Do random variables always integrate to 1?
In my experience, a classic exam question for stats students is to ask them to prove that a given function is a random variable. Typically, they are expected to give an answer which includes integrating (or summing) over the sample space (which is almost always $mathbb{R}$) and pointing out that the result is $1$. However, does this trick always work? What if, for example, we're working with a suspected random variable that's exotic enough to require Lebesgue integration? Or what if we're in a rather nasty probability space?
probability-theory random-variables lebesgue-integral
asked Nov 29 at 17:48
J. Mini
3414
add a comment |
In my experience, a classic exam question for stats students is to ask them to prove that a given function is a random variable. Typically, they are expected to give an answer which includes integrating (or summing) over the sample space (which is almost always $mathbb{R}$) and pointing out that the result is $1$. However, does this trick always work? What if, for example, we're working with a suspected random variable that's exotic enough to require Lebesgue integration? Or what if we're in a rather nasty probability space?
probability-theory random-variables lebesgue-integral
asked Nov 29 at 17:48
J. Mini
3414
2
It's a bit unclear what you are saying. A random variable is associated with a probability measure, which "sample space" has a total measure of $1$. Integrating the random variable (typically a real valued entity) with respect to the measure on its sample space gives the expected value (or "mean") of the random variable. This integral may not be finite, so not every random variable has an expected value.
– hardmath
Nov 29 at 17:53
add a comment |
In my experience, a classic exam question for stats students is to ask them to prove that a given function is a random variable. Typically, they are expected to give an answer which includes integrating (or summing) over the sample space (which is almost always $mathbb{R}$) and pointing out that the result is $1$. However, does this trick always work? What if, for example, we're working with a suspected random variable that's exotic enough to require Lebesgue integration? Or what if we're in a rather nasty probability space?
probability-theory random-variables lebesgue-integral
asked Nov 29 at 17:48
J. Mini
3414
In my experience, a classic exam question for stats students is to ask them to prove that a given function is a random variable. Typically, they are expected to give an answer which includes integrating (or summing) over the sample space (which is almost always $mathbb{R}$) and pointing out that the result is $1$. However, does this trick always work? What if, for example, we're working with a suspected random variable that's exotic enough to require Lebesgue integration? Or what if we're in a rather nasty probability space?
probability-theory random-variables lebesgue-integral
probability-theory random-variables lebesgue-integral
asked Nov 29 at 17:48
J. Mini
3414
asked Nov 29 at 17:48
J. Mini
3414
asked Nov 29 at 17:48
J. Mini
3414
asked Nov 29 at 17:48
J. Mini
3414
asked Nov 29 at 17:48
J. Mini
3414
3414
2
It's a bit unclear what you are saying. A random variable is associated with a probability measure, which "sample space" has a total measure of $1$. Integrating the random variable (typically a real valued entity) with respect to the measure on its sample space gives the expected value (or "mean") of the random variable. This integral may not be finite, so not every random variable has an expected value.
– hardmath
Nov 29 at 17:53
add a comment |
2
It's a bit unclear what you are saying. A random variable is associated with a probability measure, which "sample space" has a total measure of $1$. Integrating the random variable (typically a real valued entity) with respect to the measure on its sample space gives the expected value (or "mean") of the random variable. This integral may not be finite, so not every random variable has an expected value.
– hardmath
Nov 29 at 17:53
2
2
It's a bit unclear what you are saying. A random variable is associated with a probability measure, which "sample space" has a total measure of $1$. Integrating the random variable (typically a real valued entity) with respect to the measure on its sample space gives the expected value (or "mean") of the random variable. This integral may not be finite, so not every random variable has an expected value.
– hardmath
Nov 29 at 17:53
It's a bit unclear what you are saying. A random variable is associated with a probability measure, which "sample space" has a total measure of $1$. Integrating the random variable (typically a real valued entity) with respect to the measure on its sample space gives the expected value (or "mean") of the random variable. This integral may not be finite, so not every random variable has an expected value.
– hardmath
Nov 29 at 17:53
add a comment |
1 Answer
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There are random variables that don't have a density or a probability mass function, so it wouldn't make sense to ask a question of the form "show that this function is a density" or "show that this function is a PMF" if you're interested in a random variable like that.
But from the measure-theoretic point of view, if you have a function on a probability space, the only thing to be done to show it's a random variable is to show it's measurable. The "integrates to 1" part is already built into the measure.
Showing that a given measure is a probability measure would be closer to your question.
answered Nov 29 at 17:55
Ian
67.3k25285
add a comment |
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There are random variables that don't have a density or a probability mass function, so it wouldn't make sense to ask a question of the form "show that this function is a density" or "show that this function is a PMF" if you're interested in a random variable like that.
But from the measure-theoretic point of view, if you have a function on a probability space, the only thing to be done to show it's a random variable is to show it's measurable. The "integrates to 1" part is already built into the measure.
Showing that a given measure is a probability measure would be closer to your question.
answered Nov 29 at 17:55
Ian
67.3k25285
add a comment |
There are random variables that don't have a density or a probability mass function, so it wouldn't make sense to ask a question of the form "show that this function is a density" or "show that this function is a PMF" if you're interested in a random variable like that.
But from the measure-theoretic point of view, if you have a function on a probability space, the only thing to be done to show it's a random variable is to show it's measurable. The "integrates to 1" part is already built into the measure.
Showing that a given measure is a probability measure would be closer to your question.
answered Nov 29 at 17:55
Ian
67.3k25285
add a comment |
There are random variables that don't have a density or a probability mass function, so it wouldn't make sense to ask a question of the form "show that this function is a density" or "show that this function is a PMF" if you're interested in a random variable like that.
But from the measure-theoretic point of view, if you have a function on a probability space, the only thing to be done to show it's a random variable is to show it's measurable. The "integrates to 1" part is already built into the measure.
Showing that a given measure is a probability measure would be closer to your question.
answered Nov 29 at 17:55
Ian
67.3k25285
There are random variables that don't have a density or a probability mass function, so it wouldn't make sense to ask a question of the form "show that this function is a density" or "show that this function is a PMF" if you're interested in a random variable like that.
But from the measure-theoretic point of view, if you have a function on a probability space, the only thing to be done to show it's a random variable is to show it's measurable. The "integrates to 1" part is already built into the measure.
Showing that a given measure is a probability measure would be closer to your question.
answered Nov 29 at 17:55
Ian
67.3k25285
answered Nov 29 at 17:55
Ian
67.3k25285
answered Nov 29 at 17:55
Ian
67.3k25285
answered Nov 29 at 17:55
Ian
67.3k25285
67.3k25285
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add a comment |
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2
It's a bit unclear what you are saying. A random variable is associated with a probability measure, which "sample space" has a total measure of $1$. Integrating the random variable (typically a real valued entity) with respect to the measure on its sample space gives the expected value (or "mean") of the random variable. This integral may not be finite, so not every random variable has an expected value.
– hardmath
Nov 29 at 17:53