For random variables $(X_k)_{k=1}^infty$ in $mathbb{R}$, find $(c_k)_{k=1}^infty$ such that $P(lim...











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I'm stuck with the following problem:



Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$



With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
$$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
$$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.



How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?










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    I'm stuck with the following problem:



    Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
    $$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$



    With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
    $$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
    and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
    $$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
    If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.



    How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?










    share|cite|improve this question
























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      I'm stuck with the following problem:



      Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
      $$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$



      With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
      $$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
      and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
      $$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
      If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.



      How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?










      share|cite|improve this question













      I'm stuck with the following problem:



      Let $(X_k)_{k=1}^infty$ be a sequence of real-valued random variables. Show that there is a real sequence $(c_k)_{k=1}^infty$ such that
      $$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}=0right)=1.$$



      With the typical approach for proving almost sure convergence through the Borel-Cantelli-lemma you arrive at
      $$Pleft(lim_{krightarrowinfty}frac{X_k}{c_k}neq 0right)leq sum_{n=1}^infty Pleft(limsup_{krightarrowinfty}|X_k|>frac{c_k}{n}right)$$
      and it would be sufficient to find $(c_k)_{k=1}^infty$ such that
      $$sum_{k=1}^infty Pleft(|X_k|>frac{c_k}{n}right)<infty.$$
      If the $X_k$ were identically distributed, I think a construction of the form $c_k=kcdot q_{1-1/(k^2)}$ could work, where $q_alpha$ is an $alpha$-quantile of $|X_1|$, but I can't do this in the general case.



      How can I solve the general case (no further assumption about the $X_k$)? Is Borel-Cantelli even a good approach here?







      probability convergence borel-cantelli-lemmas






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      asked Nov 15 at 8:59









      Mau314

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          If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].






          share|cite|improve this answer























          • Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
            – Mau314
            Nov 15 at 9:21










          • I think you have a small typo: it should be $c_k/2^k$ in the first set
            – mathworker21
            Nov 15 at 10:51










          • @mathworker21 Thanks. You are absolutely right.
            – Kavi Rama Murthy
            Nov 15 at 11:49











          Your Answer





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          If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].






          share|cite|improve this answer























          • Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
            – Mau314
            Nov 15 at 9:21










          • I think you have a small typo: it should be $c_k/2^k$ in the first set
            – mathworker21
            Nov 15 at 10:51










          • @mathworker21 Thanks. You are absolutely right.
            – Kavi Rama Murthy
            Nov 15 at 11:49















          up vote
          2
          down vote













          If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].






          share|cite|improve this answer























          • Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
            – Mau314
            Nov 15 at 9:21










          • I think you have a small typo: it should be $c_k/2^k$ in the first set
            – mathworker21
            Nov 15 at 10:51










          • @mathworker21 Thanks. You are absolutely right.
            – Kavi Rama Murthy
            Nov 15 at 11:49













          up vote
          2
          down vote










          up vote
          2
          down vote









          If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].






          share|cite|improve this answer














          If $P{|X_k| >frac {c_k} {2^{k}}} <frac 1 {2^{k}}$ then $sum_k P{|X_k| >frac 1 {2^{k}}c_k} <infty$ and Borel Cantelli Lemma tells you that $frac {X_k} {c_k} to 0$ almost surely. You can always find $c_k$'s satisfying this condition, right? [All you need is the fact that $P{|X_k| >t} to 0$ as $t to infty$].







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 15 at 11:49

























          answered Nov 15 at 9:04









          Kavi Rama Murthy

          42.1k31751




          42.1k31751












          • Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
            – Mau314
            Nov 15 at 9:21










          • I think you have a small typo: it should be $c_k/2^k$ in the first set
            – mathworker21
            Nov 15 at 10:51










          • @mathworker21 Thanks. You are absolutely right.
            – Kavi Rama Murthy
            Nov 15 at 11:49


















          • Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
            – Mau314
            Nov 15 at 9:21










          • I think you have a small typo: it should be $c_k/2^k$ in the first set
            – mathworker21
            Nov 15 at 10:51










          • @mathworker21 Thanks. You are absolutely right.
            – Kavi Rama Murthy
            Nov 15 at 11:49
















          Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
          – Mau314
          Nov 15 at 9:21




          Yes, that's in spirit what I wanted to indicate with this quantile-construction (just that I compensate $n$ with $k$ rather than $2^k$) and now I don't know why I thought there's a problem with that. Thanks.
          – Mau314
          Nov 15 at 9:21












          I think you have a small typo: it should be $c_k/2^k$ in the first set
          – mathworker21
          Nov 15 at 10:51




          I think you have a small typo: it should be $c_k/2^k$ in the first set
          – mathworker21
          Nov 15 at 10:51












          @mathworker21 Thanks. You are absolutely right.
          – Kavi Rama Murthy
          Nov 15 at 11:49




          @mathworker21 Thanks. You are absolutely right.
          – Kavi Rama Murthy
          Nov 15 at 11:49


















           

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