IID random variables $(X_n)$ have $sum e^{X_n} c^n < infty$ a.s.












1














I'm working on the following exercise:




Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
$$
sum_{n=1}^infty e^{X_n} c^n begin{cases}
< infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
= infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
end{cases}
$$




I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?






probability probability-theory measure-theory convergence borel-cantelli-lemmas







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    1














    I'm working on the following exercise:




    Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
    $$
    sum_{n=1}^infty e^{X_n} c^n begin{cases}
    < infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
    = infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
    end{cases}
    $$




    I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?






    probability probability-theory measure-theory convergence borel-cantelli-lemmas







    share|cite|improve this question

























      1












      1








      1


      3





      I'm working on the following exercise:




      Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
      $$
      sum_{n=1}^infty e^{X_n} c^n begin{cases}
      < infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
      = infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
      end{cases}
      $$




      I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?






      probability probability-theory measure-theory convergence borel-cantelli-lemmas







      share|cite|improve this question













      I'm working on the following exercise:




      Let $X_1, X_2, ldots$ be i.i.d. nonnegative random variables. By virtue of the Borel-Cantelli lemma, show that for every $c in (0,1)$,
      $$
      sum_{n=1}^infty e^{X_n} c^n begin{cases}
      < infty textrm{ a.s.} & textrm{if } mathbb E[X_1] < infty; \
      = infty textrm{ a.s.} & textrm{if } mathbb E[X_1] = infty
      end{cases}
      $$




      I'm trying to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$ for some large $M > 0$. For then, Borel-Cantelli gives us that $$ mathbb Pleft[limsup left{ sum_{k=1}^n e^{X_k} c^k geq Mright}right] = mathbb Pleft[sum_{k=1}^infty e^{X_k} c^k geq Mright] = 0$$ and we're done. But I don't know how to show $sum_{n=1}^infty mathbb Pleft[sum_{k=1}^n e^{X_k} c^k geq Mright] < infty$. Any suggestions?






      probability probability-theory measure-theory convergence borel-cantelli-lemmas




      probability probability-theory measure-theory convergence borel-cantelli-lemmas






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Nov 18 '18 at 17:54









      D Ford

      549213




      549213






















          1 Answer
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          If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies



          $$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$



          hence



          $$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$



          Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ for fixed $cin (0,1)$, and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$




          If $mathbb{E}(X_1)=infty$ then



          $$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$



          and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.






          share|cite|improve this answer



















          • 1




            For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
            – Sangchul Lee
            Nov 18 '18 at 21:00












          • @SangchulLee Yes of course; thanks for your comment.
            – saz
            Nov 18 '18 at 21:02










          • @Did Ah right, I messed that up; thanks a lot for your comment.
            – saz
            Nov 19 '18 at 6:13











          Your Answer





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          1 Answer
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          active

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          1














          If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies



          $$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$



          hence



          $$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$



          Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ for fixed $cin (0,1)$, and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$




          If $mathbb{E}(X_1)=infty$ then



          $$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$



          and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.






          share|cite|improve this answer



















          • 1




            For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
            – Sangchul Lee
            Nov 18 '18 at 21:00












          • @SangchulLee Yes of course; thanks for your comment.
            – saz
            Nov 18 '18 at 21:02










          • @Did Ah right, I messed that up; thanks a lot for your comment.
            – saz
            Nov 19 '18 at 6:13
















          1














          If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies



          $$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$



          hence



          $$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$



          Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ for fixed $cin (0,1)$, and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$




          If $mathbb{E}(X_1)=infty$ then



          $$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$



          and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.






          share|cite|improve this answer



















          • 1




            For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
            – Sangchul Lee
            Nov 18 '18 at 21:00












          • @SangchulLee Yes of course; thanks for your comment.
            – saz
            Nov 18 '18 at 21:02










          • @Did Ah right, I messed that up; thanks a lot for your comment.
            – saz
            Nov 19 '18 at 6:13














          1












          1








          1






          If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies



          $$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$



          hence



          $$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$



          Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ for fixed $cin (0,1)$, and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$




          If $mathbb{E}(X_1)=infty$ then



          $$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$



          and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.






          share|cite|improve this answer














          If $mathbb{E}(X_1)< infty$ then it follows from the strong law of large numbers that $S_n := sum_{j=1}^n X_j$ satisfies



          $$lim_{n to infty} frac{S_n}{n} = mathbb{E}(X_1) quad text{a.s.};$$



          hence



          $$lim_{n to infty} frac{X_n}{n} = lim_{n to infty} left( frac{S_n}{n} - frac{S_{n-1}}{n} right)=0 quad text{a.s.}$$



          Consequently, there exists for almost all $omega in Omega$ some $N in mathbb{N}$ such that $$left| frac{X_n(omega)}{n} right| leq -log(sqrt{c}) quad text{for all $n geq N$}$$ for fixed $cin (0,1)$, and so $$sum_{n geq N} e^{X_n(omega)} c^n leq sum_{n geq N} sqrt{c}^n < infty.$$




          If $mathbb{E}(X_1)=infty$ then



          $$sum_{n geq 1} mathbb{P}(X_n geq n)=sum_{n geq 1} mathbb{P}(X_1 geq n) =infty,$$



          and therefore it follows from the Borel Cantelli lemma that $$mathbb{P}(X_n geq n , , text{infinitely often})=1,$$ i.e. $$e^{X_n} geq e^n quad text{for infinitely many $n$ with probability 1.}$$ This implies $sum_{n geq 1} e^{X_n} c^n = infty$ almost surely for $c:= 1/e$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 20 '18 at 18:41

























          answered Nov 18 '18 at 18:33









          saz

          78.1k758122




          78.1k758122








          • 1




            For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
            – Sangchul Lee
            Nov 18 '18 at 21:00












          • @SangchulLee Yes of course; thanks for your comment.
            – saz
            Nov 18 '18 at 21:02










          • @Did Ah right, I messed that up; thanks a lot for your comment.
            – saz
            Nov 19 '18 at 6:13














          • 1




            For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
            – Sangchul Lee
            Nov 18 '18 at 21:00












          • @SangchulLee Yes of course; thanks for your comment.
            – saz
            Nov 18 '18 at 21:02










          • @Did Ah right, I messed that up; thanks a lot for your comment.
            – saz
            Nov 19 '18 at 6:13








          1




          1




          For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
          – Sangchul Lee
          Nov 18 '18 at 21:00






          For the fact that $mathbb{E}[X_1] < infty$ implies $X_n/n to 0$ a.s., it may as well be proved by Borel-Cantelli: $$forall epsilon > 0 : quad sum_{n=1}^{infty} mathbb{P}[X_n > epsilon n] leq frac{1}{epsilon}mathbb{E}[X_1] < infty. $$ I am very certain that you already know this, but just wanted to mention in case OP needs an alternative approach.
          – Sangchul Lee
          Nov 18 '18 at 21:00














          @SangchulLee Yes of course; thanks for your comment.
          – saz
          Nov 18 '18 at 21:02




          @SangchulLee Yes of course; thanks for your comment.
          – saz
          Nov 18 '18 at 21:02












          @Did Ah right, I messed that up; thanks a lot for your comment.
          – saz
          Nov 19 '18 at 6:13




          @Did Ah right, I messed that up; thanks a lot for your comment.
          – saz
          Nov 19 '18 at 6:13


















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