Mixing
Entropy of continuous and discrete random variables
$begingroup$
If N is a continuous random variable and X a discrete random variable.
How can I calculate H(X|Y) if Y=X+N?
N is a triangular distribution between -1 and 1
X can take the values +-0.5 with equal probability
Assuming known: H(X), h(N)=h(Y|X), h(Y) and pdf's of X, N and Y.
All the entropy terms are finite.
probability random-variables information-theory entropy
asked Jan 2 at 18:46
misscamissca
83
$endgroup$
add a comment |
$begingroup$
If N is a continuous random variable and X a discrete random variable.
How can I calculate H(X|Y) if Y=X+N?
N is a triangular distribution between -1 and 1
X can take the values +-0.5 with equal probability
Assuming known: H(X), h(N)=h(Y|X), h(Y) and pdf's of X, N and Y.
All the entropy terms are finite.
probability random-variables information-theory entropy
asked Jan 2 at 18:46
misscamissca
83
$endgroup$
$begingroup$
What exactly is denoted by $mathbf{H(Xmid Y)}$?
$endgroup$
– drhab
Jan 2 at 18:49
$begingroup$
The entropy of X conditioned on Y
$endgroup$
– missca
Jan 2 at 18:54
add a comment |
$begingroup$
If N is a continuous random variable and X a discrete random variable.
How can I calculate H(X|Y) if Y=X+N?
N is a triangular distribution between -1 and 1
X can take the values +-0.5 with equal probability
Assuming known: H(X), h(N)=h(Y|X), h(Y) and pdf's of X, N and Y.
All the entropy terms are finite.
probability random-variables information-theory entropy
asked Jan 2 at 18:46
misscamissca
83
$endgroup$
If N is a continuous random variable and X a discrete random variable.
How can I calculate H(X|Y) if Y=X+N?
N is a triangular distribution between -1 and 1
X can take the values +-0.5 with equal probability
Assuming known: H(X), h(N)=h(Y|X), h(Y) and pdf's of X, N and Y.
All the entropy terms are finite.
probability random-variables information-theory entropy
probability random-variables information-theory entropy
asked Jan 2 at 18:46
misscamissca
83
asked Jan 2 at 18:46
misscamissca
83
edited Jan 3 at 22:49
missca
asked Jan 2 at 18:46
misscamissca
83
asked Jan 2 at 18:46
misscamissca
83
asked Jan 2 at 18:46
misscamissca
83
83
$begingroup$
What exactly is denoted by $mathbf{H(Xmid Y)}$?
$endgroup$
– drhab
Jan 2 at 18:49
$begingroup$
The entropy of X conditioned on Y
$endgroup$
– missca
Jan 2 at 18:54
add a comment |
$begingroup$
What exactly is denoted by $mathbf{H(Xmid Y)}$?
$endgroup$
– drhab
Jan 2 at 18:49
$begingroup$
The entropy of X conditioned on Y
$endgroup$
– missca
Jan 2 at 18:54
$begingroup$
What exactly is denoted by $mathbf{H(Xmid Y)}$?
$endgroup$
– drhab
Jan 2 at 18:49
$begingroup$
What exactly is denoted by $mathbf{H(Xmid Y)}$?
$endgroup$
– drhab
Jan 2 at 18:49
$begingroup$
The entropy of X conditioned on Y
$endgroup$
– missca
Jan 2 at 18:54
$begingroup$
The entropy of X conditioned on Y
$endgroup$
– missca
Jan 2 at 18:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's true that you shouldn't mix/confuse the "true" entropy with the differential entropy (differential entropy is not a true Shannon entropy). For one thing, the true entropy of a (non degenerate) continuous variable is infinite. But it's still true that the ("true") mutual information of any two random variables is well defined, using the difference of either true entropies or differential entropies (see for example here or here)
So, if $X$ is discrete and $Y$ continuous, we are justified in writing $I(X;Y)=H(X)-H(X|Y)=h(Y)-h(Y|X)$ and hence
$$ H(X|Y) = H(X) - I(X;Y)=H(X) -h(Y) + h(Y|X) $$
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
$endgroup$
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
add a comment |
$begingroup$
From the mutual information relation
$$H(X|Y) = H(X) - h(Y) + h(Y|X)$$
Use this to get your desired result.
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
$endgroup$
1
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's true that you shouldn't mix/confuse the "true" entropy with the differential entropy (differential entropy is not a true Shannon entropy). For one thing, the true entropy of a (non degenerate) continuous variable is infinite. But it's still true that the ("true") mutual information of any two random variables is well defined, using the difference of either true entropies or differential entropies (see for example here or here)
So, if $X$ is discrete and $Y$ continuous, we are justified in writing $I(X;Y)=H(X)-H(X|Y)=h(Y)-h(Y|X)$ and hence
$$ H(X|Y) = H(X) - I(X;Y)=H(X) -h(Y) + h(Y|X) $$
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
$endgroup$
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
add a comment |
$begingroup$
It's true that you shouldn't mix/confuse the "true" entropy with the differential entropy (differential entropy is not a true Shannon entropy). For one thing, the true entropy of a (non degenerate) continuous variable is infinite. But it's still true that the ("true") mutual information of any two random variables is well defined, using the difference of either true entropies or differential entropies (see for example here or here)
So, if $X$ is discrete and $Y$ continuous, we are justified in writing $I(X;Y)=H(X)-H(X|Y)=h(Y)-h(Y|X)$ and hence
$$ H(X|Y) = H(X) - I(X;Y)=H(X) -h(Y) + h(Y|X) $$
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
$endgroup$
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
add a comment |
$begingroup$
It's true that you shouldn't mix/confuse the "true" entropy with the differential entropy (differential entropy is not a true Shannon entropy). For one thing, the true entropy of a (non degenerate) continuous variable is infinite. But it's still true that the ("true") mutual information of any two random variables is well defined, using the difference of either true entropies or differential entropies (see for example here or here)
So, if $X$ is discrete and $Y$ continuous, we are justified in writing $I(X;Y)=H(X)-H(X|Y)=h(Y)-h(Y|X)$ and hence
$$ H(X|Y) = H(X) - I(X;Y)=H(X) -h(Y) + h(Y|X) $$
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
$endgroup$
It's true that you shouldn't mix/confuse the "true" entropy with the differential entropy (differential entropy is not a true Shannon entropy). For one thing, the true entropy of a (non degenerate) continuous variable is infinite. But it's still true that the ("true") mutual information of any two random variables is well defined, using the difference of either true entropies or differential entropies (see for example here or here)
So, if $X$ is discrete and $Y$ continuous, we are justified in writing $I(X;Y)=H(X)-H(X|Y)=h(Y)-h(Y|X)$ and hence
$$ H(X|Y) = H(X) - I(X;Y)=H(X) -h(Y) + h(Y|X) $$
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
edited Jan 4 at 15:45
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
answered Jan 3 at 15:09
leonbloyleonbloy
40.5k645107
40.5k645107
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
add a comment |
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
I understand what you say, but my teacher says we can not use: $$H(X|Y)=H(X)-h(Y)+h(Y|X)$$ because we are mixing H with h. He says that we can only use it if both are discrete or both are continuous, and in my case X is discrete and Y continuous. That's why I do not know how to calculate $H(X|Y)$
$endgroup$
– missca
Jan 3 at 22:42
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
$begingroup$
No matter how you solve it, the end result will be just mine (and Gautam Shenoy's). So, in the end you will be "mixing H and h" (which are your only data). You cannot escape that. You only need to justify how to do it. And that I have mentioned (and linked to the details)
$endgroup$
– leonbloy
Jan 3 at 23:59
add a comment |
$begingroup$
From the mutual information relation
$$H(X|Y) = H(X) - h(Y) + h(Y|X)$$
Use this to get your desired result.
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
$endgroup$
1
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
add a comment |
$begingroup$
From the mutual information relation
$$H(X|Y) = H(X) - h(Y) + h(Y|X)$$
Use this to get your desired result.
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
$endgroup$
1
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
add a comment |
$begingroup$
From the mutual information relation
$$H(X|Y) = H(X) - h(Y) + h(Y|X)$$
Use this to get your desired result.
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
$endgroup$
From the mutual information relation
$$H(X|Y) = H(X) - h(Y) + h(Y|X)$$
Use this to get your desired result.
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
answered Jan 2 at 18:55
Gautam ShenoyGautam Shenoy
7,15911545
7,15911545
1
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
add a comment |
1
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
1
1
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
My teacher says that you can not mix entropies (H, discrete variables) with relative entropies (h, continuous variables)
$endgroup$
– missca
Jan 2 at 19:18
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Are all the entropy terms in your example finite?
$endgroup$
– Gautam Shenoy
Jan 3 at 2:55
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
$begingroup$
Yes! All the entropy terms are finite.
$endgroup$
– missca
Jan 3 at 8:19
add a comment |
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$begingroup$
What exactly is denoted by $mathbf{H(Xmid Y)}$?
$endgroup$
– drhab
Jan 2 at 18:49
$begingroup$
The entropy of X conditioned on Y
$endgroup$
– missca
Jan 2 at 18:54