exponential bound for sum of geometric variables












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I am writing my bachelor thesis and I am using the book "Brownian motion" by Mörters and Perez. I have troubles with the following lemma:



Let $X_j$, $j in mathbb{N}$, geometrically distributed on ${1,2,...}$ with mean $2$. For $epsilon$ sufficiently small, for every $m in mathbb{N}$ and all $k leqslant m$ the following inequality holds.
begin{align*}
mathbb{P}left({|sum_{j=1}^k (X_j-2)| geqslant epsilon m}right) leqslant 4 exp{-tfrac{1}{5} epsilon^2 m}
end{align*}



I already showed, using the hints in the book, (without Hoeffdings inequality) the inequality



begin{align*}
mathbb{P}left({sum_{j=1}^k (X_j-2) geqslant epsilon m}right) leqslant exp{-tfrac{1}{5} epsilon^2 m}
end{align*}



Now I want to proof the other part of the inequality via the one I already showed. In his book he states that it is obvious. I cannot see that and my endless computation did not really lead anywhere.










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I am writing my bachelor thesis and I am using the book "Brownian motion" by Mörters and Perez. I have troubles with the following lemma:



    Let $X_j$, $j in mathbb{N}$, geometrically distributed on ${1,2,...}$ with mean $2$. For $epsilon$ sufficiently small, for every $m in mathbb{N}$ and all $k leqslant m$ the following inequality holds.
    begin{align*}
    mathbb{P}left({|sum_{j=1}^k (X_j-2)| geqslant epsilon m}right) leqslant 4 exp{-tfrac{1}{5} epsilon^2 m}
    end{align*}



    I already showed, using the hints in the book, (without Hoeffdings inequality) the inequality



    begin{align*}
    mathbb{P}left({sum_{j=1}^k (X_j-2) geqslant epsilon m}right) leqslant exp{-tfrac{1}{5} epsilon^2 m}
    end{align*}



    Now I want to proof the other part of the inequality via the one I already showed. In his book he states that it is obvious. I cannot see that and my endless computation did not really lead anywhere.










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I am writing my bachelor thesis and I am using the book "Brownian motion" by Mörters and Perez. I have troubles with the following lemma:



      Let $X_j$, $j in mathbb{N}$, geometrically distributed on ${1,2,...}$ with mean $2$. For $epsilon$ sufficiently small, for every $m in mathbb{N}$ and all $k leqslant m$ the following inequality holds.
      begin{align*}
      mathbb{P}left({|sum_{j=1}^k (X_j-2)| geqslant epsilon m}right) leqslant 4 exp{-tfrac{1}{5} epsilon^2 m}
      end{align*}



      I already showed, using the hints in the book, (without Hoeffdings inequality) the inequality



      begin{align*}
      mathbb{P}left({sum_{j=1}^k (X_j-2) geqslant epsilon m}right) leqslant exp{-tfrac{1}{5} epsilon^2 m}
      end{align*}



      Now I want to proof the other part of the inequality via the one I already showed. In his book he states that it is obvious. I cannot see that and my endless computation did not really lead anywhere.










      share|cite|improve this question









      $endgroup$




      I am writing my bachelor thesis and I am using the book "Brownian motion" by Mörters and Perez. I have troubles with the following lemma:



      Let $X_j$, $j in mathbb{N}$, geometrically distributed on ${1,2,...}$ with mean $2$. For $epsilon$ sufficiently small, for every $m in mathbb{N}$ and all $k leqslant m$ the following inequality holds.
      begin{align*}
      mathbb{P}left({|sum_{j=1}^k (X_j-2)| geqslant epsilon m}right) leqslant 4 exp{-tfrac{1}{5} epsilon^2 m}
      end{align*}



      I already showed, using the hints in the book, (without Hoeffdings inequality) the inequality



      begin{align*}
      mathbb{P}left({sum_{j=1}^k (X_j-2) geqslant epsilon m}right) leqslant exp{-tfrac{1}{5} epsilon^2 m}
      end{align*}



      Now I want to proof the other part of the inequality via the one I already showed. In his book he states that it is obvious. I cannot see that and my endless computation did not really lead anywhere.







      probability probability-theory






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      asked Jan 15 at 17:43









      user631620user631620

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          $begingroup$

          Apply the same proof to $-sum_{j=1}^k (X_j-2)$ and notice that
          $$
          mathsf{E}e^{-lambda(X_j-2)}=frac{e^{2lambda}}{2e^lambda-1}le 1+lambda^2le e^{lambda^2}.
          $$






          share|cite|improve this answer









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            1 Answer
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            active

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            active

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            $begingroup$

            Apply the same proof to $-sum_{j=1}^k (X_j-2)$ and notice that
            $$
            mathsf{E}e^{-lambda(X_j-2)}=frac{e^{2lambda}}{2e^lambda-1}le 1+lambda^2le e^{lambda^2}.
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Apply the same proof to $-sum_{j=1}^k (X_j-2)$ and notice that
              $$
              mathsf{E}e^{-lambda(X_j-2)}=frac{e^{2lambda}}{2e^lambda-1}le 1+lambda^2le e^{lambda^2}.
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Apply the same proof to $-sum_{j=1}^k (X_j-2)$ and notice that
                $$
                mathsf{E}e^{-lambda(X_j-2)}=frac{e^{2lambda}}{2e^lambda-1}le 1+lambda^2le e^{lambda^2}.
                $$






                share|cite|improve this answer









                $endgroup$



                Apply the same proof to $-sum_{j=1}^k (X_j-2)$ and notice that
                $$
                mathsf{E}e^{-lambda(X_j-2)}=frac{e^{2lambda}}{2e^lambda-1}le 1+lambda^2le e^{lambda^2}.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 6:09









                d.k.o.d.k.o.

                10.6k730




                10.6k730






























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