Mutual information: Proving $I(X;Y)=underset{r(x),s(y)}{min}D_{KL}[p(x,y)midmid r(x)cdot s(y)]$












1












$begingroup$


I was asked to show that for two discrete RVs $X,Y$ with joint distribution $pleft(x,yright)$ and marginal distributions $pleft(xright)$ and $pleft(yright)$, it holds that for any two distributions $rleft(xright)$ and $sleft(yright)$ it holds that



$$Ileft(X;Yright)=underset{rleft(xright),sleft(yright)}{min}D_{KL}left[pleft(x,yright)midmid rleft(xright)cdot sleft(yright)right]$$



Now,



$$Ileft(X;Yright)=D_{KL}left[pleft(x,yright)midmid pleft(xright)cdot pleft(yright)right]=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



so this amounts to showing $$underset{rleft(xright),sleft(yright)}{min}sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{rleft(xright)sleft(yright)}=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



but how does one show this? Do I need to differentiate w.r.t. $rleft(xright)$ and $sleft(yright)$? Is that possible?






probability-theory derivatives information-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    This amounts to showing that for every pair of distributions $r,s$ it holds that $$ sum_{x,y} p_{xy} log frac{p_{x}p_{y}}{r_x s_y} ge 0.$$ Breaking the log immediately yields that the LHS is $D(p_X|r) + D(p_Y|s)$ and the inequality is then obvious, as is the equality condition.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 6:11










  • $begingroup$
    @stochasticboy321 I don't follow, how does this inequality yield the requested result?
    $endgroup$
    – H.Rappeport
    Dec 18 '18 at 10:18










  • $begingroup$
    $$D(p_{XY}|r_X s_Y) - D(p_{XY}|p_Xp_Y) = sum p_{xy} log frac{p_{xy}}{r_x s_y} - sum p_{xy} log frac{p_{xy}}{p_x p_y}$$ equals the expression in the comment. So, the inequality shows that $D(p_{XY}|r_X s_Y) ge I(X;Y)$ for every $(r,s).$ I suppose you do have to show that $I$ is the minimum, but this is trivial - just set $r,s$ to be the appropriate marginals.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 20:30










  • $begingroup$
    @stochasticboy321 I see, thanks! Do you want to post this as an answer? I'd be happy to accept.
    $endgroup$
    – H.Rappeport
    Dec 19 '18 at 20:13


















1












$begingroup$


I was asked to show that for two discrete RVs $X,Y$ with joint distribution $pleft(x,yright)$ and marginal distributions $pleft(xright)$ and $pleft(yright)$, it holds that for any two distributions $rleft(xright)$ and $sleft(yright)$ it holds that



$$Ileft(X;Yright)=underset{rleft(xright),sleft(yright)}{min}D_{KL}left[pleft(x,yright)midmid rleft(xright)cdot sleft(yright)right]$$



Now,



$$Ileft(X;Yright)=D_{KL}left[pleft(x,yright)midmid pleft(xright)cdot pleft(yright)right]=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



so this amounts to showing $$underset{rleft(xright),sleft(yright)}{min}sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{rleft(xright)sleft(yright)}=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



but how does one show this? Do I need to differentiate w.r.t. $rleft(xright)$ and $sleft(yright)$? Is that possible?






probability-theory derivatives information-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    This amounts to showing that for every pair of distributions $r,s$ it holds that $$ sum_{x,y} p_{xy} log frac{p_{x}p_{y}}{r_x s_y} ge 0.$$ Breaking the log immediately yields that the LHS is $D(p_X|r) + D(p_Y|s)$ and the inequality is then obvious, as is the equality condition.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 6:11










  • $begingroup$
    @stochasticboy321 I don't follow, how does this inequality yield the requested result?
    $endgroup$
    – H.Rappeport
    Dec 18 '18 at 10:18










  • $begingroup$
    $$D(p_{XY}|r_X s_Y) - D(p_{XY}|p_Xp_Y) = sum p_{xy} log frac{p_{xy}}{r_x s_y} - sum p_{xy} log frac{p_{xy}}{p_x p_y}$$ equals the expression in the comment. So, the inequality shows that $D(p_{XY}|r_X s_Y) ge I(X;Y)$ for every $(r,s).$ I suppose you do have to show that $I$ is the minimum, but this is trivial - just set $r,s$ to be the appropriate marginals.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 20:30










  • $begingroup$
    @stochasticboy321 I see, thanks! Do you want to post this as an answer? I'd be happy to accept.
    $endgroup$
    – H.Rappeport
    Dec 19 '18 at 20:13
















1












1








1





$begingroup$


I was asked to show that for two discrete RVs $X,Y$ with joint distribution $pleft(x,yright)$ and marginal distributions $pleft(xright)$ and $pleft(yright)$, it holds that for any two distributions $rleft(xright)$ and $sleft(yright)$ it holds that



$$Ileft(X;Yright)=underset{rleft(xright),sleft(yright)}{min}D_{KL}left[pleft(x,yright)midmid rleft(xright)cdot sleft(yright)right]$$



Now,



$$Ileft(X;Yright)=D_{KL}left[pleft(x,yright)midmid pleft(xright)cdot pleft(yright)right]=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



so this amounts to showing $$underset{rleft(xright),sleft(yright)}{min}sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{rleft(xright)sleft(yright)}=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



but how does one show this? Do I need to differentiate w.r.t. $rleft(xright)$ and $sleft(yright)$? Is that possible?






probability-theory derivatives information-theory







share|cite|improve this question











$endgroup$




I was asked to show that for two discrete RVs $X,Y$ with joint distribution $pleft(x,yright)$ and marginal distributions $pleft(xright)$ and $pleft(yright)$, it holds that for any two distributions $rleft(xright)$ and $sleft(yright)$ it holds that



$$Ileft(X;Yright)=underset{rleft(xright),sleft(yright)}{min}D_{KL}left[pleft(x,yright)midmid rleft(xright)cdot sleft(yright)right]$$



Now,



$$Ileft(X;Yright)=D_{KL}left[pleft(x,yright)midmid pleft(xright)cdot pleft(yright)right]=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



so this amounts to showing $$underset{rleft(xright),sleft(yright)}{min}sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{rleft(xright)sleft(yright)}=sum_{x,y}pleft(x,yright)logfrac{pleft(x,yright)}{pleft(xright)pleft(yright)}$$



but how does one show this? Do I need to differentiate w.r.t. $rleft(xright)$ and $sleft(yright)$? Is that possible?






probability-theory derivatives information-theory




probability-theory derivatives information-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 21:32







H.Rappeport

















asked Dec 17 '18 at 13:46









H.RappeportH.Rappeport

6801510




6801510












  • $begingroup$
    This amounts to showing that for every pair of distributions $r,s$ it holds that $$ sum_{x,y} p_{xy} log frac{p_{x}p_{y}}{r_x s_y} ge 0.$$ Breaking the log immediately yields that the LHS is $D(p_X|r) + D(p_Y|s)$ and the inequality is then obvious, as is the equality condition.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 6:11










  • $begingroup$
    @stochasticboy321 I don't follow, how does this inequality yield the requested result?
    $endgroup$
    – H.Rappeport
    Dec 18 '18 at 10:18










  • $begingroup$
    $$D(p_{XY}|r_X s_Y) - D(p_{XY}|p_Xp_Y) = sum p_{xy} log frac{p_{xy}}{r_x s_y} - sum p_{xy} log frac{p_{xy}}{p_x p_y}$$ equals the expression in the comment. So, the inequality shows that $D(p_{XY}|r_X s_Y) ge I(X;Y)$ for every $(r,s).$ I suppose you do have to show that $I$ is the minimum, but this is trivial - just set $r,s$ to be the appropriate marginals.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 20:30










  • $begingroup$
    @stochasticboy321 I see, thanks! Do you want to post this as an answer? I'd be happy to accept.
    $endgroup$
    – H.Rappeport
    Dec 19 '18 at 20:13




















  • $begingroup$
    This amounts to showing that for every pair of distributions $r,s$ it holds that $$ sum_{x,y} p_{xy} log frac{p_{x}p_{y}}{r_x s_y} ge 0.$$ Breaking the log immediately yields that the LHS is $D(p_X|r) + D(p_Y|s)$ and the inequality is then obvious, as is the equality condition.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 6:11










  • $begingroup$
    @stochasticboy321 I don't follow, how does this inequality yield the requested result?
    $endgroup$
    – H.Rappeport
    Dec 18 '18 at 10:18










  • $begingroup$
    $$D(p_{XY}|r_X s_Y) - D(p_{XY}|p_Xp_Y) = sum p_{xy} log frac{p_{xy}}{r_x s_y} - sum p_{xy} log frac{p_{xy}}{p_x p_y}$$ equals the expression in the comment. So, the inequality shows that $D(p_{XY}|r_X s_Y) ge I(X;Y)$ for every $(r,s).$ I suppose you do have to show that $I$ is the minimum, but this is trivial - just set $r,s$ to be the appropriate marginals.
    $endgroup$
    – stochasticboy321
    Dec 18 '18 at 20:30










  • $begingroup$
    @stochasticboy321 I see, thanks! Do you want to post this as an answer? I'd be happy to accept.
    $endgroup$
    – H.Rappeport
    Dec 19 '18 at 20:13


















$begingroup$
This amounts to showing that for every pair of distributions $r,s$ it holds that $$ sum_{x,y} p_{xy} log frac{p_{x}p_{y}}{r_x s_y} ge 0.$$ Breaking the log immediately yields that the LHS is $D(p_X|r) + D(p_Y|s)$ and the inequality is then obvious, as is the equality condition.
$endgroup$
– stochasticboy321
Dec 18 '18 at 6:11




$begingroup$
This amounts to showing that for every pair of distributions $r,s$ it holds that $$ sum_{x,y} p_{xy} log frac{p_{x}p_{y}}{r_x s_y} ge 0.$$ Breaking the log immediately yields that the LHS is $D(p_X|r) + D(p_Y|s)$ and the inequality is then obvious, as is the equality condition.
$endgroup$
– stochasticboy321
Dec 18 '18 at 6:11












$begingroup$
@stochasticboy321 I don't follow, how does this inequality yield the requested result?
$endgroup$
– H.Rappeport
Dec 18 '18 at 10:18




$begingroup$
@stochasticboy321 I don't follow, how does this inequality yield the requested result?
$endgroup$
– H.Rappeport
Dec 18 '18 at 10:18












$begingroup$
$$D(p_{XY}|r_X s_Y) - D(p_{XY}|p_Xp_Y) = sum p_{xy} log frac{p_{xy}}{r_x s_y} - sum p_{xy} log frac{p_{xy}}{p_x p_y}$$ equals the expression in the comment. So, the inequality shows that $D(p_{XY}|r_X s_Y) ge I(X;Y)$ for every $(r,s).$ I suppose you do have to show that $I$ is the minimum, but this is trivial - just set $r,s$ to be the appropriate marginals.
$endgroup$
– stochasticboy321
Dec 18 '18 at 20:30




$begingroup$
$$D(p_{XY}|r_X s_Y) - D(p_{XY}|p_Xp_Y) = sum p_{xy} log frac{p_{xy}}{r_x s_y} - sum p_{xy} log frac{p_{xy}}{p_x p_y}$$ equals the expression in the comment. So, the inequality shows that $D(p_{XY}|r_X s_Y) ge I(X;Y)$ for every $(r,s).$ I suppose you do have to show that $I$ is the minimum, but this is trivial - just set $r,s$ to be the appropriate marginals.
$endgroup$
– stochasticboy321
Dec 18 '18 at 20:30












$begingroup$
@stochasticboy321 I see, thanks! Do you want to post this as an answer? I'd be happy to accept.
$endgroup$
– H.Rappeport
Dec 19 '18 at 20:13






$begingroup$
@stochasticboy321 I see, thanks! Do you want to post this as an answer? I'd be happy to accept.
$endgroup$
– H.Rappeport
Dec 19 '18 at 20:13












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