Cauchy convergence in probability implies the existence of a (finite a.e.) limit $X$












7















Cauchy convergence of a sequence $X_n$ of random variables in probability implies the existence of an $X$ (finite a.e.), such that $X_n$ converges to $X$ in probability.




The problem's hint suggests constructing a subsequence $n_k$ so that $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) < infty$, and this I have accomplished, showing that in fact I have a subsequence with $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) leq sum_{k=1}^infty 1/2^k = 1 < infty$. But now that I have this subsequence, I'm not clear on how it implies that a limit random variable $X$ exists. I feel like I must be missing something obvious here, but I just can't put it together.






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    7















    Cauchy convergence of a sequence $X_n$ of random variables in probability implies the existence of an $X$ (finite a.e.), such that $X_n$ converges to $X$ in probability.




    The problem's hint suggests constructing a subsequence $n_k$ so that $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) < infty$, and this I have accomplished, showing that in fact I have a subsequence with $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) leq sum_{k=1}^infty 1/2^k = 1 < infty$. But now that I have this subsequence, I'm not clear on how it implies that a limit random variable $X$ exists. I feel like I must be missing something obvious here, but I just can't put it together.






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      Cauchy convergence of a sequence $X_n$ of random variables in probability implies the existence of an $X$ (finite a.e.), such that $X_n$ converges to $X$ in probability.




      The problem's hint suggests constructing a subsequence $n_k$ so that $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) < infty$, and this I have accomplished, showing that in fact I have a subsequence with $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) leq sum_{k=1}^infty 1/2^k = 1 < infty$. But now that I have this subsequence, I'm not clear on how it implies that a limit random variable $X$ exists. I feel like I must be missing something obvious here, but I just can't put it together.






      probability-theory measure-theory convergence cauchy-sequences







      share|cite|improve this question
















      Cauchy convergence of a sequence $X_n$ of random variables in probability implies the existence of an $X$ (finite a.e.), such that $X_n$ converges to $X$ in probability.




      The problem's hint suggests constructing a subsequence $n_k$ so that $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) < infty$, and this I have accomplished, showing that in fact I have a subsequence with $sum_{k=1}^infty Pleft(|X_{n_{k-1}} - X_{n_k}| > 1/2^kright) leq sum_{k=1}^infty 1/2^k = 1 < infty$. But now that I have this subsequence, I'm not clear on how it implies that a limit random variable $X$ exists. I feel like I must be missing something obvious here, but I just can't put it together.






      probability-theory measure-theory convergence cauchy-sequences




      probability-theory measure-theory convergence cauchy-sequences






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 11 '18 at 13:08









      Davide Giraudo

      125k16150260




      125k16150260










      asked Oct 6 '14 at 15:22









      Xindaris

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          You have $sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$, we are going to show almost surely ${X_{n_k}}$ is a Cauchy sequence in $mathbb{R}$ (or $mathbb{C}$ and any other complete metric space).



          It's easy to see
          $$Esum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}} = sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$$



          which implies $sum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}}$ is finite almost surely. This is to say that almost surely $exists N(omega)$ such that for all $k > N(omega)$, we have $|X_{n_{k-1}} - X_{n_k}| leq dfrac{1}{2^k}$



          For any $epsilon >0$, take $K$ such that $dfrac{1}{2^K} < epsilon$ and $M = max{K, N(omega)}$. Then for any $l>k >M$, we have



          $$|X_{n_l} - X_{n_k}| leq sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| leq sum_{i=k+1}^l dfrac{1}{2^i} leq sum_{i=k+1}^{infty} dfrac{1}{2^i} = frac{1}{2^k} < frac{1}{2^K} <epsilon.$$



          So we see ${X_{n_k}}$ are almost surely a Cauchy sequence, define $X$ as its almost surely limit, of course $X$ is almost surely finite.



          Then use the fact
          begin{align}
          P(|X_n - X| > epsilon) &<Pleft(|X_n - X_{n_k}| + |X_{n_k} - X| >epsilonright) \
          &< Pleft(|X_n - X_{n_k}| > frac{epsilon}{2}) + P(|X_{n_k} - X|>dfrac{epsilon}{2}right)
          end{align}



          and $X_{n}$'s Cauchy convergence in probability and $X_{n_k} to X$ almost surely to conclude.






          share|cite|improve this answer































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            By Borel-Cantelli Lemma $$ Pleft(cap_{i=1}^infty cup_{k=i}^inftyleft{mid X_{n_{k-1}} - X_{n_k}mid>2^{-k}right}right) =0, $$ so for all $omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(omega)$. So $X_{n_k}$ converges almost surely to $X$.






            share|cite|improve this answer





















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              2 Answers
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              You have $sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$, we are going to show almost surely ${X_{n_k}}$ is a Cauchy sequence in $mathbb{R}$ (or $mathbb{C}$ and any other complete metric space).



              It's easy to see
              $$Esum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}} = sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$$



              which implies $sum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}}$ is finite almost surely. This is to say that almost surely $exists N(omega)$ such that for all $k > N(omega)$, we have $|X_{n_{k-1}} - X_{n_k}| leq dfrac{1}{2^k}$



              For any $epsilon >0$, take $K$ such that $dfrac{1}{2^K} < epsilon$ and $M = max{K, N(omega)}$. Then for any $l>k >M$, we have



              $$|X_{n_l} - X_{n_k}| leq sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| leq sum_{i=k+1}^l dfrac{1}{2^i} leq sum_{i=k+1}^{infty} dfrac{1}{2^i} = frac{1}{2^k} < frac{1}{2^K} <epsilon.$$



              So we see ${X_{n_k}}$ are almost surely a Cauchy sequence, define $X$ as its almost surely limit, of course $X$ is almost surely finite.



              Then use the fact
              begin{align}
              P(|X_n - X| > epsilon) &<Pleft(|X_n - X_{n_k}| + |X_{n_k} - X| >epsilonright) \
              &< Pleft(|X_n - X_{n_k}| > frac{epsilon}{2}) + P(|X_{n_k} - X|>dfrac{epsilon}{2}right)
              end{align}



              and $X_{n}$'s Cauchy convergence in probability and $X_{n_k} to X$ almost surely to conclude.






              share|cite|improve this answer




























                4














                You have $sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$, we are going to show almost surely ${X_{n_k}}$ is a Cauchy sequence in $mathbb{R}$ (or $mathbb{C}$ and any other complete metric space).



                It's easy to see
                $$Esum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}} = sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$$



                which implies $sum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}}$ is finite almost surely. This is to say that almost surely $exists N(omega)$ such that for all $k > N(omega)$, we have $|X_{n_{k-1}} - X_{n_k}| leq dfrac{1}{2^k}$



                For any $epsilon >0$, take $K$ such that $dfrac{1}{2^K} < epsilon$ and $M = max{K, N(omega)}$. Then for any $l>k >M$, we have



                $$|X_{n_l} - X_{n_k}| leq sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| leq sum_{i=k+1}^l dfrac{1}{2^i} leq sum_{i=k+1}^{infty} dfrac{1}{2^i} = frac{1}{2^k} < frac{1}{2^K} <epsilon.$$



                So we see ${X_{n_k}}$ are almost surely a Cauchy sequence, define $X$ as its almost surely limit, of course $X$ is almost surely finite.



                Then use the fact
                begin{align}
                P(|X_n - X| > epsilon) &<Pleft(|X_n - X_{n_k}| + |X_{n_k} - X| >epsilonright) \
                &< Pleft(|X_n - X_{n_k}| > frac{epsilon}{2}) + P(|X_{n_k} - X|>dfrac{epsilon}{2}right)
                end{align}



                and $X_{n}$'s Cauchy convergence in probability and $X_{n_k} to X$ almost surely to conclude.






                share|cite|improve this answer


























                  4












                  4








                  4






                  You have $sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$, we are going to show almost surely ${X_{n_k}}$ is a Cauchy sequence in $mathbb{R}$ (or $mathbb{C}$ and any other complete metric space).



                  It's easy to see
                  $$Esum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}} = sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$$



                  which implies $sum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}}$ is finite almost surely. This is to say that almost surely $exists N(omega)$ such that for all $k > N(omega)$, we have $|X_{n_{k-1}} - X_{n_k}| leq dfrac{1}{2^k}$



                  For any $epsilon >0$, take $K$ such that $dfrac{1}{2^K} < epsilon$ and $M = max{K, N(omega)}$. Then for any $l>k >M$, we have



                  $$|X_{n_l} - X_{n_k}| leq sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| leq sum_{i=k+1}^l dfrac{1}{2^i} leq sum_{i=k+1}^{infty} dfrac{1}{2^i} = frac{1}{2^k} < frac{1}{2^K} <epsilon.$$



                  So we see ${X_{n_k}}$ are almost surely a Cauchy sequence, define $X$ as its almost surely limit, of course $X$ is almost surely finite.



                  Then use the fact
                  begin{align}
                  P(|X_n - X| > epsilon) &<Pleft(|X_n - X_{n_k}| + |X_{n_k} - X| >epsilonright) \
                  &< Pleft(|X_n - X_{n_k}| > frac{epsilon}{2}) + P(|X_{n_k} - X|>dfrac{epsilon}{2}right)
                  end{align}



                  and $X_{n}$'s Cauchy convergence in probability and $X_{n_k} to X$ almost surely to conclude.






                  share|cite|improve this answer














                  You have $sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$, we are going to show almost surely ${X_{n_k}}$ is a Cauchy sequence in $mathbb{R}$ (or $mathbb{C}$ and any other complete metric space).



                  It's easy to see
                  $$Esum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}} = sum_{k=1}^infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < infty$$



                  which implies $sum_{k=1}^infty 1_{{|X_{n_{k-1}} - X_{n_k}| > 1/2^k}}$ is finite almost surely. This is to say that almost surely $exists N(omega)$ such that for all $k > N(omega)$, we have $|X_{n_{k-1}} - X_{n_k}| leq dfrac{1}{2^k}$



                  For any $epsilon >0$, take $K$ such that $dfrac{1}{2^K} < epsilon$ and $M = max{K, N(omega)}$. Then for any $l>k >M$, we have



                  $$|X_{n_l} - X_{n_k}| leq sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| leq sum_{i=k+1}^l dfrac{1}{2^i} leq sum_{i=k+1}^{infty} dfrac{1}{2^i} = frac{1}{2^k} < frac{1}{2^K} <epsilon.$$



                  So we see ${X_{n_k}}$ are almost surely a Cauchy sequence, define $X$ as its almost surely limit, of course $X$ is almost surely finite.



                  Then use the fact
                  begin{align}
                  P(|X_n - X| > epsilon) &<Pleft(|X_n - X_{n_k}| + |X_{n_k} - X| >epsilonright) \
                  &< Pleft(|X_n - X_{n_k}| > frac{epsilon}{2}) + P(|X_{n_k} - X|>dfrac{epsilon}{2}right)
                  end{align}



                  and $X_{n}$'s Cauchy convergence in probability and $X_{n_k} to X$ almost surely to conclude.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 11 '18 at 13:10









                  Davide Giraudo

                  125k16150260




                  125k16150260










                  answered Oct 6 '14 at 15:59









                  Petite Etincelle

                  12.3k12147




                  12.3k12147























                      3














                      By Borel-Cantelli Lemma $$ Pleft(cap_{i=1}^infty cup_{k=i}^inftyleft{mid X_{n_{k-1}} - X_{n_k}mid>2^{-k}right}right) =0, $$ so for all $omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(omega)$. So $X_{n_k}$ converges almost surely to $X$.






                      share|cite|improve this answer


























                        3














                        By Borel-Cantelli Lemma $$ Pleft(cap_{i=1}^infty cup_{k=i}^inftyleft{mid X_{n_{k-1}} - X_{n_k}mid>2^{-k}right}right) =0, $$ so for all $omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(omega)$. So $X_{n_k}$ converges almost surely to $X$.






                        share|cite|improve this answer
























                          3












                          3








                          3






                          By Borel-Cantelli Lemma $$ Pleft(cap_{i=1}^infty cup_{k=i}^inftyleft{mid X_{n_{k-1}} - X_{n_k}mid>2^{-k}right}right) =0, $$ so for all $omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(omega)$. So $X_{n_k}$ converges almost surely to $X$.






                          share|cite|improve this answer












                          By Borel-Cantelli Lemma $$ Pleft(cap_{i=1}^infty cup_{k=i}^inftyleft{mid X_{n_{k-1}} - X_{n_k}mid>2^{-k}right}right) =0, $$ so for all $omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(omega)$. So $X_{n_k}$ converges almost surely to $X$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Oct 6 '14 at 16:30









                          ir7

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