Does $| E[X*1{Xmu}] |$?
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Does mean of a random variable $mu=E[X]$ devide conditional mean in half, in a sense that
$$ | E]X*1{X<mu}] | = | E[X*1{X>mu}] |$$
or is this only true for symmetric distributions?
probability-distributions exponentiation
asked Nov 23 at 9:04
Confounded
1408
add a comment |
up vote
0
down vote
favorite
Does mean of a random variable $mu=E[X]$ devide conditional mean in half, in a sense that
$$ | E]X*1{X<mu}] | = | E[X*1{X>mu}] |$$
or is this only true for symmetric distributions?
probability-distributions exponentiation
asked Nov 23 at 9:04
Confounded
1408
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Does mean of a random variable $mu=E[X]$ devide conditional mean in half, in a sense that
$$ | E]X*1{X<mu}] | = | E[X*1{X>mu}] |$$
or is this only true for symmetric distributions?
probability-distributions exponentiation
asked Nov 23 at 9:04
Confounded
1408
Does mean of a random variable $mu=E[X]$ devide conditional mean in half, in a sense that
$$ | E]X*1{X<mu}] | = | E[X*1{X>mu}] |$$
or is this only true for symmetric distributions?
probability-distributions exponentiation
probability-distributions exponentiation
asked Nov 23 at 9:04
Confounded
1408
asked Nov 23 at 9:04
Confounded
1408
edited Nov 23 at 10:13
asked Nov 23 at 9:04
Confounded
1408
asked Nov 23 at 9:04
Confounded
1408
asked Nov 23 at 9:04
Confounded
1408
1408
add a comment |
add a comment |
2 Answers
2
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1
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Not true in general. Let $X$ take values $1$ and $2$ with ptobabilites $frac 1 3$ and $frac 2 3$. Then EX=$frac 5 3$, LHS is $frac 1 3$ and RHS is $frac 4 3$
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
add a comment |
up vote
1
down vote
This is not even true for symmetric distributions. Suppose that $P(X=-1)=P(X=1)=1/2$. Then
$$
operatorname E[X1{X<0}]=-frac12
$$
but
$$
operatorname E[X1{X>0}]=frac12.
$$
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
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
Not true in general. Let $X$ take values $1$ and $2$ with ptobabilites $frac 1 3$ and $frac 2 3$. Then EX=$frac 5 3$, LHS is $frac 1 3$ and RHS is $frac 4 3$
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
add a comment |
up vote
1
down vote
Not true in general. Let $X$ take values $1$ and $2$ with ptobabilites $frac 1 3$ and $frac 2 3$. Then EX=$frac 5 3$, LHS is $frac 1 3$ and RHS is $frac 4 3$
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
add a comment |
up vote
1
down vote
up vote
1
down vote
Not true in general. Let $X$ take values $1$ and $2$ with ptobabilites $frac 1 3$ and $frac 2 3$. Then EX=$frac 5 3$, LHS is $frac 1 3$ and RHS is $frac 4 3$
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
Not true in general. Let $X$ take values $1$ and $2$ with ptobabilites $frac 1 3$ and $frac 2 3$. Then EX=$frac 5 3$, LHS is $frac 1 3$ and RHS is $frac 4 3$
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
answered Nov 23 at 9:08
Kavi Rama Murthy
43k31751
43k31751
add a comment |
add a comment |
up vote
1
down vote
This is not even true for symmetric distributions. Suppose that $P(X=-1)=P(X=1)=1/2$. Then
$$
operatorname E[X1{X<0}]=-frac12
$$
but
$$
operatorname E[X1{X>0}]=frac12.
$$
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
add a comment |
up vote
1
down vote
This is not even true for symmetric distributions. Suppose that $P(X=-1)=P(X=1)=1/2$. Then
$$
operatorname E[X1{X<0}]=-frac12
$$
but
$$
operatorname E[X1{X>0}]=frac12.
$$
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
add a comment |
up vote
1
down vote
up vote
1
down vote
This is not even true for symmetric distributions. Suppose that $P(X=-1)=P(X=1)=1/2$. Then
$$
operatorname E[X1{X<0}]=-frac12
$$
but
$$
operatorname E[X1{X>0}]=frac12.
$$
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
This is not even true for symmetric distributions. Suppose that $P(X=-1)=P(X=1)=1/2$. Then
$$
operatorname E[X1{X<0}]=-frac12
$$
but
$$
operatorname E[X1{X>0}]=frac12.
$$
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
answered Nov 23 at 9:55
Cm7F7Bb
12.3k32142
12.3k32142
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
add a comment |
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
Sorry, I forgot to apply the absolute to the expectations on both sides of the mean. But thank you for your reply in any case.
– Confounded
Nov 23 at 10:09
add a comment |
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