Inequality in proof of 2nd Borel-Cantelli Lemma











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At some point in the proof of the second Borel-Cantelli Lemma the the following inequality is mentioned:



$$...=expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq exp bigl (-sum_{m=n}^kP(A_m) bigr)$$



How do I this inequality? Some simple logarithmic calculation rules?






probability-theory logarithms borel-cantelli-lemmas







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    At some point in the proof of the second Borel-Cantelli Lemma the the following inequality is mentioned:



    $$...=expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq exp bigl (-sum_{m=n}^kP(A_m) bigr)$$



    How do I this inequality? Some simple logarithmic calculation rules?






    probability-theory logarithms borel-cantelli-lemmas







    share|cite|improve this question
























      up vote
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      favorite
      1









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      At some point in the proof of the second Borel-Cantelli Lemma the the following inequality is mentioned:



      $$...=expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq exp bigl (-sum_{m=n}^kP(A_m) bigr)$$



      How do I this inequality? Some simple logarithmic calculation rules?






      probability-theory logarithms borel-cantelli-lemmas







      share|cite|improve this question













      At some point in the proof of the second Borel-Cantelli Lemma the the following inequality is mentioned:



      $$...=expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq exp bigl (-sum_{m=n}^kP(A_m) bigr)$$



      How do I this inequality? Some simple logarithmic calculation rules?






      probability-theory logarithms borel-cantelli-lemmas




      probability-theory logarithms borel-cantelli-lemmas






      share|cite|improve this question













      share|cite|improve this question











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          Note that





          • $log (1-x) =- sum_{n=1}^{infty}frac{x^n}{n} leq -x$ for $0 leq x<1$
            $$Rightarrow expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq expbigl ( sum_{m=n}^k -P(A_m)bigr ) = exp bigl (-sum_{m=n}^kP(A_m) bigr)$$






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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Note that





            • $log (1-x) =- sum_{n=1}^{infty}frac{x^n}{n} leq -x$ for $0 leq x<1$
              $$Rightarrow expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq expbigl ( sum_{m=n}^k -P(A_m)bigr ) = exp bigl (-sum_{m=n}^kP(A_m) bigr)$$






            share|cite|improve this answer



























              up vote
              2
              down vote



              accepted










              Note that





              • $log (1-x) =- sum_{n=1}^{infty}frac{x^n}{n} leq -x$ for $0 leq x<1$
                $$Rightarrow expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq expbigl ( sum_{m=n}^k -P(A_m)bigr ) = exp bigl (-sum_{m=n}^kP(A_m) bigr)$$






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Note that





                • $log (1-x) =- sum_{n=1}^{infty}frac{x^n}{n} leq -x$ for $0 leq x<1$
                  $$Rightarrow expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq expbigl ( sum_{m=n}^k -P(A_m)bigr ) = exp bigl (-sum_{m=n}^kP(A_m) bigr)$$






                share|cite|improve this answer














                Note that





                • $log (1-x) =- sum_{n=1}^{infty}frac{x^n}{n} leq -x$ for $0 leq x<1$
                  $$Rightarrow expbigl ( sum_{m=n}^klog(1-P(A_m)bigr ) leq expbigl ( sum_{m=n}^k -P(A_m)bigr ) = exp bigl (-sum_{m=n}^kP(A_m) bigr)$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








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