD|X,D=1]cdot P(D=1|X)+mathbb E[YD|X,D=0]cdot P(D=0|X)\&overset{(1)}=mathbb E[Y| X, D=1]cdotmathbb E[D|X]end{align}
and the conditional equation follows by rearranging terms and assuming that $P(D=1|X)neq0$.
answered Nov 20 at 4:59
J.Mike
313110
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
First observe that $$mathbb E[D]=1cdot P(D=1)+0cdot P(D=0)=P(D=1)tag{1}$$
So, begin{align}mathbb E[YD]&=mathbb E[Ycdot 1mid D=1]P(D=1)+mathbb E[Ycdot 0mid D=0]P(D=0)\&overset{(1)}=mathbb E[Ymid D=1]mathbb E[D]+0end{align} and the result follows by rearranging terms and assuming that $P(D=1)>0$. Can you do the same for the $X$ part?
answered Nov 20 at 4:36
Jimmy R.
33k42157
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
add a comment |
up vote
1
down vote
accepted
First observe that $$mathbb E[D]=1cdot P(D=1)+0cdot P(D=0)=P(D=1)tag{1}$$
So, begin{align}mathbb E[YD]&=mathbb E[Ycdot 1mid D=1]P(D=1)+mathbb E[Ycdot 0mid D=0]P(D=0)\&overset{(1)}=mathbb E[Ymid D=1]mathbb E[D]+0end{align} and the result follows by rearranging terms and assuming that $P(D=1)>0$. Can you do the same for the $X$ part?
answered Nov 20 at 4:36
Jimmy R.
33k42157
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
First observe that $$mathbb E[D]=1cdot P(D=1)+0cdot P(D=0)=P(D=1)tag{1}$$
So, begin{align}mathbb E[YD]&=mathbb E[Ycdot 1mid D=1]P(D=1)+mathbb E[Ycdot 0mid D=0]P(D=0)\&overset{(1)}=mathbb E[Ymid D=1]mathbb E[D]+0end{align} and the result follows by rearranging terms and assuming that $P(D=1)>0$. Can you do the same for the $X$ part?
answered Nov 20 at 4:36
Jimmy R.
33k42157
First observe that $$mathbb E[D]=1cdot P(D=1)+0cdot P(D=0)=P(D=1)tag{1}$$
So, begin{align}mathbb E[YD]&=mathbb E[Ycdot 1mid D=1]P(D=1)+mathbb E[Ycdot 0mid D=0]P(D=0)\&overset{(1)}=mathbb E[Ymid D=1]mathbb E[D]+0end{align} and the result follows by rearranging terms and assuming that $P(D=1)>0$. Can you do the same for the $X$ part?
answered Nov 20 at 4:36
Jimmy R.
33k42157
answered Nov 20 at 4:36
Jimmy R.
33k42157
answered Nov 20 at 4:36
Jimmy R.
33k42157
answered Nov 20 at 4:36
Jimmy R.
33k42157
33k42157
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
add a comment |
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
1
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
Nov 20 at 5:03
add a comment |
up vote
1
down vote
The unconditional case is shown as above. For the conditional case, first observe that
$$mathbb E[D|X]=1cdot P(D=1|X)+0cdot P(D=0|X)=P(D=1|X)tag{1}$$
So,
begin{align}mathbb E[YD|X]&=mathbb E[YD|X,D=1]cdot P(D=1|X)+mathbb E[YD|X,D=0]cdot P(D=0|X)\&overset{(1)}=mathbb E[Y| X, D=1]cdotmathbb E[D|X]end{align}
and the conditional equation follows by rearranging terms and assuming that $P(D=1|X)neq0$.
answered Nov 20 at 4:59
J.Mike
313110
add a comment |
up vote
1
down vote
The unconditional case is shown as above. For the conditional case, first observe that
$$mathbb E[D|X]=1cdot P(D=1|X)+0cdot P(D=0|X)=P(D=1|X)tag{1}$$
So,
begin{align}mathbb E[YD|X]&=mathbb E[YD|X,D=1]cdot P(D=1|X)+mathbb E[YD|X,D=0]cdot P(D=0|X)\&overset{(1)}=mathbb E[Y| X, D=1]cdotmathbb E[D|X]end{align}
and the conditional equation follows by rearranging terms and assuming that $P(D=1|X)neq0$.
answered Nov 20 at 4:59
J.Mike
313110
add a comment |
up vote
1
down vote
up vote
1
down vote
The unconditional case is shown as above. For the conditional case, first observe that
$$mathbb E[D|X]=1cdot P(D=1|X)+0cdot P(D=0|X)=P(D=1|X)tag{1}$$
So,
begin{align}mathbb E[YD|X]&=mathbb E[YD|X,D=1]cdot P(D=1|X)+mathbb E[YD|X,D=0]cdot P(D=0|X)\&overset{(1)}=mathbb E[Y| X, D=1]cdotmathbb E[D|X]end{align}
and the conditional equation follows by rearranging terms and assuming that $P(D=1|X)neq0$.
answered Nov 20 at 4:59
J.Mike
313110
The unconditional case is shown as above. For the conditional case, first observe that
$$mathbb E[D|X]=1cdot P(D=1|X)+0cdot P(D=0|X)=P(D=1|X)tag{1}$$
So,
begin{align}mathbb E[YD|X]&=mathbb E[YD|X,D=1]cdot P(D=1|X)+mathbb E[YD|X,D=0]cdot P(D=0|X)\&overset{(1)}=mathbb E[Y| X, D=1]cdotmathbb E[D|X]end{align}
and the conditional equation follows by rearranging terms and assuming that $P(D=1|X)neq0$.
answered Nov 20 at 4:59
J.Mike
313110
edited Nov 20 at 5:05
answered Nov 20 at 4:59
J.Mike
313110
answered Nov 20 at 4:59
J.Mike
313110
answered Nov 20 at 4:59
J.Mike
313110
313110
add a comment |
add a comment |
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Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
Nov 19 at 17:07
1
This is not true, in general. Say, $Y=D$, $E(D|D=1)=1$, and the r.h.s. $E(D^2)/E(D)$ need not to be $1$.
– NCh
Nov 20 at 1:56
Thank you for your comment. Sorry, I forgot to give the definition of $D$. $D$ is a binary variable which takes value of 0 and 1. So $D^2=D$, then $E(D^2)/E(D)=E(D)/E(D)=1$
– J.Mike
Nov 20 at 4:26