Expected value: Is $E|X| = |E(X)|$?
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 at 14:47
Especially Lime
21.6k22758
asked Dec 7 at 14:44
bob
1089
add a comment |
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 at 14:47
Especially Lime
21.6k22758
asked Dec 7 at 14:44
bob
1089
add a comment |
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 at 14:47
Especially Lime
21.6k22758
asked Dec 7 at 14:44
bob
1089
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 at 14:47
Especially Lime
21.6k22758
asked Dec 7 at 14:44
bob
1089
edited Dec 7 at 14:47
Especially Lime
21.6k22758
asked Dec 7 at 14:44
bob
1089
edited Dec 7 at 14:47
Especially Lime
21.6k22758
edited Dec 7 at 14:47
Especially Lime
21.6k22758
edited Dec 7 at 14:47
Especially Lime
21.6k22758
21.6k22758
asked Dec 7 at 14:44
bob
1089
asked Dec 7 at 14:44
bob
1089
asked Dec 7 at 14:44
bob
1089
1089
add a comment |
add a comment |
3 Answers
3
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oldest
votes
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 at 14:46
A. Pongrácz
5,216725
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
2
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
add a comment |
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 at 14:55
N.Beck
1816
add a comment |
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 at 14:48
drhab
97.1k544128
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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active
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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 at 14:46
A. Pongrácz
5,216725
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
2
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
add a comment |
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 at 14:46
A. Pongrácz
5,216725
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
2
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
add a comment |
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 at 14:46
A. Pongrácz
5,216725
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 at 14:46
A. Pongrácz
5,216725
answered Dec 7 at 14:46
A. Pongrácz
5,216725
answered Dec 7 at 14:46
A. Pongrácz
5,216725
answered Dec 7 at 14:46
A. Pongrácz
5,216725
5,216725
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
2
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
add a comment |
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
2
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
If the reader does not find the latter it trivial, it is a result of the triangular inequality.
– Jonas
Dec 7 at 15:04
2
2
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
@Jonas Not needed, simply use $-|X|leqslant Xleqslant|X|$.
– Did
Dec 7 at 15:30
add a comment |
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 at 14:55
N.Beck
1816
add a comment |
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 at 14:55
N.Beck
1816
add a comment |
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 at 14:55
N.Beck
1816
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 at 14:55
N.Beck
1816
answered Dec 7 at 14:55
N.Beck
1816
answered Dec 7 at 14:55
N.Beck
1816
answered Dec 7 at 14:55
N.Beck
1816
1816
add a comment |
add a comment |
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 at 14:48
drhab
97.1k544128
add a comment |
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 at 14:48
drhab
97.1k544128
add a comment |
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 at 14:48
drhab
97.1k544128
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 at 14:48
drhab
97.1k544128
answered Dec 7 at 14:48
drhab
97.1k544128
answered Dec 7 at 14:48
drhab
97.1k544128
answered Dec 7 at 14:48
drhab
97.1k544128
97.1k544128
add a comment |
add a comment |
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