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Signle variable integration with respect to a function
Lebesgue vs. Riemann integrable functionMeasurable functions - integrationIs integration with respect to spherical measure equivalent to manifold integration over sphere?What is the difference between integrating with respect to dS and dx in the context of scalar field line integrals?Function with finite integral is finite a.e.Integrating volume integral with respect to a single variableIntegration with respect to the empirical process?Rigorous definition of characteristic function of random variable as Lebesgue integralIntegration with the gradient of an indicator functionIntegration by parts for multivariable functions using the Divergence Theorem
$begingroup$
I guess this is a trivial problem. I was reading about expected value on wiki and I came across a notation of an integral I don't understand. There is a statement that a general case of expected value has this form:
$$E[X]=int_Omega X(omega),dP(omega)$$
with a comment that this is a Lebesgue integral. I was taught to calculate integrals or multi integrals with respect to a number of variables, not functions. When I see the term $dP(omega)$, I am confused! I know an expected value can also be expressed in this form
$$E[X] = int_X x,p(x),dx$$
because it is simply a weighted sum / integral of a random variable over probabilities associated with its realizations.
How to understand an integral when it is calculate with respect to a function?
integration
asked Mar 12 at 10:48
CeldorCeldor
31439
$endgroup$
add a comment |
$begingroup$
I guess this is a trivial problem. I was reading about expected value on wiki and I came across a notation of an integral I don't understand. There is a statement that a general case of expected value has this form:
$$E[X]=int_Omega X(omega),dP(omega)$$
with a comment that this is a Lebesgue integral. I was taught to calculate integrals or multi integrals with respect to a number of variables, not functions. When I see the term $dP(omega)$, I am confused! I know an expected value can also be expressed in this form
$$E[X] = int_X x,p(x),dx$$
because it is simply a weighted sum / integral of a random variable over probabilities associated with its realizations.
How to understand an integral when it is calculate with respect to a function?
integration
asked Mar 12 at 10:48
CeldorCeldor
31439
$endgroup$
add a comment |
$begingroup$
I guess this is a trivial problem. I was reading about expected value on wiki and I came across a notation of an integral I don't understand. There is a statement that a general case of expected value has this form:
$$E[X]=int_Omega X(omega),dP(omega)$$
with a comment that this is a Lebesgue integral. I was taught to calculate integrals or multi integrals with respect to a number of variables, not functions. When I see the term $dP(omega)$, I am confused! I know an expected value can also be expressed in this form
$$E[X] = int_X x,p(x),dx$$
because it is simply a weighted sum / integral of a random variable over probabilities associated with its realizations.
How to understand an integral when it is calculate with respect to a function?
integration
asked Mar 12 at 10:48
CeldorCeldor
31439
$endgroup$
I guess this is a trivial problem. I was reading about expected value on wiki and I came across a notation of an integral I don't understand. There is a statement that a general case of expected value has this form:
$$E[X]=int_Omega X(omega),dP(omega)$$
with a comment that this is a Lebesgue integral. I was taught to calculate integrals or multi integrals with respect to a number of variables, not functions. When I see the term $dP(omega)$, I am confused! I know an expected value can also be expressed in this form
$$E[X] = int_X x,p(x),dx$$
because it is simply a weighted sum / integral of a random variable over probabilities associated with its realizations.
How to understand an integral when it is calculate with respect to a function?
integration
integration
asked Mar 12 at 10:48
CeldorCeldor
31439
asked Mar 12 at 10:48
CeldorCeldor
31439
edited Mar 12 at 11:35
Celdor
asked Mar 12 at 10:48
CeldorCeldor
31439
asked Mar 12 at 10:48
CeldorCeldor
31439
asked Mar 12 at 10:48
CeldorCeldor
31439
31439
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Formally speaking, a probability space is a measure space. Given a set of possible outcomes, we can find the measure of that set - the probability of being in that set.
That integral "$dP(omega)$" is simply the integral with respect to that measure.
In practice, how will we evaluate it? We'll find a density function $rho$, or a probability mass function $p$, and convert it to something like
$$E(X) = int_Omega xrho(x),dx$$
in the density case (where $Omega$ is the space of possible values), or
$$E(X) = sum_xin Omegaxp(x)$$
Writing it in terms of the probability measure $P(omega)$ allows us to unify those two expressions, as well as more complicated cases (that hardly ever come up in practice).
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
$endgroup$
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Formally speaking, a probability space is a measure space. Given a set of possible outcomes, we can find the measure of that set - the probability of being in that set.
That integral "$dP(omega)$" is simply the integral with respect to that measure.
In practice, how will we evaluate it? We'll find a density function $rho$, or a probability mass function $p$, and convert it to something like
$$E(X) = int_Omega xrho(x),dx$$
in the density case (where $Omega$ is the space of possible values), or
$$E(X) = sum_xin Omegaxp(x)$$
Writing it in terms of the probability measure $P(omega)$ allows us to unify those two expressions, as well as more complicated cases (that hardly ever come up in practice).
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
$endgroup$
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
add a comment |
$begingroup$
Formally speaking, a probability space is a measure space. Given a set of possible outcomes, we can find the measure of that set - the probability of being in that set.
That integral "$dP(omega)$" is simply the integral with respect to that measure.
In practice, how will we evaluate it? We'll find a density function $rho$, or a probability mass function $p$, and convert it to something like
$$E(X) = int_Omega xrho(x),dx$$
in the density case (where $Omega$ is the space of possible values), or
$$E(X) = sum_xin Omegaxp(x)$$
Writing it in terms of the probability measure $P(omega)$ allows us to unify those two expressions, as well as more complicated cases (that hardly ever come up in practice).
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
$endgroup$
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
add a comment |
$begingroup$
Formally speaking, a probability space is a measure space. Given a set of possible outcomes, we can find the measure of that set - the probability of being in that set.
That integral "$dP(omega)$" is simply the integral with respect to that measure.
In practice, how will we evaluate it? We'll find a density function $rho$, or a probability mass function $p$, and convert it to something like
$$E(X) = int_Omega xrho(x),dx$$
in the density case (where $Omega$ is the space of possible values), or
$$E(X) = sum_xin Omegaxp(x)$$
Writing it in terms of the probability measure $P(omega)$ allows us to unify those two expressions, as well as more complicated cases (that hardly ever come up in practice).
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
$endgroup$
Formally speaking, a probability space is a measure space. Given a set of possible outcomes, we can find the measure of that set - the probability of being in that set.
That integral "$dP(omega)$" is simply the integral with respect to that measure.
In practice, how will we evaluate it? We'll find a density function $rho$, or a probability mass function $p$, and convert it to something like
$$E(X) = int_Omega xrho(x),dx$$
in the density case (where $Omega$ is the space of possible values), or
$$E(X) = sum_xin Omegaxp(x)$$
Writing it in terms of the probability measure $P(omega)$ allows us to unify those two expressions, as well as more complicated cases (that hardly ever come up in practice).
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
answered Mar 12 at 11:32
jmerryjmerry
14.4k1629
14.4k1629
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
add a comment |
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
I am afraid don't know what a measure or a measure of set is but when I read your explanation I kind of understand that $P(omega)$ is one of the measures we could use. In this example, it is Probability but it could be something esle. When we integrate with respect to $dP(omega)$, we take numbers a function returns (probability) and not what variables are. Could integration be considered as a operand over product of terms and this time the "term" is not a variable but a value of a function? I am confused :p but thanks for the answer.
$endgroup$
– Celdor
Mar 12 at 12:04
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
At your level of understanding? Ignore that talk about measures. We know how to find the expected value if it's a "continuous" random variable with a density function (the integral expression), and we know how to find the expected value if it's a discrete random variable with a probability mass function (the sum expression). That's all you really need. Later, when you've got more of a real analysis background, you can come back and update the theory behind things with the idea of measures.
$endgroup$
– jmerry
Mar 12 at 12:22
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
$begingroup$
Thanks for your time. I am not going to ignore it as I figured out my question is also about that part which as it turned out is a measure. This is what confuses me. Cheers!
$endgroup$
– Celdor
Mar 12 at 12:30
add a comment |
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