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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 yesterday
J.Mike
315110
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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 yesterday
J.Mike
315110
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
yesterday
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
yesterday
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
yesterday
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 yesterday
J.Mike
315110
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 yesterday
J.Mike
315110
asked yesterday
J.Mike
315110
edited yesterday
asked yesterday
J.Mike
315110
asked yesterday
J.Mike
315110
asked yesterday
J.Mike
315110
315110
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
yesterday
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
yesterday
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
yesterday
add a comment |
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
yesterday
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
yesterday
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
yesterday
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
yesterday
Have you tried Bayes' Theorem? That's the first thing that pops to mind
– kcborys
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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 yesterday
Jimmy R.
32.8k42156
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
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 yesterday
J.Mike
315110
add a comment |
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 yesterday
Jimmy R.
32.8k42156
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
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 yesterday
Jimmy R.
32.8k42156
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
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 yesterday
Jimmy R.
32.8k42156
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 yesterday
Jimmy R.
32.8k42156
answered yesterday
Jimmy R.
32.8k42156
answered yesterday
Jimmy R.
32.8k42156
answered yesterday
Jimmy R.
32.8k42156
32.8k42156
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
add a comment |
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
1
1
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
Thank you very much for your kind help. Following your solution, I prove the conditional case as below.
– J.Mike
yesterday
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 yesterday
J.Mike
315110
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 yesterday
J.Mike
315110
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 yesterday
J.Mike
315110
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 yesterday
J.Mike
315110
edited yesterday
answered yesterday
J.Mike
315110
answered yesterday
J.Mike
315110
answered yesterday
J.Mike
315110
315110
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
yesterday
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
yesterday
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
yesterday