Mixing
Mutual information between dependent and independent variables
$begingroup$
This question is a follow-up for the previously asked question. Assume we have three random variables $A$, $B$, and $C$, where $A$ and $B$ are independent (i.e. $I(A ,; B)=0$), but the relation between $C$ and $A$, $B$ is arbitrary. The question is how to simplify the following:
$$
I(A ,; B,C).
$$
information-theory entropy
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
add a comment |
$begingroup$
This question is a follow-up for the previously asked question. Assume we have three random variables $A$, $B$, and $C$, where $A$ and $B$ are independent (i.e. $I(A ,; B)=0$), but the relation between $C$ and $A$, $B$ is arbitrary. The question is how to simplify the following:
$$
I(A ,; B,C).
$$
information-theory entropy
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
add a comment |
$begingroup$
This question is a follow-up for the previously asked question. Assume we have three random variables $A$, $B$, and $C$, where $A$ and $B$ are independent (i.e. $I(A ,; B)=0$), but the relation between $C$ and $A$, $B$ is arbitrary. The question is how to simplify the following:
$$
I(A ,; B,C).
$$
information-theory entropy
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
This question is a follow-up for the previously asked question. Assume we have three random variables $A$, $B$, and $C$, where $A$ and $B$ are independent (i.e. $I(A ,; B)=0$), but the relation between $C$ and $A$, $B$ is arbitrary. The question is how to simplify the following:
$$
I(A ,; B,C).
$$
information-theory entropy
information-theory entropy
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
asked Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
16210
add a comment |
add a comment |
1 Answer
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$begingroup$
Independence of $A$ and $B$ also means that $H(A mid B) = H(A)$. Next, by using definition of mutual information and chain rule for entropies:
$$
I(A ,; B, C) = H(A) - H(A mid B, C) = H(A mid B) - H(A mid B, C) = I(A ,; C mid B).
$$
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Independence of $A$ and $B$ also means that $H(A mid B) = H(A)$. Next, by using definition of mutual information and chain rule for entropies:
$$
I(A ,; B, C) = H(A) - H(A mid B, C) = H(A mid B) - H(A mid B, C) = I(A ,; C mid B).
$$
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
add a comment |
$begingroup$
Independence of $A$ and $B$ also means that $H(A mid B) = H(A)$. Next, by using definition of mutual information and chain rule for entropies:
$$
I(A ,; B, C) = H(A) - H(A mid B, C) = H(A mid B) - H(A mid B, C) = I(A ,; C mid B).
$$
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
add a comment |
$begingroup$
Independence of $A$ and $B$ also means that $H(A mid B) = H(A)$. Next, by using definition of mutual information and chain rule for entropies:
$$
I(A ,; B, C) = H(A) - H(A mid B, C) = H(A mid B) - H(A mid B, C) = I(A ,; C mid B).
$$
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
$endgroup$
Independence of $A$ and $B$ also means that $H(A mid B) = H(A)$. Next, by using definition of mutual information and chain rule for entropies:
$$
I(A ,; B, C) = H(A) - H(A mid B, C) = H(A mid B) - H(A mid B, C) = I(A ,; C mid B).
$$
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
answered Jan 29 at 9:37
Yauhen YakimenkaYauhen Yakimenka
16210
16210
add a comment |
add a comment |
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