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Signle variable integration with respect to a function

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0

$begingroup$

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

share|cite|improve this question

edited Mar 12 at 11:35

Celdor

asked Mar 12 at 10:48

CeldorCeldor

31439

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this question

edited Mar 12 at 11:35

Celdor

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

share|cite|improve this question

edited Mar 12 at 11:35

Celdor

asked Mar 12 at 10:48

CeldorCeldor

31439

$endgroup$

share|cite|improve this question

edited Mar 12 at 11:35

Celdor

asked Mar 12 at 10:48

CeldorCeldor

31439

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

0

$begingroup$

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).

share|cite|improve this answer

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 | 

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0

$begingroup$

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).

share|cite|improve this answer

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 | 

0

$begingroup$

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).

share|cite|improve this answer

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).

share|cite|improve this answer

answered Mar 12 at 11:32

jmerryjmerry

14.4k1629

$endgroup$

share|cite|improve this answer

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