Expectation of absolute value of the difference between a random variable and its mean
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Suppose $Y$ is a random variable with finite mean $mu$. Then is $mathbb{E}(|Y-mu|)$ finite or not? Why?
probability
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
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$begingroup$
Suppose $Y$ is a random variable with finite mean $mu$. Then is $mathbb{E}(|Y-mu|)$ finite or not? Why?
probability
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
$endgroup$
add a comment |
$begingroup$
Suppose $Y$ is a random variable with finite mean $mu$. Then is $mathbb{E}(|Y-mu|)$ finite or not? Why?
probability
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
$endgroup$
Suppose $Y$ is a random variable with finite mean $mu$. Then is $mathbb{E}(|Y-mu|)$ finite or not? Why?
probability
probability
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
asked Dec 5 '18 at 20:05
Will.ZWill.Z
133
133
add a comment |
add a comment |
1 Answer
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The expected value of a random variable $X$ exists and is finite if both $operatorname EX^+$ and $operatorname EX^{-}$ are finite, where $X^+(omega)=max{X(omega),0}$ and $X^-(omega)=-min{X(omega),0}$ for $omegainOmega$. We have that $operatorname EX=operatorname EX^+-operatorname EX^{-}$. If the expected value is finite, the first absolute moment is also finite since $E|X|=operatorname EX^++operatorname EX^-$. It follows that
$$
operatorname E|X-mu|leoperatorname E|X|+|mu|<infty.
$$
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
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1 Answer
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1 Answer
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$begingroup$
The expected value of a random variable $X$ exists and is finite if both $operatorname EX^+$ and $operatorname EX^{-}$ are finite, where $X^+(omega)=max{X(omega),0}$ and $X^-(omega)=-min{X(omega),0}$ for $omegainOmega$. We have that $operatorname EX=operatorname EX^+-operatorname EX^{-}$. If the expected value is finite, the first absolute moment is also finite since $E|X|=operatorname EX^++operatorname EX^-$. It follows that
$$
operatorname E|X-mu|leoperatorname E|X|+|mu|<infty.
$$
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
add a comment |
$begingroup$
The expected value of a random variable $X$ exists and is finite if both $operatorname EX^+$ and $operatorname EX^{-}$ are finite, where $X^+(omega)=max{X(omega),0}$ and $X^-(omega)=-min{X(omega),0}$ for $omegainOmega$. We have that $operatorname EX=operatorname EX^+-operatorname EX^{-}$. If the expected value is finite, the first absolute moment is also finite since $E|X|=operatorname EX^++operatorname EX^-$. It follows that
$$
operatorname E|X-mu|leoperatorname E|X|+|mu|<infty.
$$
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
add a comment |
$begingroup$
The expected value of a random variable $X$ exists and is finite if both $operatorname EX^+$ and $operatorname EX^{-}$ are finite, where $X^+(omega)=max{X(omega),0}$ and $X^-(omega)=-min{X(omega),0}$ for $omegainOmega$. We have that $operatorname EX=operatorname EX^+-operatorname EX^{-}$. If the expected value is finite, the first absolute moment is also finite since $E|X|=operatorname EX^++operatorname EX^-$. It follows that
$$
operatorname E|X-mu|leoperatorname E|X|+|mu|<infty.
$$
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
$endgroup$
The expected value of a random variable $X$ exists and is finite if both $operatorname EX^+$ and $operatorname EX^{-}$ are finite, where $X^+(omega)=max{X(omega),0}$ and $X^-(omega)=-min{X(omega),0}$ for $omegainOmega$. We have that $operatorname EX=operatorname EX^+-operatorname EX^{-}$. If the expected value is finite, the first absolute moment is also finite since $E|X|=operatorname EX^++operatorname EX^-$. It follows that
$$
operatorname E|X-mu|leoperatorname E|X|+|mu|<infty.
$$
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
answered Dec 5 '18 at 20:11
Cm7F7BbCm7F7Bb
12.5k32243
12.5k32243
add a comment |
add a comment |
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