A.S. convergence of sum of square-integrable independent random variables with summable variation












2












$begingroup$


I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (exercise 6.1.4):




Let $X_1, X_2, ldots$ be independent, square integrable, centered random variables with $sum_{i=1}^infty mathbf{Var}[X_i] < infty$. Show that there exists a square integrable $X$ with $X = lim_{n to infty} sum_{i=1}^n X_i$ almost surely.




Chebyshev's inequality gives us
$$
mathbf Pleft[|S_m - S_n| > epsilonright] leq epsilon^{-2} mathbf{Var}left[ sum_{i=m+1}^n X_iright] = epsilon^{-2} sum_{i=m+1}^n mathbf{Var}left[X_iright] xrightarrow{m,n to infty} 0.
$$

whence $(S_n)_{n in mathbb N}$ is a Cauchy sequence in probability. Thus $S_n xrightarrow{mathbf P} X$. Using a similar strategy, we can in fact show that $S_n to X$ in $L^2$.



Now, to prove almost sure convergence, I'd like to use the following result (Corollary 6.13 in Klenke):




Let $(E,d)$ be a separable metric space. Let $f, f_1, f_2, ldots$ be measurable maps $Omega to E$. Then the following statements are equivalent.



(i)$quad f_n to f$ in measure as $n to infty$.



(ii)$quad$For any subsequence of $(f_n)_{n in mathbb N}$, there exists a sub-subsequence that converges to $f$ almost everywhere.




and somehow use the fact that we're working with a sum of centered random variables to show that in fact every subsequence converges a.s. But I'm not sure how to do this since our $X_i$ are not nonnegative. I tried reconstructing the proof of this theorem, but I've only been able to show once again that there are a.e. convergent subsequences.



My other thought was to apply the Borel-Cantelli lemma to the events $B_n(epsilon) := left{ |X - S_n| > epsilonright}$ and prove that $limsup_{n to infty} B_n(epsilon) =: B(epsilon)$ has probability $0$, but in the latter case I don't know how to approximate the probability of $B_n(epsilon)$. Chebyshev doesn't seem available to us since strictly speaking we don't know what $X$ looks like, only that $S_n$ converges in $L^2$ to it. Even if we could say $X - S_n = sum_{i=n+1}^infty X_i$, the above approximation using Chebyshev with $|X - S_n|$ instead of $|S_m - S_n|$ would work out to
$$
mathbf Pleft[|X - S_n| > epsilonright] leq epsilon^{-2} sum_{i=n+1}^infty mathbf{Var}[X_i]
$$

which would sum to $epsilon^{-2} sum_{n=1}^infty nmathbf{Var}[X_n]$, but I don't see why this series converges.



Any thoughts on how to prove $S_n to X$ almost surely?










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (exercise 6.1.4):




    Let $X_1, X_2, ldots$ be independent, square integrable, centered random variables with $sum_{i=1}^infty mathbf{Var}[X_i] < infty$. Show that there exists a square integrable $X$ with $X = lim_{n to infty} sum_{i=1}^n X_i$ almost surely.




    Chebyshev's inequality gives us
    $$
    mathbf Pleft[|S_m - S_n| > epsilonright] leq epsilon^{-2} mathbf{Var}left[ sum_{i=m+1}^n X_iright] = epsilon^{-2} sum_{i=m+1}^n mathbf{Var}left[X_iright] xrightarrow{m,n to infty} 0.
    $$

    whence $(S_n)_{n in mathbb N}$ is a Cauchy sequence in probability. Thus $S_n xrightarrow{mathbf P} X$. Using a similar strategy, we can in fact show that $S_n to X$ in $L^2$.



    Now, to prove almost sure convergence, I'd like to use the following result (Corollary 6.13 in Klenke):




    Let $(E,d)$ be a separable metric space. Let $f, f_1, f_2, ldots$ be measurable maps $Omega to E$. Then the following statements are equivalent.



    (i)$quad f_n to f$ in measure as $n to infty$.



    (ii)$quad$For any subsequence of $(f_n)_{n in mathbb N}$, there exists a sub-subsequence that converges to $f$ almost everywhere.




    and somehow use the fact that we're working with a sum of centered random variables to show that in fact every subsequence converges a.s. But I'm not sure how to do this since our $X_i$ are not nonnegative. I tried reconstructing the proof of this theorem, but I've only been able to show once again that there are a.e. convergent subsequences.



    My other thought was to apply the Borel-Cantelli lemma to the events $B_n(epsilon) := left{ |X - S_n| > epsilonright}$ and prove that $limsup_{n to infty} B_n(epsilon) =: B(epsilon)$ has probability $0$, but in the latter case I don't know how to approximate the probability of $B_n(epsilon)$. Chebyshev doesn't seem available to us since strictly speaking we don't know what $X$ looks like, only that $S_n$ converges in $L^2$ to it. Even if we could say $X - S_n = sum_{i=n+1}^infty X_i$, the above approximation using Chebyshev with $|X - S_n|$ instead of $|S_m - S_n|$ would work out to
    $$
    mathbf Pleft[|X - S_n| > epsilonright] leq epsilon^{-2} sum_{i=n+1}^infty mathbf{Var}[X_i]
    $$

    which would sum to $epsilon^{-2} sum_{n=1}^infty nmathbf{Var}[X_n]$, but I don't see why this series converges.



    Any thoughts on how to prove $S_n to X$ almost surely?










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (exercise 6.1.4):




      Let $X_1, X_2, ldots$ be independent, square integrable, centered random variables with $sum_{i=1}^infty mathbf{Var}[X_i] < infty$. Show that there exists a square integrable $X$ with $X = lim_{n to infty} sum_{i=1}^n X_i$ almost surely.




      Chebyshev's inequality gives us
      $$
      mathbf Pleft[|S_m - S_n| > epsilonright] leq epsilon^{-2} mathbf{Var}left[ sum_{i=m+1}^n X_iright] = epsilon^{-2} sum_{i=m+1}^n mathbf{Var}left[X_iright] xrightarrow{m,n to infty} 0.
      $$

      whence $(S_n)_{n in mathbb N}$ is a Cauchy sequence in probability. Thus $S_n xrightarrow{mathbf P} X$. Using a similar strategy, we can in fact show that $S_n to X$ in $L^2$.



      Now, to prove almost sure convergence, I'd like to use the following result (Corollary 6.13 in Klenke):




      Let $(E,d)$ be a separable metric space. Let $f, f_1, f_2, ldots$ be measurable maps $Omega to E$. Then the following statements are equivalent.



      (i)$quad f_n to f$ in measure as $n to infty$.



      (ii)$quad$For any subsequence of $(f_n)_{n in mathbb N}$, there exists a sub-subsequence that converges to $f$ almost everywhere.




      and somehow use the fact that we're working with a sum of centered random variables to show that in fact every subsequence converges a.s. But I'm not sure how to do this since our $X_i$ are not nonnegative. I tried reconstructing the proof of this theorem, but I've only been able to show once again that there are a.e. convergent subsequences.



      My other thought was to apply the Borel-Cantelli lemma to the events $B_n(epsilon) := left{ |X - S_n| > epsilonright}$ and prove that $limsup_{n to infty} B_n(epsilon) =: B(epsilon)$ has probability $0$, but in the latter case I don't know how to approximate the probability of $B_n(epsilon)$. Chebyshev doesn't seem available to us since strictly speaking we don't know what $X$ looks like, only that $S_n$ converges in $L^2$ to it. Even if we could say $X - S_n = sum_{i=n+1}^infty X_i$, the above approximation using Chebyshev with $|X - S_n|$ instead of $|S_m - S_n|$ would work out to
      $$
      mathbf Pleft[|X - S_n| > epsilonright] leq epsilon^{-2} sum_{i=n+1}^infty mathbf{Var}[X_i]
      $$

      which would sum to $epsilon^{-2} sum_{n=1}^infty nmathbf{Var}[X_n]$, but I don't see why this series converges.



      Any thoughts on how to prove $S_n to X$ almost surely?










      share|cite|improve this question









      $endgroup$




      I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (exercise 6.1.4):




      Let $X_1, X_2, ldots$ be independent, square integrable, centered random variables with $sum_{i=1}^infty mathbf{Var}[X_i] < infty$. Show that there exists a square integrable $X$ with $X = lim_{n to infty} sum_{i=1}^n X_i$ almost surely.




      Chebyshev's inequality gives us
      $$
      mathbf Pleft[|S_m - S_n| > epsilonright] leq epsilon^{-2} mathbf{Var}left[ sum_{i=m+1}^n X_iright] = epsilon^{-2} sum_{i=m+1}^n mathbf{Var}left[X_iright] xrightarrow{m,n to infty} 0.
      $$

      whence $(S_n)_{n in mathbb N}$ is a Cauchy sequence in probability. Thus $S_n xrightarrow{mathbf P} X$. Using a similar strategy, we can in fact show that $S_n to X$ in $L^2$.



      Now, to prove almost sure convergence, I'd like to use the following result (Corollary 6.13 in Klenke):




      Let $(E,d)$ be a separable metric space. Let $f, f_1, f_2, ldots$ be measurable maps $Omega to E$. Then the following statements are equivalent.



      (i)$quad f_n to f$ in measure as $n to infty$.



      (ii)$quad$For any subsequence of $(f_n)_{n in mathbb N}$, there exists a sub-subsequence that converges to $f$ almost everywhere.




      and somehow use the fact that we're working with a sum of centered random variables to show that in fact every subsequence converges a.s. But I'm not sure how to do this since our $X_i$ are not nonnegative. I tried reconstructing the proof of this theorem, but I've only been able to show once again that there are a.e. convergent subsequences.



      My other thought was to apply the Borel-Cantelli lemma to the events $B_n(epsilon) := left{ |X - S_n| > epsilonright}$ and prove that $limsup_{n to infty} B_n(epsilon) =: B(epsilon)$ has probability $0$, but in the latter case I don't know how to approximate the probability of $B_n(epsilon)$. Chebyshev doesn't seem available to us since strictly speaking we don't know what $X$ looks like, only that $S_n$ converges in $L^2$ to it. Even if we could say $X - S_n = sum_{i=n+1}^infty X_i$, the above approximation using Chebyshev with $|X - S_n|$ instead of $|S_m - S_n|$ would work out to
      $$
      mathbf Pleft[|X - S_n| > epsilonright] leq epsilon^{-2} sum_{i=n+1}^infty mathbf{Var}[X_i]
      $$

      which would sum to $epsilon^{-2} sum_{n=1}^infty nmathbf{Var}[X_n]$, but I don't see why this series converges.



      Any thoughts on how to prove $S_n to X$ almost surely?







      real-analysis probability probability-theory convergence borel-cantelli-lemmas






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Dec 16 '18 at 6:15









      D FordD Ford

      655313




      655313






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Since
          $$
          lim_{ntoinfty}mathsf{P}left(sup_{kge n}|S_n-S_k|> epsilonright)=0,
          $$

          the set on which the sequence ${S_n}$ is not Cauchy,
          $$
          N=bigcup_{epsilon>0}bigcap_{nge 1}left{sup_{j,kge n}|S_j-S_k|>epsilonright}
          $$

          is a null set ($because sup_{j,kge n}|S_j-S_k|le 2sup_{kge n}|S_n-S_k|$). So you define $X:=lim_{ntoinfty} S_n1_{N^c}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:27










          • $begingroup$
            @DFord "Why does this not imply..."? Why should it?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:20










          • $begingroup$
            Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:26










          • $begingroup$
            @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:28








          • 1




            $begingroup$
            Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 19:00



















          1












          $begingroup$

          $var (sum _{i=n}^{m} X_i) =sum _{i=n}^{m} var(X_i) to 0$ as $n,m to infty$so the partial sums of $sum X_i$ form a Cauchy sequence in $L^{2}$. Hence there is a square integrable random variable $X$ such that $sum _1^{n} X_i to X$ in $L^{2}$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:28










          • $begingroup$
            It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
            $endgroup$
            – Kavi Rama Murthy
            Dec 16 '18 at 23:22



















          1












          $begingroup$

          Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables
          $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely. But we know that in general $$
          P(sum_{i=1}^infty |X_i|<infty) =1
          $$
          fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$
          P(max_{nin mathbb{N}} |S_n|>lambda) leq frac{C}{lambda^2}sum_{n=1}^infty operatorname{Var}(X_n)
          $$
          provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence
          $$
          nmapsto S_n(omega)
          $$
          for almost all $omegain Omega$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$
          S_n text{ converges a.s.} iff S_n text{ converges in probability.}
          $$
          (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely follows immediately.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:48






          • 1




            $begingroup$
            It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
            $endgroup$
            – Song
            Dec 16 '18 at 18:53













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          3 Answers
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          active

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Since
          $$
          lim_{ntoinfty}mathsf{P}left(sup_{kge n}|S_n-S_k|> epsilonright)=0,
          $$

          the set on which the sequence ${S_n}$ is not Cauchy,
          $$
          N=bigcup_{epsilon>0}bigcap_{nge 1}left{sup_{j,kge n}|S_j-S_k|>epsilonright}
          $$

          is a null set ($because sup_{j,kge n}|S_j-S_k|le 2sup_{kge n}|S_n-S_k|$). So you define $X:=lim_{ntoinfty} S_n1_{N^c}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:27










          • $begingroup$
            @DFord "Why does this not imply..."? Why should it?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:20










          • $begingroup$
            Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:26










          • $begingroup$
            @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:28








          • 1




            $begingroup$
            Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 19:00
















          2












          $begingroup$

          Since
          $$
          lim_{ntoinfty}mathsf{P}left(sup_{kge n}|S_n-S_k|> epsilonright)=0,
          $$

          the set on which the sequence ${S_n}$ is not Cauchy,
          $$
          N=bigcup_{epsilon>0}bigcap_{nge 1}left{sup_{j,kge n}|S_j-S_k|>epsilonright}
          $$

          is a null set ($because sup_{j,kge n}|S_j-S_k|le 2sup_{kge n}|S_n-S_k|$). So you define $X:=lim_{ntoinfty} S_n1_{N^c}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:27










          • $begingroup$
            @DFord "Why does this not imply..."? Why should it?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:20










          • $begingroup$
            Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:26










          • $begingroup$
            @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:28








          • 1




            $begingroup$
            Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 19:00














          2












          2








          2





          $begingroup$

          Since
          $$
          lim_{ntoinfty}mathsf{P}left(sup_{kge n}|S_n-S_k|> epsilonright)=0,
          $$

          the set on which the sequence ${S_n}$ is not Cauchy,
          $$
          N=bigcup_{epsilon>0}bigcap_{nge 1}left{sup_{j,kge n}|S_j-S_k|>epsilonright}
          $$

          is a null set ($because sup_{j,kge n}|S_j-S_k|le 2sup_{kge n}|S_n-S_k|$). So you define $X:=lim_{ntoinfty} S_n1_{N^c}$.






          share|cite|improve this answer











          $endgroup$



          Since
          $$
          lim_{ntoinfty}mathsf{P}left(sup_{kge n}|S_n-S_k|> epsilonright)=0,
          $$

          the set on which the sequence ${S_n}$ is not Cauchy,
          $$
          N=bigcup_{epsilon>0}bigcap_{nge 1}left{sup_{j,kge n}|S_j-S_k|>epsilonright}
          $$

          is a null set ($because sup_{j,kge n}|S_j-S_k|le 2sup_{kge n}|S_n-S_k|$). So you define $X:=lim_{ntoinfty} S_n1_{N^c}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 8:01

























          answered Dec 16 '18 at 7:15









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

          10.2k629




          10.2k629












          • $begingroup$
            This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:27










          • $begingroup$
            @DFord "Why does this not imply..."? Why should it?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:20










          • $begingroup$
            Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:26










          • $begingroup$
            @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:28








          • 1




            $begingroup$
            Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 19:00


















          • $begingroup$
            This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:27










          • $begingroup$
            @DFord "Why does this not imply..."? Why should it?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:20










          • $begingroup$
            Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:26










          • $begingroup$
            @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 18:28








          • 1




            $begingroup$
            Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
            $endgroup$
            – d.k.o.
            Dec 16 '18 at 19:00
















          $begingroup$
          This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
          $endgroup$
          – D Ford
          Dec 16 '18 at 14:27




          $begingroup$
          This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n sim mathrm{Ber}_{1/n}$.)
          $endgroup$
          – D Ford
          Dec 16 '18 at 14:27












          $begingroup$
          @DFord "Why does this not imply..."? Why should it?
          $endgroup$
          – d.k.o.
          Dec 16 '18 at 18:20




          $begingroup$
          @DFord "Why does this not imply..."? Why should it?
          $endgroup$
          – d.k.o.
          Dec 16 '18 at 18:20












          $begingroup$
          Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
          $endgroup$
          – D Ford
          Dec 16 '18 at 18:26




          $begingroup$
          Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which ${X_n}$ is not Cauchy is a null set.
          $endgroup$
          – D Ford
          Dec 16 '18 at 18:26












          $begingroup$
          @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
          $endgroup$
          – d.k.o.
          Dec 16 '18 at 18:28






          $begingroup$
          @DFord Does $mathsf{P}(sup_{kge n}|X_n-X_k|>epsilon)$ converge to $0$ in that case?
          $endgroup$
          – d.k.o.
          Dec 16 '18 at 18:28






          1




          1




          $begingroup$
          Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
          $endgroup$
          – d.k.o.
          Dec 16 '18 at 19:00




          $begingroup$
          Cauchy in prob. means that $mathsf{P}(|X_j-X_k|>epsilon)to 0$ as $j,kto infty$. For the a.s. convergence you need a stronger condition.
          $endgroup$
          – d.k.o.
          Dec 16 '18 at 19:00











          1












          $begingroup$

          $var (sum _{i=n}^{m} X_i) =sum _{i=n}^{m} var(X_i) to 0$ as $n,m to infty$so the partial sums of $sum X_i$ form a Cauchy sequence in $L^{2}$. Hence there is a square integrable random variable $X$ such that $sum _1^{n} X_i to X$ in $L^{2}$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:28










          • $begingroup$
            It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
            $endgroup$
            – Kavi Rama Murthy
            Dec 16 '18 at 23:22
















          1












          $begingroup$

          $var (sum _{i=n}^{m} X_i) =sum _{i=n}^{m} var(X_i) to 0$ as $n,m to infty$so the partial sums of $sum X_i$ form a Cauchy sequence in $L^{2}$. Hence there is a square integrable random variable $X$ such that $sum _1^{n} X_i to X$ in $L^{2}$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:28










          • $begingroup$
            It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
            $endgroup$
            – Kavi Rama Murthy
            Dec 16 '18 at 23:22














          1












          1








          1





          $begingroup$

          $var (sum _{i=n}^{m} X_i) =sum _{i=n}^{m} var(X_i) to 0$ as $n,m to infty$so the partial sums of $sum X_i$ form a Cauchy sequence in $L^{2}$. Hence there is a square integrable random variable $X$ such that $sum _1^{n} X_i to X$ in $L^{2}$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.






          share|cite|improve this answer









          $endgroup$



          $var (sum _{i=n}^{m} X_i) =sum _{i=n}^{m} var(X_i) to 0$ as $n,m to infty$so the partial sums of $sum X_i$ form a Cauchy sequence in $L^{2}$. Hence there is a square integrable random variable $X$ such that $sum _1^{n} X_i to X$ in $L^{2}$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 11:56









          Kavi Rama MurthyKavi Rama Murthy

          66k42867




          66k42867












          • $begingroup$
            ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:28










          • $begingroup$
            It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
            $endgroup$
            – Kavi Rama Murthy
            Dec 16 '18 at 23:22


















          • $begingroup$
            ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
            $endgroup$
            – D Ford
            Dec 16 '18 at 14:28










          • $begingroup$
            It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
            $endgroup$
            – Kavi Rama Murthy
            Dec 16 '18 at 23:22
















          $begingroup$
          ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
          $endgroup$
          – D Ford
          Dec 16 '18 at 14:28




          $begingroup$
          ``For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this.
          $endgroup$
          – D Ford
          Dec 16 '18 at 14:28












          $begingroup$
          It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
          $endgroup$
          – Kavi Rama Murthy
          Dec 16 '18 at 23:22




          $begingroup$
          It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord
          $endgroup$
          – Kavi Rama Murthy
          Dec 16 '18 at 23:22











          1












          $begingroup$

          Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables
          $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely. But we know that in general $$
          P(sum_{i=1}^infty |X_i|<infty) =1
          $$
          fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$
          P(max_{nin mathbb{N}} |S_n|>lambda) leq frac{C}{lambda^2}sum_{n=1}^infty operatorname{Var}(X_n)
          $$
          provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence
          $$
          nmapsto S_n(omega)
          $$
          for almost all $omegain Omega$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$
          S_n text{ converges a.s.} iff S_n text{ converges in probability.}
          $$
          (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely follows immediately.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:48






          • 1




            $begingroup$
            It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
            $endgroup$
            – Song
            Dec 16 '18 at 18:53


















          1












          $begingroup$

          Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables
          $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely. But we know that in general $$
          P(sum_{i=1}^infty |X_i|<infty) =1
          $$
          fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$
          P(max_{nin mathbb{N}} |S_n|>lambda) leq frac{C}{lambda^2}sum_{n=1}^infty operatorname{Var}(X_n)
          $$
          provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence
          $$
          nmapsto S_n(omega)
          $$
          for almost all $omegain Omega$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$
          S_n text{ converges a.s.} iff S_n text{ converges in probability.}
          $$
          (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely follows immediately.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:48






          • 1




            $begingroup$
            It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
            $endgroup$
            – Song
            Dec 16 '18 at 18:53
















          1












          1








          1





          $begingroup$

          Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables
          $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely. But we know that in general $$
          P(sum_{i=1}^infty |X_i|<infty) =1
          $$
          fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$
          P(max_{nin mathbb{N}} |S_n|>lambda) leq frac{C}{lambda^2}sum_{n=1}^infty operatorname{Var}(X_n)
          $$
          provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence
          $$
          nmapsto S_n(omega)
          $$
          for almost all $omegain Omega$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$
          S_n text{ converges a.s.} iff S_n text{ converges in probability.}
          $$
          (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely follows immediately.






          share|cite|improve this answer











          $endgroup$



          Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables
          $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely. But we know that in general $$
          P(sum_{i=1}^infty |X_i|<infty) =1
          $$
          fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$
          P(max_{nin mathbb{N}} |S_n|>lambda) leq frac{C}{lambda^2}sum_{n=1}^infty operatorname{Var}(X_n)
          $$
          provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence
          $$
          nmapsto S_n(omega)
          $$
          for almost all $omegain Omega$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$
          S_n text{ converges a.s.} iff S_n text{ converges in probability.}
          $$
          (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = sum_{i=1}^n X_i to S_infty
          $$
          almost surely follows immediately.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 16 '18 at 12:06

























          answered Dec 16 '18 at 7:15









          SongSong

          16.8k21145




          16.8k21145












          • $begingroup$
            Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:48






          • 1




            $begingroup$
            It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
            $endgroup$
            – Song
            Dec 16 '18 at 18:53




















          • $begingroup$
            Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
            $endgroup$
            – D Ford
            Dec 16 '18 at 18:48






          • 1




            $begingroup$
            It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
            $endgroup$
            – Song
            Dec 16 '18 at 18:53


















          $begingroup$
          Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
          $endgroup$
          – D Ford
          Dec 16 '18 at 18:48




          $begingroup$
          Is there a clear reason why $S_n$ converges a.s. $iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it.
          $endgroup$
          – D Ford
          Dec 16 '18 at 18:48




          1




          1




          $begingroup$
          It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
          $endgroup$
          – Song
          Dec 16 '18 at 18:53






          $begingroup$
          It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = max_{jleq n} P(|S_n-S_j|>a).$ Then $P(max_{jleq n}|S_j|>x) leq frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+cdots +X_n$ and $X_i$ are independent (need not be identical).
          $endgroup$
          – Song
          Dec 16 '18 at 18:53




















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