Limit distribution of infinite sum of Bernoulli random variables











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I know that the finite sum of Bernoulli i.i.d. random variables is a binomial distribution, but what is the distribution of $$lim_{n to infty}sum_{k=1}^{n} frac{x_k}{2^k}$$ where $x_k$ is a Bernoulli random variable with parameter $frac12$?






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  • Are the random variables $x_k$ supposed to be independent?
    – r.e.s.
    May 6 '15 at 0:00






  • 1




    It should be noted that the limit is suggestively written in the form $0.x_1x_2x_3...$ as the base-$2$ representation of a random real number in the unit interval.
    – r.e.s.
    May 6 '15 at 0:05






  • 1




    A related question is "If $x$ is uniformly distributed on the unit interval, then what is the distribution of the binary digits of $x$?"
    – r.e.s.
    May 6 '15 at 0:14















up vote
3
down vote

favorite
7












I know that the finite sum of Bernoulli i.i.d. random variables is a binomial distribution, but what is the distribution of $$lim_{n to infty}sum_{k=1}^{n} frac{x_k}{2^k}$$ where $x_k$ is a Bernoulli random variable with parameter $frac12$?






probability probability-theory probability-distributions







share|cite|improve this question
























  • Are the random variables $x_k$ supposed to be independent?
    – r.e.s.
    May 6 '15 at 0:00






  • 1




    It should be noted that the limit is suggestively written in the form $0.x_1x_2x_3...$ as the base-$2$ representation of a random real number in the unit interval.
    – r.e.s.
    May 6 '15 at 0:05






  • 1




    A related question is "If $x$ is uniformly distributed on the unit interval, then what is the distribution of the binary digits of $x$?"
    – r.e.s.
    May 6 '15 at 0:14













up vote
3
down vote

favorite
7









up vote
3
down vote

favorite
7






7





I know that the finite sum of Bernoulli i.i.d. random variables is a binomial distribution, but what is the distribution of $$lim_{n to infty}sum_{k=1}^{n} frac{x_k}{2^k}$$ where $x_k$ is a Bernoulli random variable with parameter $frac12$?






probability probability-theory probability-distributions







share|cite|improve this question















I know that the finite sum of Bernoulli i.i.d. random variables is a binomial distribution, but what is the distribution of $$lim_{n to infty}sum_{k=1}^{n} frac{x_k}{2^k}$$ where $x_k$ is a Bernoulli random variable with parameter $frac12$?






probability probability-theory probability-distributions




probability probability-theory probability-distributions






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share|cite|improve this question













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edited Feb 25 '16 at 9:41









tonytonov

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asked May 5 '15 at 22:43









user165576

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  • Are the random variables $x_k$ supposed to be independent?
    – r.e.s.
    May 6 '15 at 0:00






  • 1




    It should be noted that the limit is suggestively written in the form $0.x_1x_2x_3...$ as the base-$2$ representation of a random real number in the unit interval.
    – r.e.s.
    May 6 '15 at 0:05






  • 1




    A related question is "If $x$ is uniformly distributed on the unit interval, then what is the distribution of the binary digits of $x$?"
    – r.e.s.
    May 6 '15 at 0:14


















  • Are the random variables $x_k$ supposed to be independent?
    – r.e.s.
    May 6 '15 at 0:00






  • 1




    It should be noted that the limit is suggestively written in the form $0.x_1x_2x_3...$ as the base-$2$ representation of a random real number in the unit interval.
    – r.e.s.
    May 6 '15 at 0:05






  • 1




    A related question is "If $x$ is uniformly distributed on the unit interval, then what is the distribution of the binary digits of $x$?"
    – r.e.s.
    May 6 '15 at 0:14
















Are the random variables $x_k$ supposed to be independent?
– r.e.s.
May 6 '15 at 0:00




Are the random variables $x_k$ supposed to be independent?
– r.e.s.
May 6 '15 at 0:00




1




1




It should be noted that the limit is suggestively written in the form $0.x_1x_2x_3...$ as the base-$2$ representation of a random real number in the unit interval.
– r.e.s.
May 6 '15 at 0:05




It should be noted that the limit is suggestively written in the form $0.x_1x_2x_3...$ as the base-$2$ representation of a random real number in the unit interval.
– r.e.s.
May 6 '15 at 0:05




1




1




A related question is "If $x$ is uniformly distributed on the unit interval, then what is the distribution of the binary digits of $x$?"
– r.e.s.
May 6 '15 at 0:14




A related question is "If $x$ is uniformly distributed on the unit interval, then what is the distribution of the binary digits of $x$?"
– r.e.s.
May 6 '15 at 0:14










2 Answers
2






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7
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An approach using generating functions is possible.



Define $Y_k = X_k/2^k$, $k = 1, 2, ldots$. Then $$M_{Y_k}(t) = operatorname{E}[e^{tY_k}] = e^0 Pr[Y_k = 0] + e^{t/2^k} Pr[Y_k = 2^{-k}] = frac{e^{t/2^k} + 1}{2}.$$ Then defining $$S_n = sum_{k=1}^n Y_k,$$ we find that the MGF for $S_n$ is $$begin{align*} M_{S_n}(t) &= operatorname{E}left[prod_{k=1}^n e^{tY_k}right] = prod_{k=1}^n M_{Y_k}(t) \ &= 2^{-n} prod_{k=1}^n (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^n} - 1)(e^{t/2^n} + 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^{n-1}} - 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{e^t - 1}{2^n(e^{t/2^n} - 1)} .end{align*}$$ Taking the limit as $n to infty$, we easily get $$M_{S_infty}(t) = frac{e^t-1}{t},$$ which is the MGF of a $operatorname{Uniform}(0,1)$ distribution.






share|cite|improve this answer





















  • Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
    – tonytonov
    Feb 25 '16 at 9:26


















up vote
2
down vote













Let $$S_n = sum_{k=1}^n 2^{-k}X_k.$$ As $$0leqslant S_nleqslant sum_{k=1}^n 2^{-k} = 1$$ almost surely for all $n$, this inequality holds in the limit, so by dominated convergence,
$$
begin{align*}
mathbb Eleft[lim_{ntoinfty}S_nright] &= lim_{ntoinfty}mathbb E[S_n]\
&= lim_{ntoinfty}mathbb Eleft[ sum_{k=1}^n 2^{-k}X_kright]\
&= lim_{ntoinfty}sum_{k=1}^n 2^{-k}mathbb E[X_k]\
&= lim_{ntoinfty}sum_{k=1}^n 2^{-k+1}\
&= frac12sum_{k=1}^infty 2^{-k}\
&= frac12.
end{align*}
$$
However, the limiting distribution is actually continuous. For each $n$, $S_n$ is uniformly distributed over
$$E_n=left{sum_{kin S}2^{-k} : Ssubset{1,2,ldots,n}right}. $$
Given an $n$, and some $omegain E_n$, we have for $mgeqslant n$
$$mathbb P(S_m = omega)=2^{-m}.$$
Hence
$$lim_{mtoinfty} P(S_m = omega)=0.$$
@gmath suggests in this answer that the limiting distribution is actually uniform on the interval $(0,1)$.






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    2 Answers
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    2 Answers
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    active

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    up vote
    7
    down vote













    An approach using generating functions is possible.



    Define $Y_k = X_k/2^k$, $k = 1, 2, ldots$. Then $$M_{Y_k}(t) = operatorname{E}[e^{tY_k}] = e^0 Pr[Y_k = 0] + e^{t/2^k} Pr[Y_k = 2^{-k}] = frac{e^{t/2^k} + 1}{2}.$$ Then defining $$S_n = sum_{k=1}^n Y_k,$$ we find that the MGF for $S_n$ is $$begin{align*} M_{S_n}(t) &= operatorname{E}left[prod_{k=1}^n e^{tY_k}right] = prod_{k=1}^n M_{Y_k}(t) \ &= 2^{-n} prod_{k=1}^n (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^n} - 1)(e^{t/2^n} + 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^{n-1}} - 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{e^t - 1}{2^n(e^{t/2^n} - 1)} .end{align*}$$ Taking the limit as $n to infty$, we easily get $$M_{S_infty}(t) = frac{e^t-1}{t},$$ which is the MGF of a $operatorname{Uniform}(0,1)$ distribution.






    share|cite|improve this answer





















    • Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
      – tonytonov
      Feb 25 '16 at 9:26















    up vote
    7
    down vote













    An approach using generating functions is possible.



    Define $Y_k = X_k/2^k$, $k = 1, 2, ldots$. Then $$M_{Y_k}(t) = operatorname{E}[e^{tY_k}] = e^0 Pr[Y_k = 0] + e^{t/2^k} Pr[Y_k = 2^{-k}] = frac{e^{t/2^k} + 1}{2}.$$ Then defining $$S_n = sum_{k=1}^n Y_k,$$ we find that the MGF for $S_n$ is $$begin{align*} M_{S_n}(t) &= operatorname{E}left[prod_{k=1}^n e^{tY_k}right] = prod_{k=1}^n M_{Y_k}(t) \ &= 2^{-n} prod_{k=1}^n (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^n} - 1)(e^{t/2^n} + 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^{n-1}} - 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{e^t - 1}{2^n(e^{t/2^n} - 1)} .end{align*}$$ Taking the limit as $n to infty$, we easily get $$M_{S_infty}(t) = frac{e^t-1}{t},$$ which is the MGF of a $operatorname{Uniform}(0,1)$ distribution.






    share|cite|improve this answer





















    • Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
      – tonytonov
      Feb 25 '16 at 9:26













    up vote
    7
    down vote










    up vote
    7
    down vote









    An approach using generating functions is possible.



    Define $Y_k = X_k/2^k$, $k = 1, 2, ldots$. Then $$M_{Y_k}(t) = operatorname{E}[e^{tY_k}] = e^0 Pr[Y_k = 0] + e^{t/2^k} Pr[Y_k = 2^{-k}] = frac{e^{t/2^k} + 1}{2}.$$ Then defining $$S_n = sum_{k=1}^n Y_k,$$ we find that the MGF for $S_n$ is $$begin{align*} M_{S_n}(t) &= operatorname{E}left[prod_{k=1}^n e^{tY_k}right] = prod_{k=1}^n M_{Y_k}(t) \ &= 2^{-n} prod_{k=1}^n (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^n} - 1)(e^{t/2^n} + 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^{n-1}} - 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{e^t - 1}{2^n(e^{t/2^n} - 1)} .end{align*}$$ Taking the limit as $n to infty$, we easily get $$M_{S_infty}(t) = frac{e^t-1}{t},$$ which is the MGF of a $operatorname{Uniform}(0,1)$ distribution.






    share|cite|improve this answer












    An approach using generating functions is possible.



    Define $Y_k = X_k/2^k$, $k = 1, 2, ldots$. Then $$M_{Y_k}(t) = operatorname{E}[e^{tY_k}] = e^0 Pr[Y_k = 0] + e^{t/2^k} Pr[Y_k = 2^{-k}] = frac{e^{t/2^k} + 1}{2}.$$ Then defining $$S_n = sum_{k=1}^n Y_k,$$ we find that the MGF for $S_n$ is $$begin{align*} M_{S_n}(t) &= operatorname{E}left[prod_{k=1}^n e^{tY_k}right] = prod_{k=1}^n M_{Y_k}(t) \ &= 2^{-n} prod_{k=1}^n (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^n} - 1)(e^{t/2^n} + 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{2^{-n}}{e^{t/2^n} - 1} (e^{t/2^{n-1}} - 1) prod_{k=1}^{n-1} (e^{t/2^k} + 1) \ &= frac{e^t - 1}{2^n(e^{t/2^n} - 1)} .end{align*}$$ Taking the limit as $n to infty$, we easily get $$M_{S_infty}(t) = frac{e^t-1}{t},$$ which is the MGF of a $operatorname{Uniform}(0,1)$ distribution.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 6 '15 at 0:52









    heropup

    62.3k65998




    62.3k65998












    • Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
      – tonytonov
      Feb 25 '16 at 9:26


















    • Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
      – tonytonov
      Feb 25 '16 at 9:26
















    Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
    – tonytonov
    Feb 25 '16 at 9:26




    Great answer, and the proof can also be generalized for arbitrary base $b$. Do you know any source I can link to in my paper?
    – tonytonov
    Feb 25 '16 at 9:26










    up vote
    2
    down vote













    Let $$S_n = sum_{k=1}^n 2^{-k}X_k.$$ As $$0leqslant S_nleqslant sum_{k=1}^n 2^{-k} = 1$$ almost surely for all $n$, this inequality holds in the limit, so by dominated convergence,
    $$
    begin{align*}
    mathbb Eleft[lim_{ntoinfty}S_nright] &= lim_{ntoinfty}mathbb E[S_n]\
    &= lim_{ntoinfty}mathbb Eleft[ sum_{k=1}^n 2^{-k}X_kright]\
    &= lim_{ntoinfty}sum_{k=1}^n 2^{-k}mathbb E[X_k]\
    &= lim_{ntoinfty}sum_{k=1}^n 2^{-k+1}\
    &= frac12sum_{k=1}^infty 2^{-k}\
    &= frac12.
    end{align*}
    $$
    However, the limiting distribution is actually continuous. For each $n$, $S_n$ is uniformly distributed over
    $$E_n=left{sum_{kin S}2^{-k} : Ssubset{1,2,ldots,n}right}. $$
    Given an $n$, and some $omegain E_n$, we have for $mgeqslant n$
    $$mathbb P(S_m = omega)=2^{-m}.$$
    Hence
    $$lim_{mtoinfty} P(S_m = omega)=0.$$
    @gmath suggests in this answer that the limiting distribution is actually uniform on the interval $(0,1)$.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Let $$S_n = sum_{k=1}^n 2^{-k}X_k.$$ As $$0leqslant S_nleqslant sum_{k=1}^n 2^{-k} = 1$$ almost surely for all $n$, this inequality holds in the limit, so by dominated convergence,
      $$
      begin{align*}
      mathbb Eleft[lim_{ntoinfty}S_nright] &= lim_{ntoinfty}mathbb E[S_n]\
      &= lim_{ntoinfty}mathbb Eleft[ sum_{k=1}^n 2^{-k}X_kright]\
      &= lim_{ntoinfty}sum_{k=1}^n 2^{-k}mathbb E[X_k]\
      &= lim_{ntoinfty}sum_{k=1}^n 2^{-k+1}\
      &= frac12sum_{k=1}^infty 2^{-k}\
      &= frac12.
      end{align*}
      $$
      However, the limiting distribution is actually continuous. For each $n$, $S_n$ is uniformly distributed over
      $$E_n=left{sum_{kin S}2^{-k} : Ssubset{1,2,ldots,n}right}. $$
      Given an $n$, and some $omegain E_n$, we have for $mgeqslant n$
      $$mathbb P(S_m = omega)=2^{-m}.$$
      Hence
      $$lim_{mtoinfty} P(S_m = omega)=0.$$
      @gmath suggests in this answer that the limiting distribution is actually uniform on the interval $(0,1)$.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        Let $$S_n = sum_{k=1}^n 2^{-k}X_k.$$ As $$0leqslant S_nleqslant sum_{k=1}^n 2^{-k} = 1$$ almost surely for all $n$, this inequality holds in the limit, so by dominated convergence,
        $$
        begin{align*}
        mathbb Eleft[lim_{ntoinfty}S_nright] &= lim_{ntoinfty}mathbb E[S_n]\
        &= lim_{ntoinfty}mathbb Eleft[ sum_{k=1}^n 2^{-k}X_kright]\
        &= lim_{ntoinfty}sum_{k=1}^n 2^{-k}mathbb E[X_k]\
        &= lim_{ntoinfty}sum_{k=1}^n 2^{-k+1}\
        &= frac12sum_{k=1}^infty 2^{-k}\
        &= frac12.
        end{align*}
        $$
        However, the limiting distribution is actually continuous. For each $n$, $S_n$ is uniformly distributed over
        $$E_n=left{sum_{kin S}2^{-k} : Ssubset{1,2,ldots,n}right}. $$
        Given an $n$, and some $omegain E_n$, we have for $mgeqslant n$
        $$mathbb P(S_m = omega)=2^{-m}.$$
        Hence
        $$lim_{mtoinfty} P(S_m = omega)=0.$$
        @gmath suggests in this answer that the limiting distribution is actually uniform on the interval $(0,1)$.






        share|cite|improve this answer














        Let $$S_n = sum_{k=1}^n 2^{-k}X_k.$$ As $$0leqslant S_nleqslant sum_{k=1}^n 2^{-k} = 1$$ almost surely for all $n$, this inequality holds in the limit, so by dominated convergence,
        $$
        begin{align*}
        mathbb Eleft[lim_{ntoinfty}S_nright] &= lim_{ntoinfty}mathbb E[S_n]\
        &= lim_{ntoinfty}mathbb Eleft[ sum_{k=1}^n 2^{-k}X_kright]\
        &= lim_{ntoinfty}sum_{k=1}^n 2^{-k}mathbb E[X_k]\
        &= lim_{ntoinfty}sum_{k=1}^n 2^{-k+1}\
        &= frac12sum_{k=1}^infty 2^{-k}\
        &= frac12.
        end{align*}
        $$
        However, the limiting distribution is actually continuous. For each $n$, $S_n$ is uniformly distributed over
        $$E_n=left{sum_{kin S}2^{-k} : Ssubset{1,2,ldots,n}right}. $$
        Given an $n$, and some $omegain E_n$, we have for $mgeqslant n$
        $$mathbb P(S_m = omega)=2^{-m}.$$
        Hence
        $$lim_{mtoinfty} P(S_m = omega)=0.$$
        @gmath suggests in this answer that the limiting distribution is actually uniform on the interval $(0,1)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:19









        Community♦

        1




        1










        answered May 5 '15 at 23:24









        Math1000

        18.8k31645




        18.8k31645






























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            Mont Emei

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            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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