Gaussian expectation of a distribution that involves another Gaussian RV

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I would like to understand the following expectation of a PDF:

Let $thetasimmathcalN(mu, Sigma_1)$, $xsimmathcalN(theta, Sigma_2)$. Let $f(x)$ be an arbitrary function of $x$. Then how can I get rid of the expectation in $mathbbE_theta[mathbbP_x(f(x))]$?

I have seen Normal distribution with mean coming from normal distribution, and understood that $xsimmathcalN(mu, Sigma_1+Sigma_2)$. My guess for the answer is $mathbbP_xsimmathcalN(mu, Sigma_1+Sigma_2)(f(x))$, but I'm not sure how to formalize it.

Thanks a lot for any hint.

probability probability-distributions normal-distribution

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asked Jul 14 at 22:47

user3799934

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I would like to understand the following expectation of a PDF:

Let $thetasimmathcalN(mu, Sigma_1)$, $xsimmathcalN(theta, Sigma_2)$. Let $f(x)$ be an arbitrary function of $x$. Then how can I get rid of the expectation in $mathbbE_theta[mathbbP_x(f(x))]$?

I have seen Normal distribution with mean coming from normal distribution, and understood that $xsimmathcalN(mu, Sigma_1+Sigma_2)$. My guess for the answer is $mathbbP_xsimmathcalN(mu, Sigma_1+Sigma_2)(f(x))$, but I'm not sure how to formalize it.

Thanks a lot for any hint.

probability probability-distributions normal-distribution

share|cite|improve this question

asked Jul 14 at 22:47

user3799934

252

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I would like to understand the following expectation of a PDF:

Let $thetasimmathcalN(mu, Sigma_1)$, $xsimmathcalN(theta, Sigma_2)$. Let $f(x)$ be an arbitrary function of $x$. Then how can I get rid of the expectation in $mathbbE_theta[mathbbP_x(f(x))]$?

I have seen Normal distribution with mean coming from normal distribution, and understood that $xsimmathcalN(mu, Sigma_1+Sigma_2)$. My guess for the answer is $mathbbP_xsimmathcalN(mu, Sigma_1+Sigma_2)(f(x))$, but I'm not sure how to formalize it.

Thanks a lot for any hint.

probability probability-distributions normal-distribution

share|cite|improve this question

asked Jul 14 at 22:47

user3799934

252

I would like to understand the following expectation of a PDF:

Let $thetasimmathcalN(mu, Sigma_1)$, $xsimmathcalN(theta, Sigma_2)$. Let $f(x)$ be an arbitrary function of $x$. Then how can I get rid of the expectation in $mathbbE_theta[mathbbP_x(f(x))]$?

I have seen Normal distribution with mean coming from normal distribution, and understood that $xsimmathcalN(mu, Sigma_1+Sigma_2)$. My guess for the answer is $mathbbP_xsimmathcalN(mu, Sigma_1+Sigma_2)(f(x))$, but I'm not sure how to formalize it.

Thanks a lot for any hint.

probability probability-distributions normal-distribution

share|cite|improve this question

asked Jul 14 at 22:47

user3799934

252

share|cite|improve this question

share|cite|improve this question

asked Jul 14 at 22:47

user3799934

252

asked Jul 14 at 22:47

user3799934

252

asked Jul 14 at 22:47

user3799934

252

252

add a comment | 

add a comment | 

1 Answer
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Assuming your notation is intended to denote "the expectation over random variable $theta $ of the marginal probability density $p_xleft( x right)$ (I apologize, I have trouble creating your specific notation) subsequently evaluated at value $fleft( x right)$", then you are correct, since the marginal density $p_xleft( x right)$ does not depend on $theta $ (this is called a nuisance parameter, and the operation of marginalizing it away is what leads to the marginal variance/covariance $Sigma _1 + Sigma _2$), so that the expectation operation is effectively a no op, and just integrates out to unity.

There are, however, other possible ways to interpret your notation, depending upon the level of precision you employed. In particular, if it is intended to denote "the expectation over random variable $theta $ of the conditional probability density $p_xleft( left. x right right)$ subsequently evaluated at value $fleft( x right)$", then the expectation does actually play, although I believe it leads to the same result here, since
$$E_theta left[ p_xleft( left. x right right) right] = p_xleft( x right)$$
which is how you obtain the marginal density from the conditional density in the first place.

If your use of $p_xleft( fleft( x right) right)$ is actually intended to denote marginal density $p_yleft( y right)$ (or conditional density $p_yleft( theta right)$) for random variable transformation $y = fleft( x right)$, rather than just subsequent evaluation of the $x$ density at value $fleft( x right)$, then things would get more complicated. But my guess is this is not what you mean.

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answered Jul 15 at 1:37

John Polcari

382111

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Assuming your notation is intended to denote "the expectation over random variable $theta $ of the marginal probability density $p_xleft( x right)$ (I apologize, I have trouble creating your specific notation) subsequently evaluated at value $fleft( x right)$", then you are correct, since the marginal density $p_xleft( x right)$ does not depend on $theta $ (this is called a nuisance parameter, and the operation of marginalizing it away is what leads to the marginal variance/covariance $Sigma _1 + Sigma _2$), so that the expectation operation is effectively a no op, and just integrates out to unity.

There are, however, other possible ways to interpret your notation, depending upon the level of precision you employed. In particular, if it is intended to denote "the expectation over random variable $theta $ of the conditional probability density $p_xleft( left. x right right)$ subsequently evaluated at value $fleft( x right)$", then the expectation does actually play, although I believe it leads to the same result here, since
$$E_theta left[ p_xleft( left. x right right) right] = p_xleft( x right)$$
which is how you obtain the marginal density from the conditional density in the first place.

If your use of $p_xleft( fleft( x right) right)$ is actually intended to denote marginal density $p_yleft( y right)$ (or conditional density $p_yleft( theta right)$) for random variable transformation $y = fleft( x right)$, rather than just subsequent evaluation of the $x$ density at value $fleft( x right)$, then things would get more complicated. But my guess is this is not what you mean.

share|cite|improve this answer

answered Jul 15 at 1:37

John Polcari

382111

add a comment | 

up vote
0
down vote



accepted

Assuming your notation is intended to denote "the expectation over random variable $theta $ of the marginal probability density $p_xleft( x right)$ (I apologize, I have trouble creating your specific notation) subsequently evaluated at value $fleft( x right)$", then you are correct, since the marginal density $p_xleft( x right)$ does not depend on $theta $ (this is called a nuisance parameter, and the operation of marginalizing it away is what leads to the marginal variance/covariance $Sigma _1 + Sigma _2$), so that the expectation operation is effectively a no op, and just integrates out to unity.

There are, however, other possible ways to interpret your notation, depending upon the level of precision you employed. In particular, if it is intended to denote "the expectation over random variable $theta $ of the conditional probability density $p_xleft( left. x right right)$ subsequently evaluated at value $fleft( x right)$", then the expectation does actually play, although I believe it leads to the same result here, since
$$E_theta left[ p_xleft( left. x right right) right] = p_xleft( x right)$$
which is how you obtain the marginal density from the conditional density in the first place.

If your use of $p_xleft( fleft( x right) right)$ is actually intended to denote marginal density $p_yleft( y right)$ (or conditional density $p_yleft( theta right)$) for random variable transformation $y = fleft( x right)$, rather than just subsequent evaluation of the $x$ density at value $fleft( x right)$, then things would get more complicated. But my guess is this is not what you mean.

share|cite|improve this answer

answered Jul 15 at 1:37

John Polcari

382111

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Assuming your notation is intended to denote "the expectation over random variable $theta $ of the marginal probability density $p_xleft( x right)$ (I apologize, I have trouble creating your specific notation) subsequently evaluated at value $fleft( x right)$", then you are correct, since the marginal density $p_xleft( x right)$ does not depend on $theta $ (this is called a nuisance parameter, and the operation of marginalizing it away is what leads to the marginal variance/covariance $Sigma _1 + Sigma _2$), so that the expectation operation is effectively a no op, and just integrates out to unity.

There are, however, other possible ways to interpret your notation, depending upon the level of precision you employed. In particular, if it is intended to denote "the expectation over random variable $theta $ of the conditional probability density $p_xleft( left. x right right)$ subsequently evaluated at value $fleft( x right)$", then the expectation does actually play, although I believe it leads to the same result here, since
$$E_theta left[ p_xleft( left. x right right) right] = p_xleft( x right)$$
which is how you obtain the marginal density from the conditional density in the first place.

If your use of $p_xleft( fleft( x right) right)$ is actually intended to denote marginal density $p_yleft( y right)$ (or conditional density $p_yleft( theta right)$) for random variable transformation $y = fleft( x right)$, rather than just subsequent evaluation of the $x$ density at value $fleft( x right)$, then things would get more complicated. But my guess is this is not what you mean.

share|cite|improve this answer

answered Jul 15 at 1:37

John Polcari

382111

Assuming your notation is intended to denote "the expectation over random variable $theta $ of the marginal probability density $p_xleft( x right)$ (I apologize, I have trouble creating your specific notation) subsequently evaluated at value $fleft( x right)$", then you are correct, since the marginal density $p_xleft( x right)$ does not depend on $theta $ (this is called a nuisance parameter, and the operation of marginalizing it away is what leads to the marginal variance/covariance $Sigma _1 + Sigma _2$), so that the expectation operation is effectively a no op, and just integrates out to unity.

There are, however, other possible ways to interpret your notation, depending upon the level of precision you employed. In particular, if it is intended to denote "the expectation over random variable $theta $ of the conditional probability density $p_xleft( left. x right right)$ subsequently evaluated at value $fleft( x right)$", then the expectation does actually play, although I believe it leads to the same result here, since
$$E_theta left[ p_xleft( left. x right right) right] = p_xleft( x right)$$
which is how you obtain the marginal density from the conditional density in the first place.

If your use of $p_xleft( fleft( x right) right)$ is actually intended to denote marginal density $p_yleft( y right)$ (or conditional density $p_yleft( theta right)$) for random variable transformation $y = fleft( x right)$, rather than just subsequent evaluation of the $x$ density at value $fleft( x right)$, then things would get more complicated. But my guess is this is not what you mean.

share|cite|improve this answer

answered Jul 15 at 1:37

John Polcari

382111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 1:37

John Polcari

382111

answered Jul 15 at 1:37

John Polcari

382111

answered Jul 15 at 1:37

John Polcari

382111

382111

add a comment | 

add a comment | 

 

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