Under what conditions does the function C = f(A,B) satisfy H(C|A) = H(B)?












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







hklel

















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














  • 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










2 Answers
2






active

oldest

votes


















4












$begingroup$

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:
$$
H(B) = H(f(A,B)|A) = operatorname*{mathbb{E}}_{a sim A} H(f(a,B)).
$$
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 $f(a,b_1) neq f(a,b_2)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:03





















8












$begingroup$

Note



begin{align}
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),
end{align}



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 ${bmid mathrm{Pr}{A=a, B=b}>0}$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.






share|cite|improve this answer











$endgroup$









  • 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:04










  • $begingroup$
    @EmilJeřábek Thanks, fixed.
    $endgroup$
    – xskxzr
    Mar 28 at 10:34












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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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:
$$
H(B) = H(f(A,B)|A) = operatorname*{mathbb{E}}_{a sim A} H(f(a,B)).
$$
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 $f(a,b_1) neq f(a,b_2)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:03


















4












$begingroup$

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:
$$
H(B) = H(f(A,B)|A) = operatorname*{mathbb{E}}_{a sim A} H(f(a,B)).
$$
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 $f(a,b_1) neq f(a,b_2)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    $H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:03
















4












4








4





$begingroup$

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:
$$
H(B) = H(f(A,B)|A) = operatorname*{mathbb{E}}_{a sim A} H(f(a,B)).
$$
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 $f(a,b_1) neq f(a,b_2)$.






share|cite|improve this answer











$endgroup$



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:
$$
H(B) = H(f(A,B)|A) = operatorname*{mathbb{E}}_{a sim A} H(f(a,B)).
$$
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 $f(a,b_1) neq f(a,b_2)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 28 at 10:12

























answered Mar 28 at 8:28









Yuval FilmusYuval Filmus

195k15184349




195k15184349








  • 1




    $begingroup$
    $H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:03
















  • 1




    $begingroup$
    $H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:03










1




1




$begingroup$
$H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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






$begingroup$
$H(f(A,B)|A)=mathbb{E}_aH(f(a,B)|A=a)$, 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 $(a,b)inoperatorname{dom}(f)$.
$endgroup$
– 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 $(a,b)inoperatorname{dom}(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:03













8












$begingroup$

Note



begin{align}
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),
end{align}



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 ${bmid mathrm{Pr}{A=a, B=b}>0}$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.






share|cite|improve this answer











$endgroup$









  • 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:04










  • $begingroup$
    @EmilJeřábek Thanks, fixed.
    $endgroup$
    – xskxzr
    Mar 28 at 10:34
















8












$begingroup$

Note



begin{align}
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),
end{align}



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 ${bmid mathrm{Pr}{A=a, B=b}>0}$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.






share|cite|improve this answer











$endgroup$









  • 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:04










  • $begingroup$
    @EmilJeřábek Thanks, fixed.
    $endgroup$
    – xskxzr
    Mar 28 at 10:34














8












8








8





$begingroup$

Note



begin{align}
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),
end{align}



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 ${bmid mathrm{Pr}{A=a, B=b}>0}$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.






share|cite|improve this answer











$endgroup$



Note



begin{align}
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),
end{align}



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 ${bmid mathrm{Pr}{A=a, B=b}>0}$ is an injection. $H(B)=H(B|A)$ means $A$ and $B$ are independent of each other.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 28 at 10:34

























answered Mar 28 at 8:38









xskxzrxskxzr

4,07921033




4,07921033








  • 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – Emil Jeřábek
    Mar 28 at 10:04










  • $begingroup$
    @EmilJeřábek Thanks, fixed.
    $endgroup$
    – xskxzr
    Mar 28 at 10:34














  • 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 $(a,b)inoperatorname{dom}(f)$.
    $endgroup$
    – 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 $(a,b)inoperatorname{dom}(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04




$begingroup$
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 $(a,b)inoperatorname{dom}(f)$.
$endgroup$
– Emil Jeřábek
Mar 28 at 10:04












$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34




$begingroup$
@EmilJeřábek Thanks, fixed.
$endgroup$
– xskxzr
Mar 28 at 10:34


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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