Relative entropy is non-negative












6












$begingroup$


Let $p=(p_1,dotsc,p_r), q=(q_1,dotsc,q_r)$ be two different probability distributions. Define the relative entropy $$h(p||q) = sum_{i=1}^r p_i (ln p_i - ln q_i)$$ Show $h(p||q)geq 0$. I'm given the hint that I should show $-xln x$ is concave and then show for any concave function $f(y)-f(x)leq (y-x)f'(x)$ holds. I rewritten the relative entropy as $$h(p||q)=sum_{i=1}^r p_i ln left(frac{p_i}{q_i}right)= -sum_{i=1}^r p_i ln left(frac{q_i}{p_i}right)$$ which sort of looks like $-xln x$, and I did show that $-xln x$ is concave, but I don't really understand what I'm supposed to do, or even if this hint is helpful.






probability







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$endgroup$








  • 1




    $begingroup$
    This is called Gibb's inequality.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:24
















6












$begingroup$


Let $p=(p_1,dotsc,p_r), q=(q_1,dotsc,q_r)$ be two different probability distributions. Define the relative entropy $$h(p||q) = sum_{i=1}^r p_i (ln p_i - ln q_i)$$ Show $h(p||q)geq 0$. I'm given the hint that I should show $-xln x$ is concave and then show for any concave function $f(y)-f(x)leq (y-x)f'(x)$ holds. I rewritten the relative entropy as $$h(p||q)=sum_{i=1}^r p_i ln left(frac{p_i}{q_i}right)= -sum_{i=1}^r p_i ln left(frac{q_i}{p_i}right)$$ which sort of looks like $-xln x$, and I did show that $-xln x$ is concave, but I don't really understand what I'm supposed to do, or even if this hint is helpful.






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This is called Gibb's inequality.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:24














6












6








6


4



$begingroup$


Let $p=(p_1,dotsc,p_r), q=(q_1,dotsc,q_r)$ be two different probability distributions. Define the relative entropy $$h(p||q) = sum_{i=1}^r p_i (ln p_i - ln q_i)$$ Show $h(p||q)geq 0$. I'm given the hint that I should show $-xln x$ is concave and then show for any concave function $f(y)-f(x)leq (y-x)f'(x)$ holds. I rewritten the relative entropy as $$h(p||q)=sum_{i=1}^r p_i ln left(frac{p_i}{q_i}right)= -sum_{i=1}^r p_i ln left(frac{q_i}{p_i}right)$$ which sort of looks like $-xln x$, and I did show that $-xln x$ is concave, but I don't really understand what I'm supposed to do, or even if this hint is helpful.






probability







share|cite|improve this question









$endgroup$




Let $p=(p_1,dotsc,p_r), q=(q_1,dotsc,q_r)$ be two different probability distributions. Define the relative entropy $$h(p||q) = sum_{i=1}^r p_i (ln p_i - ln q_i)$$ Show $h(p||q)geq 0$. I'm given the hint that I should show $-xln x$ is concave and then show for any concave function $f(y)-f(x)leq (y-x)f'(x)$ holds. I rewritten the relative entropy as $$h(p||q)=sum_{i=1}^r p_i ln left(frac{p_i}{q_i}right)= -sum_{i=1}^r p_i ln left(frac{q_i}{p_i}right)$$ which sort of looks like $-xln x$, and I did show that $-xln x$ is concave, but I don't really understand what I'm supposed to do, or even if this hint is helpful.






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 4 '11 at 20:15









anonanon

5514




5514








  • 1




    $begingroup$
    This is called Gibb's inequality.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:24














  • 1




    $begingroup$
    This is called Gibb's inequality.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:24








1




1




$begingroup$
This is called Gibb's inequality.
$endgroup$
– Srivatsan
Oct 4 '11 at 20:24




$begingroup$
This is called Gibb's inequality.
$endgroup$
– Srivatsan
Oct 4 '11 at 20:24










2 Answers
2






active

oldest

votes


















6












$begingroup$

Assume that the random variable $X$ is such that $X=dfrac{p_i}{q_i}$ with probability $q_i$, for every $i$. Then,
$$
h(pmidmid q)=sumlimits_iq_ifrac{p_i}{q_i}lnleft(frac{p_i}{q_i}right)=mathrm E(Xln X).
$$
Since the function $xmapsto xln x$ is convex, $mathrm E(Xln X)geqslant mathrm E(X)lnmathrm E(X)$ by Jensen inequality.



To complete the proof one must simply compute $mathrm E(X)$, and I will let you do that.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:35








  • 1




    $begingroup$
    @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
    $endgroup$
    – Did
    Oct 4 '11 at 20:43





















0












$begingroup$

I think the proof in the textbook is an awesome one:



$D(p||q) = sum p(x) log frac{p(x)}{q(x)}
= -sum p(x) log frac{q(x)}{p(x)} le -sum log p(x) frac {q(x)}{p(x)} = -sum log q(x) = 0$



Then $D(p||q) ge 0$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    6












    $begingroup$

    Assume that the random variable $X$ is such that $X=dfrac{p_i}{q_i}$ with probability $q_i$, for every $i$. Then,
    $$
    h(pmidmid q)=sumlimits_iq_ifrac{p_i}{q_i}lnleft(frac{p_i}{q_i}right)=mathrm E(Xln X).
    $$
    Since the function $xmapsto xln x$ is convex, $mathrm E(Xln X)geqslant mathrm E(X)lnmathrm E(X)$ by Jensen inequality.



    To complete the proof one must simply compute $mathrm E(X)$, and I will let you do that.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
      $endgroup$
      – Srivatsan
      Oct 4 '11 at 20:35








    • 1




      $begingroup$
      @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
      $endgroup$
      – Did
      Oct 4 '11 at 20:43


















    6












    $begingroup$

    Assume that the random variable $X$ is such that $X=dfrac{p_i}{q_i}$ with probability $q_i$, for every $i$. Then,
    $$
    h(pmidmid q)=sumlimits_iq_ifrac{p_i}{q_i}lnleft(frac{p_i}{q_i}right)=mathrm E(Xln X).
    $$
    Since the function $xmapsto xln x$ is convex, $mathrm E(Xln X)geqslant mathrm E(X)lnmathrm E(X)$ by Jensen inequality.



    To complete the proof one must simply compute $mathrm E(X)$, and I will let you do that.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
      $endgroup$
      – Srivatsan
      Oct 4 '11 at 20:35








    • 1




      $begingroup$
      @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
      $endgroup$
      – Did
      Oct 4 '11 at 20:43
















    6












    6








    6





    $begingroup$

    Assume that the random variable $X$ is such that $X=dfrac{p_i}{q_i}$ with probability $q_i$, for every $i$. Then,
    $$
    h(pmidmid q)=sumlimits_iq_ifrac{p_i}{q_i}lnleft(frac{p_i}{q_i}right)=mathrm E(Xln X).
    $$
    Since the function $xmapsto xln x$ is convex, $mathrm E(Xln X)geqslant mathrm E(X)lnmathrm E(X)$ by Jensen inequality.



    To complete the proof one must simply compute $mathrm E(X)$, and I will let you do that.






    share|cite|improve this answer









    $endgroup$



    Assume that the random variable $X$ is such that $X=dfrac{p_i}{q_i}$ with probability $q_i$, for every $i$. Then,
    $$
    h(pmidmid q)=sumlimits_iq_ifrac{p_i}{q_i}lnleft(frac{p_i}{q_i}right)=mathrm E(Xln X).
    $$
    Since the function $xmapsto xln x$ is convex, $mathrm E(Xln X)geqslant mathrm E(X)lnmathrm E(X)$ by Jensen inequality.



    To complete the proof one must simply compute $mathrm E(X)$, and I will let you do that.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 4 '11 at 20:27









    DidDid

    248k23225463




    248k23225463








    • 2




      $begingroup$
      @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
      $endgroup$
      – Srivatsan
      Oct 4 '11 at 20:35








    • 1




      $begingroup$
      @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
      $endgroup$
      – Did
      Oct 4 '11 at 20:43
















    • 2




      $begingroup$
      @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
      $endgroup$
      – Srivatsan
      Oct 4 '11 at 20:35








    • 1




      $begingroup$
      @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
      $endgroup$
      – Did
      Oct 4 '11 at 20:43










    2




    2




    $begingroup$
    @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:35






    $begingroup$
    @anon There's another, but similar, way. Show that $x mapsto ln x$ is concave, so that $mathbf E[ln Y] leq ln (mathbf E[Y])$. Now, let $Y$ be a random variable that takes the value $(q_i/p_i)$ with probability $p_i$. Apply Jensen to $Y$.
    $endgroup$
    – Srivatsan
    Oct 4 '11 at 20:35






    1




    1




    $begingroup$
    @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
    $endgroup$
    – Did
    Oct 4 '11 at 20:43






    $begingroup$
    @Srivatsan, thanks, this might be considered as a (slightly...) simpler approach.
    $endgroup$
    – Did
    Oct 4 '11 at 20:43













    0












    $begingroup$

    I think the proof in the textbook is an awesome one:



    $D(p||q) = sum p(x) log frac{p(x)}{q(x)}
    = -sum p(x) log frac{q(x)}{p(x)} le -sum log p(x) frac {q(x)}{p(x)} = -sum log q(x) = 0$



    Then $D(p||q) ge 0$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      I think the proof in the textbook is an awesome one:



      $D(p||q) = sum p(x) log frac{p(x)}{q(x)}
      = -sum p(x) log frac{q(x)}{p(x)} le -sum log p(x) frac {q(x)}{p(x)} = -sum log q(x) = 0$



      Then $D(p||q) ge 0$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        I think the proof in the textbook is an awesome one:



        $D(p||q) = sum p(x) log frac{p(x)}{q(x)}
        = -sum p(x) log frac{q(x)}{p(x)} le -sum log p(x) frac {q(x)}{p(x)} = -sum log q(x) = 0$



        Then $D(p||q) ge 0$






        share|cite|improve this answer











        $endgroup$



        I think the proof in the textbook is an awesome one:



        $D(p||q) = sum p(x) log frac{p(x)}{q(x)}
        = -sum p(x) log frac{q(x)}{p(x)} le -sum log p(x) frac {q(x)}{p(x)} = -sum log q(x) = 0$



        Then $D(p||q) ge 0$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 7:32

























        answered Dec 16 '18 at 6:53









        Lerner ZhangLerner Zhang

        314217




        314217






























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