Mixing
Does conditional expectation imply anything about the expected value of the product of two random variables?
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 Nov 20 '18 at 19:10
mitmath514
163
add a comment |
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 Nov 20 '18 at 19:10
mitmath514
163
First true, second false. Your motivation is wrong
– Federico
Nov 20 '18 at 19:19
@Federico Could you explain the correct way to think about this?
– mitmath514
Nov 20 '18 at 19:23
add a comment |
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 Nov 20 '18 at 19:10
mitmath514
163
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 Nov 20 '18 at 19:10
mitmath514
163
asked Nov 20 '18 at 19:10
mitmath514
163
asked Nov 20 '18 at 19:10
mitmath514
163
asked Nov 20 '18 at 19:10
mitmath514
163
asked Nov 20 '18 at 19:10
mitmath514
163
163
First true, second false. Your motivation is wrong
– Federico
Nov 20 '18 at 19:19
@Federico Could you explain the correct way to think about this?
– mitmath514
Nov 20 '18 at 19:23
add a comment |
First true, second false. Your motivation is wrong
– Federico
Nov 20 '18 at 19:19
@Federico Could you explain the correct way to think about this?
– mitmath514
Nov 20 '18 at 19:23
First true, second false. Your motivation is wrong
– Federico
Nov 20 '18 at 19:19
First true, second false. Your motivation is wrong
– Federico
Nov 20 '18 at 19:19
@Federico Could you explain the correct way to think about this?
– mitmath514
Nov 20 '18 at 19:23
@Federico Could you explain the correct way to think about this?
– mitmath514
Nov 20 '18 at 19:23
add a comment |
2 Answers
2
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oldest
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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 Nov 20 '18 at 19:49
Ramiro Scorolli
655113
add a comment |
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 Nov 20 '18 at 23:43
LaserPineapple
403
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
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 Nov 20 '18 at 19:49
Ramiro Scorolli
655113
add a comment |
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 Nov 20 '18 at 19:49
Ramiro Scorolli
655113
add a comment |
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 Nov 20 '18 at 19:49
Ramiro Scorolli
655113
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 Nov 20 '18 at 19:49
Ramiro Scorolli
655113
answered Nov 20 '18 at 19:49
Ramiro Scorolli
655113
answered Nov 20 '18 at 19:49
Ramiro Scorolli
655113
answered Nov 20 '18 at 19:49
Ramiro Scorolli
655113
655113
add a comment |
add a comment |
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 Nov 20 '18 at 23:43
LaserPineapple
403
add a comment |
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 Nov 20 '18 at 23:43
LaserPineapple
403
add a comment |
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 Nov 20 '18 at 23:43
LaserPineapple
403
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 Nov 20 '18 at 23:43
LaserPineapple
403
answered Nov 20 '18 at 23:43
LaserPineapple
403
answered Nov 20 '18 at 23:43
LaserPineapple
403
answered Nov 20 '18 at 23:43
LaserPineapple
403
403
add a comment |
add a comment |
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First true, second false. Your motivation is wrong
– Federico
Nov 20 '18 at 19:19
@Federico Could you explain the correct way to think about this?
– mitmath514
Nov 20 '18 at 19:23