Does conditional expectation imply anything about the expected value of the product of two random variables?
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If $E[U|X]=0$ then $[XU] = 0$
If $E[XU]=0$ then $[U|X] = 0$
Which of the two statements above are true? This is my thought process for the first one:
if $E[U|X]=0$ then $E[U]=0$ and if U and X are independent, then $E[XU]=E[X]E[U]=E[X]*0=0$
which means the first statement is true if X and U are independent.
Is this the right way to think about it? What about the second one?
random-variables conditional-expectation expected-value
asked 5 hours ago
mitmath514
162
add a comment |
up vote
0
down vote
favorite
If $E[U|X]=0$ then $[XU] = 0$
If $E[XU]=0$ then $[U|X] = 0$
Which of the two statements above are true? This is my thought process for the first one:
if $E[U|X]=0$ then $E[U]=0$ and if U and X are independent, then $E[XU]=E[X]E[U]=E[X]*0=0$
which means the first statement is true if X and U are independent.
Is this the right way to think about it? What about the second one?
random-variables conditional-expectation expected-value
asked 5 hours ago
mitmath514
162
First true, second false. Your motivation is wrong
– Federico
5 hours ago
@Federico Could you explain the correct way to think about this?
– mitmath514
5 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $E[U|X]=0$ then $[XU] = 0$
If $E[XU]=0$ then $[U|X] = 0$
Which of the two statements above are true? This is my thought process for the first one:
if $E[U|X]=0$ then $E[U]=0$ and if U and X are independent, then $E[XU]=E[X]E[U]=E[X]*0=0$
which means the first statement is true if X and U are independent.
Is this the right way to think about it? What about the second one?
random-variables conditional-expectation expected-value
asked 5 hours ago
mitmath514
162
If $E[U|X]=0$ then $[XU] = 0$
If $E[XU]=0$ then $[U|X] = 0$
Which of the two statements above are true? This is my thought process for the first one:
if $E[U|X]=0$ then $E[U]=0$ and if U and X are independent, then $E[XU]=E[X]E[U]=E[X]*0=0$
which means the first statement is true if X and U are independent.
Is this the right way to think about it? What about the second one?
random-variables conditional-expectation expected-value
random-variables conditional-expectation expected-value
asked 5 hours ago
mitmath514
162
asked 5 hours ago
mitmath514
162
asked 5 hours ago
mitmath514
162
asked 5 hours ago
mitmath514
162
asked 5 hours ago
mitmath514
162
162
First true, second false. Your motivation is wrong
– Federico
5 hours ago
@Federico Could you explain the correct way to think about this?
– mitmath514
5 hours ago
add a comment |
First true, second false. Your motivation is wrong
– Federico
5 hours ago
@Federico Could you explain the correct way to think about this?
– mitmath514
5 hours ago
First true, second false. Your motivation is wrong
– Federico
5 hours ago
First true, second false. Your motivation is wrong
– Federico
5 hours ago
@Federico Could you explain the correct way to think about this?
– mitmath514
5 hours ago
@Federico Could you explain the correct way to think about this?
– mitmath514
5 hours ago
add a comment |
2 Answers
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For the first statement you should apply the Law of Iterated Expectations.
If $E[U|X]=0$ then $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]=E[Xcdot0]=0$
The second statement doesn't hold.
answered 5 hours ago
Ramiro Scorolli
54911
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To see why the first statement is true, I would follow Ramiro Scorolli's excellent answer and use the Law of Iterated Expectations: $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]$, so $E[U|X]] = 0$ implies that $E[XU]=0$. For the second statement, a very simple counterexample would be to take $X = 0$ and $U = 1$ with probability one; i.e. $X$ and $U$ are constant random variables. Then $E[XU] = E[0 cdot 1] = 0$, but $E[U | X] = E[U] = 1$.
answered 1 hour ago
LaserPineapple
383
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For the first statement you should apply the Law of Iterated Expectations.
If $E[U|X]=0$ then $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]=E[Xcdot0]=0$
The second statement doesn't hold.
answered 5 hours ago
Ramiro Scorolli
54911
add a comment |
up vote
0
down vote
For the first statement you should apply the Law of Iterated Expectations.
If $E[U|X]=0$ then $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]=E[Xcdot0]=0$
The second statement doesn't hold.
answered 5 hours ago
Ramiro Scorolli
54911
add a comment |
up vote
0
down vote
up vote
0
down vote
For the first statement you should apply the Law of Iterated Expectations.
If $E[U|X]=0$ then $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]=E[Xcdot0]=0$
The second statement doesn't hold.
answered 5 hours ago
Ramiro Scorolli
54911
For the first statement you should apply the Law of Iterated Expectations.
If $E[U|X]=0$ then $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]=E[Xcdot0]=0$
The second statement doesn't hold.
answered 5 hours ago
Ramiro Scorolli
54911
answered 5 hours ago
Ramiro Scorolli
54911
answered 5 hours ago
Ramiro Scorolli
54911
answered 5 hours ago
Ramiro Scorolli
54911
54911
add a comment |
add a comment |
up vote
0
down vote
To see why the first statement is true, I would follow Ramiro Scorolli's excellent answer and use the Law of Iterated Expectations: $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]$, so $E[U|X]] = 0$ implies that $E[XU]=0$. For the second statement, a very simple counterexample would be to take $X = 0$ and $U = 1$ with probability one; i.e. $X$ and $U$ are constant random variables. Then $E[XU] = E[0 cdot 1] = 0$, but $E[U | X] = E[U] = 1$.
answered 1 hour ago
LaserPineapple
383
add a comment |
up vote
0
down vote
To see why the first statement is true, I would follow Ramiro Scorolli's excellent answer and use the Law of Iterated Expectations: $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]$, so $E[U|X]] = 0$ implies that $E[XU]=0$. For the second statement, a very simple counterexample would be to take $X = 0$ and $U = 1$ with probability one; i.e. $X$ and $U$ are constant random variables. Then $E[XU] = E[0 cdot 1] = 0$, but $E[U | X] = E[U] = 1$.
answered 1 hour ago
LaserPineapple
383
add a comment |
up vote
0
down vote
up vote
0
down vote
To see why the first statement is true, I would follow Ramiro Scorolli's excellent answer and use the Law of Iterated Expectations: $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]$, so $E[U|X]] = 0$ implies that $E[XU]=0$. For the second statement, a very simple counterexample would be to take $X = 0$ and $U = 1$ with probability one; i.e. $X$ and $U$ are constant random variables. Then $E[XU] = E[0 cdot 1] = 0$, but $E[U | X] = E[U] = 1$.
answered 1 hour ago
LaserPineapple
383
To see why the first statement is true, I would follow Ramiro Scorolli's excellent answer and use the Law of Iterated Expectations: $E[UX]=E[E[UX|X]]=E[Xcdot E[U|X]]$, so $E[U|X]] = 0$ implies that $E[XU]=0$. For the second statement, a very simple counterexample would be to take $X = 0$ and $U = 1$ with probability one; i.e. $X$ and $U$ are constant random variables. Then $E[XU] = E[0 cdot 1] = 0$, but $E[U | X] = E[U] = 1$.
answered 1 hour ago
LaserPineapple
383
answered 1 hour ago
LaserPineapple
383
answered 1 hour ago
LaserPineapple
383
answered 1 hour ago
LaserPineapple
383
383
add a comment |
add a comment |
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First true, second false. Your motivation is wrong
– Federico
5 hours ago
@Federico Could you explain the correct way to think about this?
– mitmath514
5 hours ago