Mixing
Can we estimate the entropy of data from its moments?
1
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I would like to measure the entropy of a set of real numbers. I am wondering if can come up with a reasonable estimate by using the first $n$ moments.
Edit 1: It looks like the InfoGAN network estimates mutual information by way of "variational arguments." What does that mean? I think this is getting to my problem which is how to compute entropy when the data is not discrete. Any thoughts?
stochastic-processes information-theory
asked Jan 30 at 23:04
AvedisAvedis
667
$endgroup$
add a comment |
1
$begingroup$
I would like to measure the entropy of a set of real numbers. I am wondering if can come up with a reasonable estimate by using the first $n$ moments.
Edit 1: It looks like the InfoGAN network estimates mutual information by way of "variational arguments." What does that mean? I think this is getting to my problem which is how to compute entropy when the data is not discrete. Any thoughts?
stochastic-processes information-theory
asked Jan 30 at 23:04
AvedisAvedis
667
$endgroup$
$begingroup$
Entropy has little to do with moments, I doubt you can get something useful that way. BTW, you are speaking here of differential entropy, no ? (differential entropy is not really entropy) THe most natural way is to estimate first $p(x)$ by a standard density estimation procedure (parametric or not parametric), see eg here people.cs.umass.edu/~elm/Teaching/650_F14/est_IT.pdf
$endgroup$
– leonbloy
Feb 1 at 21:02
$begingroup$
@leonbloy I am not talking about differential entropy. Unfortunately, i cannot estimate p(x). I need a differentiable way to estimate p(x) and histogramming will not work.
$endgroup$
– Avedis
Feb 2 at 0:08
1
$begingroup$
If the variable is continuous, then its entropy is infinite. I still guess you are speaking of differential entropy. Histogramming is not the only non parametric estimation of a continuous function $p(x)$. See eg Parzen windows.
$endgroup$
– leonbloy
Feb 2 at 1:21
add a comment |
1
1
1
$begingroup$
I would like to measure the entropy of a set of real numbers. I am wondering if can come up with a reasonable estimate by using the first $n$ moments.
Edit 1: It looks like the InfoGAN network estimates mutual information by way of "variational arguments." What does that mean? I think this is getting to my problem which is how to compute entropy when the data is not discrete. Any thoughts?
stochastic-processes information-theory
asked Jan 30 at 23:04
AvedisAvedis
667
$endgroup$
I would like to measure the entropy of a set of real numbers. I am wondering if can come up with a reasonable estimate by using the first $n$ moments.
Edit 1: It looks like the InfoGAN network estimates mutual information by way of "variational arguments." What does that mean? I think this is getting to my problem which is how to compute entropy when the data is not discrete. Any thoughts?
stochastic-processes information-theory
stochastic-processes information-theory
asked Jan 30 at 23:04
AvedisAvedis
667
asked Jan 30 at 23:04
AvedisAvedis
667
edited Jan 31 at 20:27
Avedis
asked Jan 30 at 23:04
AvedisAvedis
667
asked Jan 30 at 23:04
AvedisAvedis
667
asked Jan 30 at 23:04
AvedisAvedis
667
667
$begingroup$
Entropy has little to do with moments, I doubt you can get something useful that way. BTW, you are speaking here of differential entropy, no ? (differential entropy is not really entropy) THe most natural way is to estimate first $p(x)$ by a standard density estimation procedure (parametric or not parametric), see eg here people.cs.umass.edu/~elm/Teaching/650_F14/est_IT.pdf
$endgroup$
– leonbloy
Feb 1 at 21:02
$begingroup$
@leonbloy I am not talking about differential entropy. Unfortunately, i cannot estimate p(x). I need a differentiable way to estimate p(x) and histogramming will not work.
$endgroup$
– Avedis
Feb 2 at 0:08
1
$begingroup$
If the variable is continuous, then its entropy is infinite. I still guess you are speaking of differential entropy. Histogramming is not the only non parametric estimation of a continuous function $p(x)$. See eg Parzen windows.
$endgroup$
– leonbloy
Feb 2 at 1:21
add a comment |
$begingroup$
Entropy has little to do with moments, I doubt you can get something useful that way. BTW, you are speaking here of differential entropy, no ? (differential entropy is not really entropy) THe most natural way is to estimate first $p(x)$ by a standard density estimation procedure (parametric or not parametric), see eg here people.cs.umass.edu/~elm/Teaching/650_F14/est_IT.pdf
$endgroup$
– leonbloy
Feb 1 at 21:02
$begingroup$
@leonbloy I am not talking about differential entropy. Unfortunately, i cannot estimate p(x). I need a differentiable way to estimate p(x) and histogramming will not work.
$endgroup$
– Avedis
Feb 2 at 0:08
1
$begingroup$
If the variable is continuous, then its entropy is infinite. I still guess you are speaking of differential entropy. Histogramming is not the only non parametric estimation of a continuous function $p(x)$. See eg Parzen windows.
$endgroup$
– leonbloy
Feb 2 at 1:21
$begingroup$
Entropy has little to do with moments, I doubt you can get something useful that way. BTW, you are speaking here of differential entropy, no ? (differential entropy is not really entropy) THe most natural way is to estimate first $p(x)$ by a standard density estimation procedure (parametric or not parametric), see eg here people.cs.umass.edu/~elm/Teaching/650_F14/est_IT.pdf
$endgroup$
– leonbloy
Feb 1 at 21:02
$begingroup$
Entropy has little to do with moments, I doubt you can get something useful that way. BTW, you are speaking here of differential entropy, no ? (differential entropy is not really entropy) THe most natural way is to estimate first $p(x)$ by a standard density estimation procedure (parametric or not parametric), see eg here people.cs.umass.edu/~elm/Teaching/650_F14/est_IT.pdf
$endgroup$
– leonbloy
Feb 1 at 21:02
$begingroup$
@leonbloy I am not talking about differential entropy. Unfortunately, i cannot estimate p(x). I need a differentiable way to estimate p(x) and histogramming will not work.
$endgroup$
– Avedis
Feb 2 at 0:08
$begingroup$
@leonbloy I am not talking about differential entropy. Unfortunately, i cannot estimate p(x). I need a differentiable way to estimate p(x) and histogramming will not work.
$endgroup$
– Avedis
Feb 2 at 0:08
1
1
$begingroup$
If the variable is continuous, then its entropy is infinite. I still guess you are speaking of differential entropy. Histogramming is not the only non parametric estimation of a continuous function $p(x)$. See eg Parzen windows.
$endgroup$
– leonbloy
Feb 2 at 1:21
$begingroup$
If the variable is continuous, then its entropy is infinite. I still guess you are speaking of differential entropy. Histogramming is not the only non parametric estimation of a continuous function $p(x)$. See eg Parzen windows.
$endgroup$
– leonbloy
Feb 2 at 1:21
add a comment |
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$begingroup$
Entropy has little to do with moments, I doubt you can get something useful that way. BTW, you are speaking here of differential entropy, no ? (differential entropy is not really entropy) THe most natural way is to estimate first $p(x)$ by a standard density estimation procedure (parametric or not parametric), see eg here people.cs.umass.edu/~elm/Teaching/650_F14/est_IT.pdf
$endgroup$
– leonbloy
Feb 1 at 21:02
$begingroup$
@leonbloy I am not talking about differential entropy. Unfortunately, i cannot estimate p(x). I need a differentiable way to estimate p(x) and histogramming will not work.
$endgroup$
– Avedis
Feb 2 at 0:08
1
$begingroup$
If the variable is continuous, then its entropy is infinite. I still guess you are speaking of differential entropy. Histogramming is not the only non parametric estimation of a continuous function $p(x)$. See eg Parzen windows.
$endgroup$
– leonbloy
Feb 2 at 1:21