Chernoff Bound: A special case












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According to the current version of the Wikipedia article entitled "Chernoff bound", the probability of simultaneous occurrence of more than $frac{n}{2}$ successes in $n$ independent Bernoulli trials each with a probability of success $p>frac{1}{2}$ is bounded from below by
$$1-e^{-2nleft(p-frac{1}{2}right)^2}$$



How is this result derived from the Chernoff bound?






probability inequality







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    According to the current version of the Wikipedia article entitled "Chernoff bound", the probability of simultaneous occurrence of more than $frac{n}{2}$ successes in $n$ independent Bernoulli trials each with a probability of success $p>frac{1}{2}$ is bounded from below by
    $$1-e^{-2nleft(p-frac{1}{2}right)^2}$$



    How is this result derived from the Chernoff bound?






    probability inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      According to the current version of the Wikipedia article entitled "Chernoff bound", the probability of simultaneous occurrence of more than $frac{n}{2}$ successes in $n$ independent Bernoulli trials each with a probability of success $p>frac{1}{2}$ is bounded from below by
      $$1-e^{-2nleft(p-frac{1}{2}right)^2}$$



      How is this result derived from the Chernoff bound?






      probability inequality







      share|cite|improve this question











      $endgroup$




      According to the current version of the Wikipedia article entitled "Chernoff bound", the probability of simultaneous occurrence of more than $frac{n}{2}$ successes in $n$ independent Bernoulli trials each with a probability of success $p>frac{1}{2}$ is bounded from below by
      $$1-e^{-2nleft(p-frac{1}{2}right)^2}$$



      How is this result derived from the Chernoff bound?






      probability inequality




      probability inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '12 at 14:52







      Evan Aad

















      asked Dec 25 '12 at 18:44









      Evan AadEvan Aad

      5,60911854




      5,60911854






















          1 Answer
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          $begingroup$

          The following lower bound can be readily established
          $$1-e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$
          Indeed, let $ninleft{1,2,dotsright}$ be fixed and let $X_0, X_1, dots, X_n$ be i.i.d. $mathrm{Bernouli}(p)$, $pinleft(frac{1}{2},1right)$. Set $S:=sum_{i=0}^n X_i$, $mu:=E[S]$. Denote by $q$ the probability of simultaneous occurrence of more than $frac{n}{2}$ successes, where success in the $i$th trial is the event $left{X_i=1right}$. Then $q=1-Pleft(Sleqfrac{n}{2}right)$ and, noticing that $mu=np$, we have by the multiplicative form of the Chernoff bound
          $$Pleft(Sleqfrac{n}{2}right)=Pleft(Sleqleft(1-left(1-frac{1}{2p}right)right)muright)leq e^{-frac{mu}{2}left(1-frac{1}{2p}right)^2}=e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$



          Now, since Wikipedia can be edited by anyone, simply replace the lower bound in the entry with the one derived above and you're all set.






          share|cite|improve this answer









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            $begingroup$

            The following lower bound can be readily established
            $$1-e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$
            Indeed, let $ninleft{1,2,dotsright}$ be fixed and let $X_0, X_1, dots, X_n$ be i.i.d. $mathrm{Bernouli}(p)$, $pinleft(frac{1}{2},1right)$. Set $S:=sum_{i=0}^n X_i$, $mu:=E[S]$. Denote by $q$ the probability of simultaneous occurrence of more than $frac{n}{2}$ successes, where success in the $i$th trial is the event $left{X_i=1right}$. Then $q=1-Pleft(Sleqfrac{n}{2}right)$ and, noticing that $mu=np$, we have by the multiplicative form of the Chernoff bound
            $$Pleft(Sleqfrac{n}{2}right)=Pleft(Sleqleft(1-left(1-frac{1}{2p}right)right)muright)leq e^{-frac{mu}{2}left(1-frac{1}{2p}right)^2}=e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$



            Now, since Wikipedia can be edited by anyone, simply replace the lower bound in the entry with the one derived above and you're all set.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The following lower bound can be readily established
              $$1-e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$
              Indeed, let $ninleft{1,2,dotsright}$ be fixed and let $X_0, X_1, dots, X_n$ be i.i.d. $mathrm{Bernouli}(p)$, $pinleft(frac{1}{2},1right)$. Set $S:=sum_{i=0}^n X_i$, $mu:=E[S]$. Denote by $q$ the probability of simultaneous occurrence of more than $frac{n}{2}$ successes, where success in the $i$th trial is the event $left{X_i=1right}$. Then $q=1-Pleft(Sleqfrac{n}{2}right)$ and, noticing that $mu=np$, we have by the multiplicative form of the Chernoff bound
              $$Pleft(Sleqfrac{n}{2}right)=Pleft(Sleqleft(1-left(1-frac{1}{2p}right)right)muright)leq e^{-frac{mu}{2}left(1-frac{1}{2p}right)^2}=e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$



              Now, since Wikipedia can be edited by anyone, simply replace the lower bound in the entry with the one derived above and you're all set.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The following lower bound can be readily established
                $$1-e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$
                Indeed, let $ninleft{1,2,dotsright}$ be fixed and let $X_0, X_1, dots, X_n$ be i.i.d. $mathrm{Bernouli}(p)$, $pinleft(frac{1}{2},1right)$. Set $S:=sum_{i=0}^n X_i$, $mu:=E[S]$. Denote by $q$ the probability of simultaneous occurrence of more than $frac{n}{2}$ successes, where success in the $i$th trial is the event $left{X_i=1right}$. Then $q=1-Pleft(Sleqfrac{n}{2}right)$ and, noticing that $mu=np$, we have by the multiplicative form of the Chernoff bound
                $$Pleft(Sleqfrac{n}{2}right)=Pleft(Sleqleft(1-left(1-frac{1}{2p}right)right)muright)leq e^{-frac{mu}{2}left(1-frac{1}{2p}right)^2}=e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$



                Now, since Wikipedia can be edited by anyone, simply replace the lower bound in the entry with the one derived above and you're all set.






                share|cite|improve this answer









                $endgroup$



                The following lower bound can be readily established
                $$1-e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$
                Indeed, let $ninleft{1,2,dotsright}$ be fixed and let $X_0, X_1, dots, X_n$ be i.i.d. $mathrm{Bernouli}(p)$, $pinleft(frac{1}{2},1right)$. Set $S:=sum_{i=0}^n X_i$, $mu:=E[S]$. Denote by $q$ the probability of simultaneous occurrence of more than $frac{n}{2}$ successes, where success in the $i$th trial is the event $left{X_i=1right}$. Then $q=1-Pleft(Sleqfrac{n}{2}right)$ and, noticing that $mu=np$, we have by the multiplicative form of the Chernoff bound
                $$Pleft(Sleqfrac{n}{2}right)=Pleft(Sleqleft(1-left(1-frac{1}{2p}right)right)muright)leq e^{-frac{mu}{2}left(1-frac{1}{2p}right)^2}=e^{-frac{1}{2p}nleft(p-frac{1}{2}right)^2}$$



                Now, since Wikipedia can be edited by anyone, simply replace the lower bound in the entry with the one derived above and you're all set.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 26 '12 at 14:51









                Evan AadEvan Aad

                5,60911854




                5,60911854






























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