Expected value: Is $E|X| = |E(X)|$?
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Is the expectation of absolute value equal to the absolute value of the expectation? $E|X| = |E(X)|$ seems intuitively true to me, but I couldn't find it online. I wanted to check whether this is true.
probability statistics expected-value
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
asked Dec 7 '18 at 14:44
bobbob
1089
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add a comment |
$begingroup$
Is the expectation of absolute value equal to the absolute value of the expectation? $E|X| = |E(X)|$ seems intuitively true to me, but I couldn't find it online. I wanted to check whether this is true.
probability statistics expected-value
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
asked Dec 7 '18 at 14:44
bobbob
1089
$endgroup$
add a comment |
$begingroup$
Is the expectation of absolute value equal to the absolute value of the expectation? $E|X| = |E(X)|$ seems intuitively true to me, but I couldn't find it online. I wanted to check whether this is true.
probability statistics expected-value
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
asked Dec 7 '18 at 14:44
bobbob
1089
$endgroup$
Is the expectation of absolute value equal to the absolute value of the expectation? $E|X| = |E(X)|$ seems intuitively true to me, but I couldn't find it online. I wanted to check whether this is true.
probability statistics expected-value
probability statistics expected-value
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
asked Dec 7 '18 at 14:44
bobbob
1089
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
asked Dec 7 '18 at 14:44
bobbob
1089
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
edited Dec 7 '18 at 14:47
Especially Lime
21.9k22858
21.9k22858
asked Dec 7 '18 at 14:44
bobbob
1089
asked Dec 7 '18 at 14:44
bobbob
1089
asked Dec 7 '18 at 14:44
bobbob
1089
1089
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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Not true in general.
E.g., let $X$ be the random variable whose value is $-1$ or $+1$, both with probability $1/2$. Then $|X|$ is $+1$ with probability $1$, so $|E(X)|= |0| = 0$ and $E(|X|)=1$.
The only thing you can claim is $|E(X)|leq E(|X|)$, and it is trivial.
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
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If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
2
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@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
add a comment |
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Let $(Omega,mathcal F, mathbb P)$ be a probability space and let $X$ be a random variable. Then
begin{align*}
mathbb E[X] := int_{Omega} X, dmathbb P
end{align*}
This is the Lebsegue integral of $X$ over the measure space $(Omega,mathcal F, mathbb P)$. Now the answer should be obvious. In general
begin{align*}
int_{Omega} |X|, dmathbb P neq left|int_{Omega} X, dmathbb Pright|
end{align*}
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
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add a comment |
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It is not true in general.
Counter example: Let $X$ take value $+1$ with probability $0.5$ and $-1$ with probability $0.5$.
Then $|X|=1$ so that $mathbb E|X|=1$.
But $|mathbb EX|=|0|=0$.
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not true in general.
E.g., let $X$ be the random variable whose value is $-1$ or $+1$, both with probability $1/2$. Then $|X|$ is $+1$ with probability $1$, so $|E(X)|= |0| = 0$ and $E(|X|)=1$.
The only thing you can claim is $|E(X)|leq E(|X|)$, and it is trivial.
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
2
$begingroup$
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
add a comment |
$begingroup$
Not true in general.
E.g., let $X$ be the random variable whose value is $-1$ or $+1$, both with probability $1/2$. Then $|X|$ is $+1$ with probability $1$, so $|E(X)|= |0| = 0$ and $E(|X|)=1$.
The only thing you can claim is $|E(X)|leq E(|X|)$, and it is trivial.
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
$endgroup$
$begingroup$
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
2
$begingroup$
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
add a comment |
$begingroup$
Not true in general.
E.g., let $X$ be the random variable whose value is $-1$ or $+1$, both with probability $1/2$. Then $|X|$ is $+1$ with probability $1$, so $|E(X)|= |0| = 0$ and $E(|X|)=1$.
The only thing you can claim is $|E(X)|leq E(|X|)$, and it is trivial.
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
$endgroup$
Not true in general.
E.g., let $X$ be the random variable whose value is $-1$ or $+1$, both with probability $1/2$. Then $|X|$ is $+1$ with probability $1$, so $|E(X)|= |0| = 0$ and $E(|X|)=1$.
The only thing you can claim is $|E(X)|leq E(|X|)$, and it is trivial.
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
answered Dec 7 '18 at 14:46
A. PongráczA. Pongrácz
5,9631929
5,9631929
$begingroup$
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
2
$begingroup$
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
add a comment |
$begingroup$
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
2
$begingroup$
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
$begingroup$
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
$begingroup$
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
$endgroup$
– Jonas
Dec 7 '18 at 15:04
2
2
$begingroup$
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
$begingroup$
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
$endgroup$
– Did
Dec 7 '18 at 15:30
add a comment |
$begingroup$
Let $(Omega,mathcal F, mathbb P)$ be a probability space and let $X$ be a random variable. Then
begin{align*}
mathbb E[X] := int_{Omega} X, dmathbb P
end{align*}
This is the Lebsegue integral of $X$ over the measure space $(Omega,mathcal F, mathbb P)$. Now the answer should be obvious. In general
begin{align*}
int_{Omega} |X|, dmathbb P neq left|int_{Omega} X, dmathbb Pright|
end{align*}
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal F, mathbb P)$ be a probability space and let $X$ be a random variable. Then
begin{align*}
mathbb E[X] := int_{Omega} X, dmathbb P
end{align*}
This is the Lebsegue integral of $X$ over the measure space $(Omega,mathcal F, mathbb P)$. Now the answer should be obvious. In general
begin{align*}
int_{Omega} |X|, dmathbb P neq left|int_{Omega} X, dmathbb Pright|
end{align*}
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal F, mathbb P)$ be a probability space and let $X$ be a random variable. Then
begin{align*}
mathbb E[X] := int_{Omega} X, dmathbb P
end{align*}
This is the Lebsegue integral of $X$ over the measure space $(Omega,mathcal F, mathbb P)$. Now the answer should be obvious. In general
begin{align*}
int_{Omega} |X|, dmathbb P neq left|int_{Omega} X, dmathbb Pright|
end{align*}
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
$endgroup$
Let $(Omega,mathcal F, mathbb P)$ be a probability space and let $X$ be a random variable. Then
begin{align*}
mathbb E[X] := int_{Omega} X, dmathbb P
end{align*}
This is the Lebsegue integral of $X$ over the measure space $(Omega,mathcal F, mathbb P)$. Now the answer should be obvious. In general
begin{align*}
int_{Omega} |X|, dmathbb P neq left|int_{Omega} X, dmathbb Pright|
end{align*}
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
answered Dec 7 '18 at 14:55
N.BeckN.Beck
2987
2987
add a comment |
add a comment |
$begingroup$
It is not true in general.
Counter example: Let $X$ take value $+1$ with probability $0.5$ and $-1$ with probability $0.5$.
Then $|X|=1$ so that $mathbb E|X|=1$.
But $|mathbb EX|=|0|=0$.
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
It is not true in general.
Counter example: Let $X$ take value $+1$ with probability $0.5$ and $-1$ with probability $0.5$.
Then $|X|=1$ so that $mathbb E|X|=1$.
But $|mathbb EX|=|0|=0$.
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
It is not true in general.
Counter example: Let $X$ take value $+1$ with probability $0.5$ and $-1$ with probability $0.5$.
Then $|X|=1$ so that $mathbb E|X|=1$.
But $|mathbb EX|=|0|=0$.
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
$endgroup$
It is not true in general.
Counter example: Let $X$ take value $+1$ with probability $0.5$ and $-1$ with probability $0.5$.
Then $|X|=1$ so that $mathbb E|X|=1$.
But $|mathbb EX|=|0|=0$.
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
answered Dec 7 '18 at 14:48
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
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