Proving Pinsker's inequality











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I wanna show that for $a$ and $b >0$ and $<1$. Then
$$2(a-b)^2leq (a) log (a) + (1-a)log(1-a) -(a)log(b)-(1-a)log(1-b).$$
If I wanna go from left to right, then I think I gotta use Taylor's but I don't know how. Also, the right hand side I figured out is equal to
$$(a)logfrac ab+(1-a)logfrac{1-a}{1-b}.$$










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    I wanna show that for $a$ and $b >0$ and $<1$. Then
    $$2(a-b)^2leq (a) log (a) + (1-a)log(1-a) -(a)log(b)-(1-a)log(1-b).$$
    If I wanna go from left to right, then I think I gotta use Taylor's but I don't know how. Also, the right hand side I figured out is equal to
    $$(a)logfrac ab+(1-a)logfrac{1-a}{1-b}.$$










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I wanna show that for $a$ and $b >0$ and $<1$. Then
      $$2(a-b)^2leq (a) log (a) + (1-a)log(1-a) -(a)log(b)-(1-a)log(1-b).$$
      If I wanna go from left to right, then I think I gotta use Taylor's but I don't know how. Also, the right hand side I figured out is equal to
      $$(a)logfrac ab+(1-a)logfrac{1-a}{1-b}.$$










      share|cite|improve this question















      I wanna show that for $a$ and $b >0$ and $<1$. Then
      $$2(a-b)^2leq (a) log (a) + (1-a)log(1-a) -(a)log(b)-(1-a)log(1-b).$$
      If I wanna go from left to right, then I think I gotta use Taylor's but I don't know how. Also, the right hand side I figured out is equal to
      $$(a)logfrac ab+(1-a)logfrac{1-a}{1-b}.$$







      real-analysis inequality logarithms taylor-expansion






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      edited Nov 16 at 18:32









      Angelo Lucia

      578213




      578213










      asked Nov 15 at 0:16









      Jack

      1017




      1017






















          2 Answers
          2






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          up vote
          4
          down vote



          accepted










          As already mentioned, this is a special case of Pinsker's inequality.
          The following proof for this special case is taken from Pinsker’s inequality and its applications to lower bounds:



          For fixed $a in (0, 1)$ define $f:(0, 1) to Bbb R$ as
          $$
          f(b) = alogfrac ab+(1-a)logfrac{1-a}{1-b} - 2(a-b)^2 , .
          $$

          Then
          $$
          f'(b) = -frac ab + frac{1-a}{1-b} + 4(a-b) = (b-a) left(frac{1}{b(1-b)} - 4right) , .
          $$

          The second factor is $ge 0$, with equality only at a single point ($b= frac 12$), so that $f$ is strictly decreasing on $(0, a]$ and
          strictly increasing on $[a, 1)$. It follows that
          $$
          f(b) ge f(a) = 0 , ,
          $$

          which is the desired inequality. Equality holds only if and only if $b=a$.






          share|cite|improve this answer






























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            3
            down vote













            You are trying to prove Pinsker's inequality. Since both $a$ and $b$ are between 0 and 1, we can think of $(a,1-a)$ and $(b,1-b)$ as binary probability distributions. Let me call them $P$ and $Q$ respectively, so that $P(0)=a$, $P(1)=1-a$, $Q(0)=b$ and $Q(1)=1-b$.



            Now it is easy to see that your left hand side is $2delta(P,Q)^2$, where $delta$ is the total variation distance between $P$ and $Q$, while the right hand side is the relative entropy, or Kullback-Leibler divergence, $D(P||Q)$.



            You can find the proof of Pinsker's inequality in many places, for example here: Proof of Pinsker's inequality.






            share|cite|improve this answer





















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote



              accepted










              As already mentioned, this is a special case of Pinsker's inequality.
              The following proof for this special case is taken from Pinsker’s inequality and its applications to lower bounds:



              For fixed $a in (0, 1)$ define $f:(0, 1) to Bbb R$ as
              $$
              f(b) = alogfrac ab+(1-a)logfrac{1-a}{1-b} - 2(a-b)^2 , .
              $$

              Then
              $$
              f'(b) = -frac ab + frac{1-a}{1-b} + 4(a-b) = (b-a) left(frac{1}{b(1-b)} - 4right) , .
              $$

              The second factor is $ge 0$, with equality only at a single point ($b= frac 12$), so that $f$ is strictly decreasing on $(0, a]$ and
              strictly increasing on $[a, 1)$. It follows that
              $$
              f(b) ge f(a) = 0 , ,
              $$

              which is the desired inequality. Equality holds only if and only if $b=a$.






              share|cite|improve this answer



























                up vote
                4
                down vote



                accepted










                As already mentioned, this is a special case of Pinsker's inequality.
                The following proof for this special case is taken from Pinsker’s inequality and its applications to lower bounds:



                For fixed $a in (0, 1)$ define $f:(0, 1) to Bbb R$ as
                $$
                f(b) = alogfrac ab+(1-a)logfrac{1-a}{1-b} - 2(a-b)^2 , .
                $$

                Then
                $$
                f'(b) = -frac ab + frac{1-a}{1-b} + 4(a-b) = (b-a) left(frac{1}{b(1-b)} - 4right) , .
                $$

                The second factor is $ge 0$, with equality only at a single point ($b= frac 12$), so that $f$ is strictly decreasing on $(0, a]$ and
                strictly increasing on $[a, 1)$. It follows that
                $$
                f(b) ge f(a) = 0 , ,
                $$

                which is the desired inequality. Equality holds only if and only if $b=a$.






                share|cite|improve this answer

























                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  As already mentioned, this is a special case of Pinsker's inequality.
                  The following proof for this special case is taken from Pinsker’s inequality and its applications to lower bounds:



                  For fixed $a in (0, 1)$ define $f:(0, 1) to Bbb R$ as
                  $$
                  f(b) = alogfrac ab+(1-a)logfrac{1-a}{1-b} - 2(a-b)^2 , .
                  $$

                  Then
                  $$
                  f'(b) = -frac ab + frac{1-a}{1-b} + 4(a-b) = (b-a) left(frac{1}{b(1-b)} - 4right) , .
                  $$

                  The second factor is $ge 0$, with equality only at a single point ($b= frac 12$), so that $f$ is strictly decreasing on $(0, a]$ and
                  strictly increasing on $[a, 1)$. It follows that
                  $$
                  f(b) ge f(a) = 0 , ,
                  $$

                  which is the desired inequality. Equality holds only if and only if $b=a$.






                  share|cite|improve this answer














                  As already mentioned, this is a special case of Pinsker's inequality.
                  The following proof for this special case is taken from Pinsker’s inequality and its applications to lower bounds:



                  For fixed $a in (0, 1)$ define $f:(0, 1) to Bbb R$ as
                  $$
                  f(b) = alogfrac ab+(1-a)logfrac{1-a}{1-b} - 2(a-b)^2 , .
                  $$

                  Then
                  $$
                  f'(b) = -frac ab + frac{1-a}{1-b} + 4(a-b) = (b-a) left(frac{1}{b(1-b)} - 4right) , .
                  $$

                  The second factor is $ge 0$, with equality only at a single point ($b= frac 12$), so that $f$ is strictly decreasing on $(0, a]$ and
                  strictly increasing on $[a, 1)$. It follows that
                  $$
                  f(b) ge f(a) = 0 , ,
                  $$

                  which is the desired inequality. Equality holds only if and only if $b=a$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 18 at 10:26

























                  answered Nov 17 at 17:34









                  Martin R

                  26.1k33046




                  26.1k33046






















                      up vote
                      3
                      down vote













                      You are trying to prove Pinsker's inequality. Since both $a$ and $b$ are between 0 and 1, we can think of $(a,1-a)$ and $(b,1-b)$ as binary probability distributions. Let me call them $P$ and $Q$ respectively, so that $P(0)=a$, $P(1)=1-a$, $Q(0)=b$ and $Q(1)=1-b$.



                      Now it is easy to see that your left hand side is $2delta(P,Q)^2$, where $delta$ is the total variation distance between $P$ and $Q$, while the right hand side is the relative entropy, or Kullback-Leibler divergence, $D(P||Q)$.



                      You can find the proof of Pinsker's inequality in many places, for example here: Proof of Pinsker's inequality.






                      share|cite|improve this answer

























                        up vote
                        3
                        down vote













                        You are trying to prove Pinsker's inequality. Since both $a$ and $b$ are between 0 and 1, we can think of $(a,1-a)$ and $(b,1-b)$ as binary probability distributions. Let me call them $P$ and $Q$ respectively, so that $P(0)=a$, $P(1)=1-a$, $Q(0)=b$ and $Q(1)=1-b$.



                        Now it is easy to see that your left hand side is $2delta(P,Q)^2$, where $delta$ is the total variation distance between $P$ and $Q$, while the right hand side is the relative entropy, or Kullback-Leibler divergence, $D(P||Q)$.



                        You can find the proof of Pinsker's inequality in many places, for example here: Proof of Pinsker's inequality.






                        share|cite|improve this answer























                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          You are trying to prove Pinsker's inequality. Since both $a$ and $b$ are between 0 and 1, we can think of $(a,1-a)$ and $(b,1-b)$ as binary probability distributions. Let me call them $P$ and $Q$ respectively, so that $P(0)=a$, $P(1)=1-a$, $Q(0)=b$ and $Q(1)=1-b$.



                          Now it is easy to see that your left hand side is $2delta(P,Q)^2$, where $delta$ is the total variation distance between $P$ and $Q$, while the right hand side is the relative entropy, or Kullback-Leibler divergence, $D(P||Q)$.



                          You can find the proof of Pinsker's inequality in many places, for example here: Proof of Pinsker's inequality.






                          share|cite|improve this answer












                          You are trying to prove Pinsker's inequality. Since both $a$ and $b$ are between 0 and 1, we can think of $(a,1-a)$ and $(b,1-b)$ as binary probability distributions. Let me call them $P$ and $Q$ respectively, so that $P(0)=a$, $P(1)=1-a$, $Q(0)=b$ and $Q(1)=1-b$.



                          Now it is easy to see that your left hand side is $2delta(P,Q)^2$, where $delta$ is the total variation distance between $P$ and $Q$, while the right hand side is the relative entropy, or Kullback-Leibler divergence, $D(P||Q)$.



                          You can find the proof of Pinsker's inequality in many places, for example here: Proof of Pinsker's inequality.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 15 at 0:54









                          Angelo Lucia

                          578213




                          578213






























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