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Under what conditions does the function C = f(A,B) satisfy H(C|A) = H(B)?
Measuring entropy for a table (e.g., SQL results)Information of a stream of bitsCan a transcendental number like $e$ or $pi$ be compressed as not algorithmically random?How to compare conditional entropy and mutual information?Why the alphabet of the digital information is composed of 2 elements?One-shot Private Randomness ExtractorFind minimum conditional entropyHigher order empirical entropy is not the entropy of the empirical distribution?Conceptual overview: Self-information, Mutual information, uncertainty, entropyHow realistic is the i.i.d assumption in the definition of Shannon's entropy?
$begingroup$
Suppose we have a function $f$,
$$
C = f(A,B),
$$
where $A$, $B$ and $C$ are random variables.
I notice that when the random variables are binary ($0, 1$) and $f$ is the XOR operation, we have the following identity:
$$
H(C|A) = H(B),
$$
where $H(B)$ is the entropy of $B$ and $H(C|A)$ is the conditional entropy of $C$ given $A$.
Obviously this is not true for a general $f$. What I am interested to know is, is there a set of conditions on $f$ and $A,B,C$, under which the identity above holds.
information-theory
asked Mar 28 at 7:20
hklelhklel
1255
$endgroup$
add a comment |
$begingroup$
Suppose we have a function $f$,
$$
C = f(A,B),
$$
where $A$, $B$ and $C$ are random variables.
I notice that when the random variables are binary ($0, 1$) and $f$ is the XOR operation, we have the following identity:
$$
H(C|A) = H(B),
$$
where $H(B)$ is the entropy of $B$ and $H(C|A)$ is the conditional entropy of $C$ given $A$.
Obviously this is not true for a general $f$. What I am interested to know is, is there a set of conditions on $f$ and $A,B,C$, under which the identity above holds.
information-theory
asked Mar 28 at 7:20
hklelhklel
1255
$endgroup$
1
$begingroup$
The function needs to be injective with respect to its second argument.
$endgroup$
– Yuval Filmus
Mar 28 at 8:04
$begingroup$
@YuvalFilmus Ah that makes sense! I didn't know the term "injective". Do you want to elaborate a bit and write an answer so I can upvote it?
$endgroup$
– hklel
Mar 28 at 8:11
add a comment |
$begingroup$
Suppose we have a function $f$,
$$
C = f(A,B),
$$
where $A$, $B$ and $C$ are random variables.
I notice that when the random variables are binary ($0, 1$) and $f$ is the XOR operation, we have the following identity:
$$
H(C|A) = H(B),
$$
where $H(B)$ is the entropy of $B$ and $H(C|A)$ is the conditional entropy of $C$ given $A$.
Obviously this is not true for a general $f$. What I am interested to know is, is there a set of conditions on $f$ and $A,B,C$, under which the identity above holds.
information-theory
asked Mar 28 at 7:20
hklelhklel
1255
$endgroup$
Suppose we have a function $f$,
$$
C = f(A,B),
$$
where $A$, $B$ and $C$ are random variables.
I notice that when the random variables are binary ($0, 1$) and $f$ is the XOR operation, we have the following identity:
$$
H(C|A) = H(B),
$$
where $H(B)$ is the entropy of $B$ and $H(C|A)$ is the conditional entropy of $C$ given $A$.
Obviously this is not true for a general $f$. What I am interested to know is, is there a set of conditions on $f$ and $A,B,C$, under which the identity above holds.
information-theory
information-theory
asked Mar 28 at 7:20
hklelhklel
1255
asked Mar 28 at 7:20
hklelhklel
1255
edited Apr 1 at 5:51
hklel
asked Mar 28 at 7:20
hklelhklel
1255
asked Mar 28 at 7:20
hklelhklel
1255
asked Mar 28 at 7:20
hklelhklel
1255
1255
1
$begingroup$
The function needs to be injective with respect to its second argument.
$endgroup$
– Yuval Filmus
Mar 28 at 8:04
$begingroup$
@YuvalFilmus Ah that makes sense! I didn't know the term "injective". Do you want to elaborate a bit and write an answer so I can upvote it?
$endgroup$
– hklel
Mar 28 at 8:11
add a comment |
1
$begingroup$
The function needs to be injective with respect to its second argument.
$endgroup$
– Yuval Filmus
Mar 28 at 8:04
$begingroup$
@YuvalFilmus Ah that makes sense! I didn't know the term "injective". Do you want to elaborate a bit and write an answer so I can upvote it?
$endgroup$
– hklel
Mar 28 at 8:11
1
1
$begingroup$
The function needs to be injective with respect to its second argument.
$endgroup$
– Yuval Filmus
Mar 28 at 8:04
$begingroup$
The function needs to be injective with respect to its second argument.
$endgroup$
– Yuval Filmus
Mar 28 at 8:04
$begingroup$
@YuvalFilmus Ah that makes sense! I didn't know the term "injective". Do you want to elaborate a bit and write an answer so I can upvote it?
$endgroup$
– hklel
Mar 28 at 8:11
$begingroup$
@YuvalFilmus Ah that makes sense! I didn't know the term "injective". Do you want to elaborate a bit and write an answer so I can upvote it?
$endgroup$
– hklel
Mar 28 at 8:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The following answer assumes that $A,B$ are independent, and that $A,B$ have full support on their respective domains (the latter is without loss of generality). For the general case, see the other answer.
Let's write your equation in a slightly different way:
$$
H(B) = H(f(A,B)|A) = operatorname*mathbbE_a sim A H(f(a,B)).
$$
Clearly $H(f(a,B)) leq H(B)$, with equality if and only if $f(a,b_1) neq f(a,b_2)$ whenever $b_1 neq b_2$. We deduce that $H(B) = H(f(A,B)|A)$ if and only if $f$ is injective in its second argument, i.e., for all $a$ and $b_1 neq b_2$, we have $f(a,b_1) neq f(a,b_2)$.
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
1
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
3
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
add a comment |
$begingroup$
Note
beginalign
0&=H(C|A,B)\
&=H(A,B,C)-H(A,B)\
&=H(B|A,C)+H(C|A)+H(A)-H(A,B)quadtext(chain rule)\
&=H(B|A,C)+H(C|A)-H(B|A),
endalign
so $H(C|A)=H(B)$ is equivalently $H(B|A,C)+H(B)-H(B|A)=0$. Also note $H(B|A,C)ge 0$ and $H(B)ge H(B|A)$, your condition is equivalently $H(B|A,C)=0wedge H(B)=H(B|A)$.
For a human-readable explanation, $H(B|A,C)=0$ means $B$ is determined by $A$ and $C$, that is, for any fixed $a$ in the support of $A$, $f(a,b)$ as a function of $b$ with domain $bmid mathrmPrA=a, B=b>0$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
$endgroup$
2
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04
$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The following answer assumes that $A,B$ are independent, and that $A,B$ have full support on their respective domains (the latter is without loss of generality). For the general case, see the other answer.
Let's write your equation in a slightly different way:
$$
H(B) = H(f(A,B)|A) = operatorname*mathbbE_a sim A H(f(a,B)).
$$
Clearly $H(f(a,B)) leq H(B)$, with equality if and only if $f(a,b_1) neq f(a,b_2)$ whenever $b_1 neq b_2$. We deduce that $H(B) = H(f(A,B)|A)$ if and only if $f$ is injective in its second argument, i.e., for all $a$ and $b_1 neq b_2$, we have $f(a,b_1) neq f(a,b_2)$.
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
1
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
3
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
add a comment |
$begingroup$
The following answer assumes that $A,B$ are independent, and that $A,B$ have full support on their respective domains (the latter is without loss of generality). For the general case, see the other answer.
Let's write your equation in a slightly different way:
$$
H(B) = H(f(A,B)|A) = operatorname*mathbbE_a sim A H(f(a,B)).
$$
Clearly $H(f(a,B)) leq H(B)$, with equality if and only if $f(a,b_1) neq f(a,b_2)$ whenever $b_1 neq b_2$. We deduce that $H(B) = H(f(A,B)|A)$ if and only if $f$ is injective in its second argument, i.e., for all $a$ and $b_1 neq b_2$, we have $f(a,b_1) neq f(a,b_2)$.
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
1
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
3
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
add a comment |
$begingroup$
The following answer assumes that $A,B$ are independent, and that $A,B$ have full support on their respective domains (the latter is without loss of generality). For the general case, see the other answer.
Let's write your equation in a slightly different way:
$$
H(B) = H(f(A,B)|A) = operatorname*mathbbE_a sim A H(f(a,B)).
$$
Clearly $H(f(a,B)) leq H(B)$, with equality if and only if $f(a,b_1) neq f(a,b_2)$ whenever $b_1 neq b_2$. We deduce that $H(B) = H(f(A,B)|A)$ if and only if $f$ is injective in its second argument, i.e., for all $a$ and $b_1 neq b_2$, we have $f(a,b_1) neq f(a,b_2)$.
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
The following answer assumes that $A,B$ are independent, and that $A,B$ have full support on their respective domains (the latter is without loss of generality). For the general case, see the other answer.
Let's write your equation in a slightly different way:
$$
H(B) = H(f(A,B)|A) = operatorname*mathbbE_a sim A H(f(a,B)).
$$
Clearly $H(f(a,B)) leq H(B)$, with equality if and only if $f(a,b_1) neq f(a,b_2)$ whenever $b_1 neq b_2$. We deduce that $H(B) = H(f(A,B)|A)$ if and only if $f$ is injective in its second argument, i.e., for all $a$ and $b_1 neq b_2$, we have $f(a,b_1) neq f(a,b_2)$.
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
edited Mar 28 at 10:12
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
answered Mar 28 at 8:28
Yuval FilmusYuval Filmus
196k15184349
196k15184349
1
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
3
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
add a comment |
1
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
3
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
1
1
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
$begingroup$
$H(f(A,B)|A)=mathbbE_aH(f(a,B)|A=a)$, and $H(f(a,B)|A=a)$ is different from $H(f(a,B))$ since $A$ and $B$ may be dependent.
$endgroup$
– xskxzr
Mar 28 at 8:45
3
3
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03
add a comment |
$begingroup$
Note
beginalign
0&=H(C|A,B)\
&=H(A,B,C)-H(A,B)\
&=H(B|A,C)+H(C|A)+H(A)-H(A,B)quadtext(chain rule)\
&=H(B|A,C)+H(C|A)-H(B|A),
endalign
so $H(C|A)=H(B)$ is equivalently $H(B|A,C)+H(B)-H(B|A)=0$. Also note $H(B|A,C)ge 0$ and $H(B)ge H(B|A)$, your condition is equivalently $H(B|A,C)=0wedge H(B)=H(B|A)$.
For a human-readable explanation, $H(B|A,C)=0$ means $B$ is determined by $A$ and $C$, that is, for any fixed $a$ in the support of $A$, $f(a,b)$ as a function of $b$ with domain $bmid mathrmPrA=a, B=b>0$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
$endgroup$
2
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04
$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34
add a comment |
$begingroup$
Note
beginalign
0&=H(C|A,B)\
&=H(A,B,C)-H(A,B)\
&=H(B|A,C)+H(C|A)+H(A)-H(A,B)quadtext(chain rule)\
&=H(B|A,C)+H(C|A)-H(B|A),
endalign
so $H(C|A)=H(B)$ is equivalently $H(B|A,C)+H(B)-H(B|A)=0$. Also note $H(B|A,C)ge 0$ and $H(B)ge H(B|A)$, your condition is equivalently $H(B|A,C)=0wedge H(B)=H(B|A)$.
For a human-readable explanation, $H(B|A,C)=0$ means $B$ is determined by $A$ and $C$, that is, for any fixed $a$ in the support of $A$, $f(a,b)$ as a function of $b$ with domain $bmid mathrmPrA=a, B=b>0$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
$endgroup$
2
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04
$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34
add a comment |
$begingroup$
Note
beginalign
0&=H(C|A,B)\
&=H(A,B,C)-H(A,B)\
&=H(B|A,C)+H(C|A)+H(A)-H(A,B)quadtext(chain rule)\
&=H(B|A,C)+H(C|A)-H(B|A),
endalign
so $H(C|A)=H(B)$ is equivalently $H(B|A,C)+H(B)-H(B|A)=0$. Also note $H(B|A,C)ge 0$ and $H(B)ge H(B|A)$, your condition is equivalently $H(B|A,C)=0wedge H(B)=H(B|A)$.
For a human-readable explanation, $H(B|A,C)=0$ means $B$ is determined by $A$ and $C$, that is, for any fixed $a$ in the support of $A$, $f(a,b)$ as a function of $b$ with domain $bmid mathrmPrA=a, B=b>0$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
$endgroup$
Note
beginalign
0&=H(C|A,B)\
&=H(A,B,C)-H(A,B)\
&=H(B|A,C)+H(C|A)+H(A)-H(A,B)quadtext(chain rule)\
&=H(B|A,C)+H(C|A)-H(B|A),
endalign
so $H(C|A)=H(B)$ is equivalently $H(B|A,C)+H(B)-H(B|A)=0$. Also note $H(B|A,C)ge 0$ and $H(B)ge H(B|A)$, your condition is equivalently $H(B|A,C)=0wedge H(B)=H(B|A)$.
For a human-readable explanation, $H(B|A,C)=0$ means $B$ is determined by $A$ and $C$, that is, for any fixed $a$ in the support of $A$, $f(a,b)$ as a function of $b$ with domain $bmid mathrmPrA=a, B=b>0$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
edited Mar 28 at 10:34
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
answered Mar 28 at 8:38
xskxzrxskxzr
4,18921033
4,18921033
2
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04
$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34
add a comment |
2
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04
$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34
2
2
$begingroup$
The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
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– Emil Jeřábek
Mar 28 at 10:04
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The conclusion that $f$ is injective in the second argument is only correct if $Pr(A=a)>0$ and $Pr(B=b)>0$ for all $(a,b)inoperatornamedom(f)$.
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– Emil Jeřábek
Mar 28 at 10:04
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@EmilJeřábek Thanks, fixed.
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– xskxzr
Mar 28 at 10:34
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@EmilJeřábek Thanks, fixed.
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– xskxzr
Mar 28 at 10:34
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The function needs to be injective with respect to its second argument.
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– Yuval Filmus
Mar 28 at 8:04
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@YuvalFilmus Ah that makes sense! I didn't know the term "injective". Do you want to elaborate a bit and write an answer so I can upvote it?
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– hklel
Mar 28 at 8:11