Convergence of sequence of uniforms












1












$begingroup$


Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.



I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$



Then:



$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$



As n goes to infinity, I think I get the following function:



$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?



To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.






probability convergence random-variables weak-convergence uniform-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
    $endgroup$
    – Henry
    Jan 18 at 16:24


















1












$begingroup$


Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.



I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$



Then:



$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$



As n goes to infinity, I think I get the following function:



$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?



To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.






probability convergence random-variables weak-convergence uniform-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
    $endgroup$
    – Henry
    Jan 18 at 16:24
















1












1








1





$begingroup$


Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.



I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$



Then:



$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$



As n goes to infinity, I think I get the following function:



$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?



To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.






probability convergence random-variables weak-convergence uniform-distribution







share|cite|improve this question











$endgroup$




Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.



I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$



Then:



$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$



As n goes to infinity, I think I get the following function:



$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?



To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.






probability convergence random-variables weak-convergence uniform-distribution




probability convergence random-variables weak-convergence uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 13:40









Davide Giraudo

127k16151265




127k16151265










asked Jan 18 at 16:11









qcc101qcc101

627213




627213












  • $begingroup$
    Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
    $endgroup$
    – Henry
    Jan 18 at 16:24




















  • $begingroup$
    Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
    $endgroup$
    – Henry
    Jan 18 at 16:24


















$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24






$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24












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Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.



Actually, the convergence $X_nto 0$ holds almost surely hence in probability
hence in distribution.






share|cite|improve this answer











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    $begingroup$

    Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.



    Actually, the convergence $X_nto 0$ holds almost surely hence in probability
    hence in distribution.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.



      Actually, the convergence $X_nto 0$ holds almost surely hence in probability
      hence in distribution.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.



        Actually, the convergence $X_nto 0$ holds almost surely hence in probability
        hence in distribution.






        share|cite|improve this answer











        $endgroup$



        Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.



        Actually, the convergence $X_nto 0$ holds almost surely hence in probability
        hence in distribution.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 24 at 10:21

























        answered Jan 19 at 10:53









        Davide GiraudoDavide Giraudo

        127k16151265




        127k16151265






























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