Mixing
Expected Value well-defined
$begingroup$
Say $X$ is an arbitrary random variable
I have been told that $X$ has a well-defined expectation if $mathbb E[|X|] < infty$. Does this mean that if $mathbb E[X]=infty$, then $X$ does not have a well-defined expectation? Or can there be cases where $mathbb E[|X|] < infty$ while $mathbb E[X]=infty$?
I am trying to think of examples on my own, but cannot find one.
In this sense, is the statement: The expectation $mathbb E[X]$ exists equivalent to the statement $mathbb E[X]$ is well-defined?
probability probability-theory random-variables
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
$endgroup$
add a comment |
$begingroup$
Say $X$ is an arbitrary random variable
I have been told that $X$ has a well-defined expectation if $mathbb E[|X|] < infty$. Does this mean that if $mathbb E[X]=infty$, then $X$ does not have a well-defined expectation? Or can there be cases where $mathbb E[|X|] < infty$ while $mathbb E[X]=infty$?
I am trying to think of examples on my own, but cannot find one.
In this sense, is the statement: The expectation $mathbb E[X]$ exists equivalent to the statement $mathbb E[X]$ is well-defined?
probability probability-theory random-variables
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
$endgroup$
1
$begingroup$
Note that $-|X| le Xle |X|$. So if $E[|X|] < infty$, then we must have $E[X] < infty$.
$endgroup$
– angryavian
Feb 2 at 21:31
$begingroup$
So, if $mathbb E[X]=infty$ then the expectation of random variable $X$ does not exist?
$endgroup$
– MinaThuma
Feb 2 at 21:35
add a comment |
$begingroup$
Say $X$ is an arbitrary random variable
I have been told that $X$ has a well-defined expectation if $mathbb E[|X|] < infty$. Does this mean that if $mathbb E[X]=infty$, then $X$ does not have a well-defined expectation? Or can there be cases where $mathbb E[|X|] < infty$ while $mathbb E[X]=infty$?
I am trying to think of examples on my own, but cannot find one.
In this sense, is the statement: The expectation $mathbb E[X]$ exists equivalent to the statement $mathbb E[X]$ is well-defined?
probability probability-theory random-variables
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
$endgroup$
Say $X$ is an arbitrary random variable
I have been told that $X$ has a well-defined expectation if $mathbb E[|X|] < infty$. Does this mean that if $mathbb E[X]=infty$, then $X$ does not have a well-defined expectation? Or can there be cases where $mathbb E[|X|] < infty$ while $mathbb E[X]=infty$?
I am trying to think of examples on my own, but cannot find one.
In this sense, is the statement: The expectation $mathbb E[X]$ exists equivalent to the statement $mathbb E[X]$ is well-defined?
probability probability-theory random-variables
probability probability-theory random-variables
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
asked Feb 2 at 21:21
MinaThumaMinaThuma
2709
2709
1
$begingroup$
Note that $-|X| le Xle |X|$. So if $E[|X|] < infty$, then we must have $E[X] < infty$.
$endgroup$
– angryavian
Feb 2 at 21:31
$begingroup$
So, if $mathbb E[X]=infty$ then the expectation of random variable $X$ does not exist?
$endgroup$
– MinaThuma
Feb 2 at 21:35
add a comment |
1
$begingroup$
Note that $-|X| le Xle |X|$. So if $E[|X|] < infty$, then we must have $E[X] < infty$.
$endgroup$
– angryavian
Feb 2 at 21:31
$begingroup$
So, if $mathbb E[X]=infty$ then the expectation of random variable $X$ does not exist?
$endgroup$
– MinaThuma
Feb 2 at 21:35
1
1
$begingroup$
Note that $-|X| le Xle |X|$. So if $E[|X|] < infty$, then we must have $E[X] < infty$.
$endgroup$
– angryavian
Feb 2 at 21:31
$begingroup$
Note that $-|X| le Xle |X|$. So if $E[|X|] < infty$, then we must have $E[X] < infty$.
$endgroup$
– angryavian
Feb 2 at 21:31
$begingroup$
So, if $mathbb E[X]=infty$ then the expectation of random variable $X$ does not exist?
$endgroup$
– MinaThuma
Feb 2 at 21:35
$begingroup$
So, if $mathbb E[X]=infty$ then the expectation of random variable $X$ does not exist?
$endgroup$
– MinaThuma
Feb 2 at 21:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $X$ is a random variable which is always nonnegative, then its expectation is always defined, though it may be infinite.
In general, letting $X^+=max(X,0)$ and $X^-=max(-X,0)$, so that $X=X^+-X^-$, then both $X^+$ and $X^-$ are always nonnegative, so their expectations are well defined. There are then four cases for $EX$:
$$
begin{array}{r|cc}
&E[X^+]<infty & E[X^+]=infty\
hline
E[X^-]<infty & EXtext{ is finite} & EX=+infty\
E[X^-]=infty & EX=-infty & EXtext{ is not well defined}
end{array}
$$
Only in the upper left hand corner of that table do we have $E|X|<infty$.
Whether $EX=infty$ implies $EX$ exists or not is a matter of convention. Some people say $lim_{xto0}frac1{x^2}$ does not exist, while others say this limit exists and is equal to $+infty$. I think most probabalists agree that $infty$ exists.
In summary, I disagree with the sentiment that $EX$ only exists when $E|X|<infty$. For example, the St. Petersburg random variable for which $P(X=2^i)=2^{-i}$ for $ige 1$ has $E|X|=infty$, and I would say that $EX$ exists and is equal to $+infty$.
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
If $X$ is a random variable which is always nonnegative, then its expectation is always defined, though it may be infinite.
In general, letting $X^+=max(X,0)$ and $X^-=max(-X,0)$, so that $X=X^+-X^-$, then both $X^+$ and $X^-$ are always nonnegative, so their expectations are well defined. There are then four cases for $EX$:
$$
begin{array}{r|cc}
&E[X^+]<infty & E[X^+]=infty\
hline
E[X^-]<infty & EXtext{ is finite} & EX=+infty\
E[X^-]=infty & EX=-infty & EXtext{ is not well defined}
end{array}
$$
Only in the upper left hand corner of that table do we have $E|X|<infty$.
Whether $EX=infty$ implies $EX$ exists or not is a matter of convention. Some people say $lim_{xto0}frac1{x^2}$ does not exist, while others say this limit exists and is equal to $+infty$. I think most probabalists agree that $infty$ exists.
In summary, I disagree with the sentiment that $EX$ only exists when $E|X|<infty$. For example, the St. Petersburg random variable for which $P(X=2^i)=2^{-i}$ for $ige 1$ has $E|X|=infty$, and I would say that $EX$ exists and is equal to $+infty$.
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
$endgroup$
add a comment |
$begingroup$
If $X$ is a random variable which is always nonnegative, then its expectation is always defined, though it may be infinite.
In general, letting $X^+=max(X,0)$ and $X^-=max(-X,0)$, so that $X=X^+-X^-$, then both $X^+$ and $X^-$ are always nonnegative, so their expectations are well defined. There are then four cases for $EX$:
$$
begin{array}{r|cc}
&E[X^+]<infty & E[X^+]=infty\
hline
E[X^-]<infty & EXtext{ is finite} & EX=+infty\
E[X^-]=infty & EX=-infty & EXtext{ is not well defined}
end{array}
$$
Only in the upper left hand corner of that table do we have $E|X|<infty$.
Whether $EX=infty$ implies $EX$ exists or not is a matter of convention. Some people say $lim_{xto0}frac1{x^2}$ does not exist, while others say this limit exists and is equal to $+infty$. I think most probabalists agree that $infty$ exists.
In summary, I disagree with the sentiment that $EX$ only exists when $E|X|<infty$. For example, the St. Petersburg random variable for which $P(X=2^i)=2^{-i}$ for $ige 1$ has $E|X|=infty$, and I would say that $EX$ exists and is equal to $+infty$.
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
$endgroup$
add a comment |
$begingroup$
If $X$ is a random variable which is always nonnegative, then its expectation is always defined, though it may be infinite.
In general, letting $X^+=max(X,0)$ and $X^-=max(-X,0)$, so that $X=X^+-X^-$, then both $X^+$ and $X^-$ are always nonnegative, so their expectations are well defined. There are then four cases for $EX$:
$$
begin{array}{r|cc}
&E[X^+]<infty & E[X^+]=infty\
hline
E[X^-]<infty & EXtext{ is finite} & EX=+infty\
E[X^-]=infty & EX=-infty & EXtext{ is not well defined}
end{array}
$$
Only in the upper left hand corner of that table do we have $E|X|<infty$.
Whether $EX=infty$ implies $EX$ exists or not is a matter of convention. Some people say $lim_{xto0}frac1{x^2}$ does not exist, while others say this limit exists and is equal to $+infty$. I think most probabalists agree that $infty$ exists.
In summary, I disagree with the sentiment that $EX$ only exists when $E|X|<infty$. For example, the St. Petersburg random variable for which $P(X=2^i)=2^{-i}$ for $ige 1$ has $E|X|=infty$, and I would say that $EX$ exists and is equal to $+infty$.
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
$endgroup$
If $X$ is a random variable which is always nonnegative, then its expectation is always defined, though it may be infinite.
In general, letting $X^+=max(X,0)$ and $X^-=max(-X,0)$, so that $X=X^+-X^-$, then both $X^+$ and $X^-$ are always nonnegative, so their expectations are well defined. There are then four cases for $EX$:
$$
begin{array}{r|cc}
&E[X^+]<infty & E[X^+]=infty\
hline
E[X^-]<infty & EXtext{ is finite} & EX=+infty\
E[X^-]=infty & EX=-infty & EXtext{ is not well defined}
end{array}
$$
Only in the upper left hand corner of that table do we have $E|X|<infty$.
Whether $EX=infty$ implies $EX$ exists or not is a matter of convention. Some people say $lim_{xto0}frac1{x^2}$ does not exist, while others say this limit exists and is equal to $+infty$. I think most probabalists agree that $infty$ exists.
In summary, I disagree with the sentiment that $EX$ only exists when $E|X|<infty$. For example, the St. Petersburg random variable for which $P(X=2^i)=2^{-i}$ for $ige 1$ has $E|X|=infty$, and I would say that $EX$ exists and is equal to $+infty$.
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
answered Feb 2 at 21:50
Mike EarnestMike Earnest
27.9k22152
27.9k22152
add a comment |
add a comment |
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1
$begingroup$
Note that $-|X| le Xle |X|$. So if $E[|X|] < infty$, then we must have $E[X] < infty$.
$endgroup$
– angryavian
Feb 2 at 21:31
$begingroup$
So, if $mathbb E[X]=infty$ then the expectation of random variable $X$ does not exist?
$endgroup$
– MinaThuma
Feb 2 at 21:35