Central Limit Theorem for symmetric Beta Distribution












2












$begingroup$



Prove that for $X_nsim Beta(n,n)$ $$2sqrt{2n}left( X_n-frac{1}{2}right)Rightarrowmathscr N(0,1)$$ in distribution as $nrightarrowinfty$.




Let $Y_n$ and $Z_n$ be $Gamma(n,1)$ independent random variables. Then using the facts, that $$X_n=frac{Y_n}{Y_n+Z_n},$$ and that both $Y_n$ and $Z_n$ are sum of independent $Exp(1)$ random variables, I can apply CLT to those, so both $Y_n$ and $Y_n+Z_n$ are normally distributed. Then I can use the delta method to show, that $X_n$ is normally distributed also.



That is the part where I get confused. Shall I apply the delta method to $Y_n$ with $g(x)=frac{x}{x+Z_n}$?






probability central-limit-theorem







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  • $begingroup$
    This is also related to this interesting question: The distribution of the inner product of two independent vectors uniformly distributed on the unit sphere in high dimensions: stats.stackexchange.com/questions/85916/…
    $endgroup$
    – passerby51
    Jan 2 at 3:01


















2












$begingroup$



Prove that for $X_nsim Beta(n,n)$ $$2sqrt{2n}left( X_n-frac{1}{2}right)Rightarrowmathscr N(0,1)$$ in distribution as $nrightarrowinfty$.




Let $Y_n$ and $Z_n$ be $Gamma(n,1)$ independent random variables. Then using the facts, that $$X_n=frac{Y_n}{Y_n+Z_n},$$ and that both $Y_n$ and $Z_n$ are sum of independent $Exp(1)$ random variables, I can apply CLT to those, so both $Y_n$ and $Y_n+Z_n$ are normally distributed. Then I can use the delta method to show, that $X_n$ is normally distributed also.



That is the part where I get confused. Shall I apply the delta method to $Y_n$ with $g(x)=frac{x}{x+Z_n}$?






probability central-limit-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is also related to this interesting question: The distribution of the inner product of two independent vectors uniformly distributed on the unit sphere in high dimensions: stats.stackexchange.com/questions/85916/…
    $endgroup$
    – passerby51
    Jan 2 at 3:01
















2












2








2


2



$begingroup$



Prove that for $X_nsim Beta(n,n)$ $$2sqrt{2n}left( X_n-frac{1}{2}right)Rightarrowmathscr N(0,1)$$ in distribution as $nrightarrowinfty$.




Let $Y_n$ and $Z_n$ be $Gamma(n,1)$ independent random variables. Then using the facts, that $$X_n=frac{Y_n}{Y_n+Z_n},$$ and that both $Y_n$ and $Z_n$ are sum of independent $Exp(1)$ random variables, I can apply CLT to those, so both $Y_n$ and $Y_n+Z_n$ are normally distributed. Then I can use the delta method to show, that $X_n$ is normally distributed also.



That is the part where I get confused. Shall I apply the delta method to $Y_n$ with $g(x)=frac{x}{x+Z_n}$?






probability central-limit-theorem







share|cite|improve this question









$endgroup$





Prove that for $X_nsim Beta(n,n)$ $$2sqrt{2n}left( X_n-frac{1}{2}right)Rightarrowmathscr N(0,1)$$ in distribution as $nrightarrowinfty$.




Let $Y_n$ and $Z_n$ be $Gamma(n,1)$ independent random variables. Then using the facts, that $$X_n=frac{Y_n}{Y_n+Z_n},$$ and that both $Y_n$ and $Z_n$ are sum of independent $Exp(1)$ random variables, I can apply CLT to those, so both $Y_n$ and $Y_n+Z_n$ are normally distributed. Then I can use the delta method to show, that $X_n$ is normally distributed also.



That is the part where I get confused. Shall I apply the delta method to $Y_n$ with $g(x)=frac{x}{x+Z_n}$?






probability central-limit-theorem




probability central-limit-theorem






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asked Mar 6 '16 at 13:45









Hodossy SzabolcsHodossy Szabolcs

21416




21416












  • $begingroup$
    This is also related to this interesting question: The distribution of the inner product of two independent vectors uniformly distributed on the unit sphere in high dimensions: stats.stackexchange.com/questions/85916/…
    $endgroup$
    – passerby51
    Jan 2 at 3:01




















  • $begingroup$
    This is also related to this interesting question: The distribution of the inner product of two independent vectors uniformly distributed on the unit sphere in high dimensions: stats.stackexchange.com/questions/85916/…
    $endgroup$
    – passerby51
    Jan 2 at 3:01


















$begingroup$
This is also related to this interesting question: The distribution of the inner product of two independent vectors uniformly distributed on the unit sphere in high dimensions: stats.stackexchange.com/questions/85916/…
$endgroup$
– passerby51
Jan 2 at 3:01






$begingroup$
This is also related to this interesting question: The distribution of the inner product of two independent vectors uniformly distributed on the unit sphere in high dimensions: stats.stackexchange.com/questions/85916/…
$endgroup$
– passerby51
Jan 2 at 3:01












2 Answers
2






active

oldest

votes


















1












$begingroup$

Maybe not the method you need/prefer, but if you define $T_n=2sqrt{2n}(X_n-1/2)$, the pdf of $T_n$ is
$$
f_{T_n}(t)=frac{1}{B(n,n)}int_0^1 dx x^{n-1}(1-x)^{n-1}delta(t-2sqrt{2n}x+sqrt{2n}) ,
$$
where $B(alpha,beta)$ is a Beta function. Therefore
$$
f_{T_n}(t)=frac{mathbf{1}_{-sqrt{2n}<t<sqrt{2n}}}{2sqrt{2n}B(n,n)}left(frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}left(1-frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}to frac{e^{-frac{t^2}{2}}}{sqrt{2 pi }} ,
$$
as $ntoinfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
    $endgroup$
    – user321627
    Oct 24 '16 at 1:53



















0












$begingroup$

One can use the multivariate version of the delta method:



Assume that $a_n to infty$ and $a_n(T_n - mu) stackrel{d}{to} U$ where $T_n,mu,U in mathbb R^d$. Then, under mild regularity assumptions on $f:mathbb R^d to mathbb R$, we have
begin{align*}
sqrt{n} big(f(T_n) - f(mu)big) stackrel{d}{to} J_mu U
end{align*}
where $J_mu$ is the Jacobian matrix of $f$ evaluated at $mu$, that is $ [J_mu]_{ij} = partial f_i / partial x_j ; big|_{x = mu}$.



In this example, let $bar Y_n = Y_n / n$ and $bar Z_n = Z_n/n$. Then, by the multivariate CLT (or arguing by independence from univariate CLT), we have
begin{align*}
sqrt{n} Big[
begin{pmatrix}
bar Y_n \ bar Z_n
end{pmatrix} -
begin{pmatrix}
1 \ 1
end{pmatrix}
Big]
stackrel{d}{to} U :=
begin{pmatrix}
U_1 \ U_2
end{pmatrix} sim N(0,I_2).
end{align*}
Let $f(y,z) = y/(y+z)$. The Jacobian can be identified with a row vector $$Big[frac{z}{(y+z)^2},; frac{-y}{(y+z)^2}Big].$$ Since $X_n = f(bar Y_n, bar Z_n)$, we have by the multivariate delta method
begin{align*}
sqrt{n} big( X_n - f((1,1))big) stackrel{d}{to}
begin{bmatrix}
1/4 & -1/4
end{bmatrix}
begin{bmatrix}
U_1 \ U_2
end{bmatrix} = frac{U_1-U_2}{4} sim N(0, frac18).
end{align*}
Rescaling, we have $sqrt{8n} (X_n - 1/2) stackrel{d}{to} N(0,1)$, the desired result.






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    active

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    1












    $begingroup$

    Maybe not the method you need/prefer, but if you define $T_n=2sqrt{2n}(X_n-1/2)$, the pdf of $T_n$ is
    $$
    f_{T_n}(t)=frac{1}{B(n,n)}int_0^1 dx x^{n-1}(1-x)^{n-1}delta(t-2sqrt{2n}x+sqrt{2n}) ,
    $$
    where $B(alpha,beta)$ is a Beta function. Therefore
    $$
    f_{T_n}(t)=frac{mathbf{1}_{-sqrt{2n}<t<sqrt{2n}}}{2sqrt{2n}B(n,n)}left(frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}left(1-frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}to frac{e^{-frac{t^2}{2}}}{sqrt{2 pi }} ,
    $$
    as $ntoinfty$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
      $endgroup$
      – user321627
      Oct 24 '16 at 1:53
















    1












    $begingroup$

    Maybe not the method you need/prefer, but if you define $T_n=2sqrt{2n}(X_n-1/2)$, the pdf of $T_n$ is
    $$
    f_{T_n}(t)=frac{1}{B(n,n)}int_0^1 dx x^{n-1}(1-x)^{n-1}delta(t-2sqrt{2n}x+sqrt{2n}) ,
    $$
    where $B(alpha,beta)$ is a Beta function. Therefore
    $$
    f_{T_n}(t)=frac{mathbf{1}_{-sqrt{2n}<t<sqrt{2n}}}{2sqrt{2n}B(n,n)}left(frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}left(1-frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}to frac{e^{-frac{t^2}{2}}}{sqrt{2 pi }} ,
    $$
    as $ntoinfty$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
      $endgroup$
      – user321627
      Oct 24 '16 at 1:53














    1












    1








    1





    $begingroup$

    Maybe not the method you need/prefer, but if you define $T_n=2sqrt{2n}(X_n-1/2)$, the pdf of $T_n$ is
    $$
    f_{T_n}(t)=frac{1}{B(n,n)}int_0^1 dx x^{n-1}(1-x)^{n-1}delta(t-2sqrt{2n}x+sqrt{2n}) ,
    $$
    where $B(alpha,beta)$ is a Beta function. Therefore
    $$
    f_{T_n}(t)=frac{mathbf{1}_{-sqrt{2n}<t<sqrt{2n}}}{2sqrt{2n}B(n,n)}left(frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}left(1-frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}to frac{e^{-frac{t^2}{2}}}{sqrt{2 pi }} ,
    $$
    as $ntoinfty$.






    share|cite|improve this answer









    $endgroup$



    Maybe not the method you need/prefer, but if you define $T_n=2sqrt{2n}(X_n-1/2)$, the pdf of $T_n$ is
    $$
    f_{T_n}(t)=frac{1}{B(n,n)}int_0^1 dx x^{n-1}(1-x)^{n-1}delta(t-2sqrt{2n}x+sqrt{2n}) ,
    $$
    where $B(alpha,beta)$ is a Beta function. Therefore
    $$
    f_{T_n}(t)=frac{mathbf{1}_{-sqrt{2n}<t<sqrt{2n}}}{2sqrt{2n}B(n,n)}left(frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}left(1-frac{t+sqrt{2n}}{2sqrt{2n}}right)^{n-1}to frac{e^{-frac{t^2}{2}}}{sqrt{2 pi }} ,
    $$
    as $ntoinfty$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 6 '16 at 14:26









    Pierpaolo VivoPierpaolo Vivo

    5,3812724




    5,3812724












    • $begingroup$
      Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
      $endgroup$
      – user321627
      Oct 24 '16 at 1:53


















    • $begingroup$
      Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
      $endgroup$
      – user321627
      Oct 24 '16 at 1:53
















    $begingroup$
    Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
    $endgroup$
    – user321627
    Oct 24 '16 at 1:53




    $begingroup$
    Can I ask why you defined the pdf as an integral of the beta? Wouldn't that result in the cdf?
    $endgroup$
    – user321627
    Oct 24 '16 at 1:53











    0












    $begingroup$

    One can use the multivariate version of the delta method:



    Assume that $a_n to infty$ and $a_n(T_n - mu) stackrel{d}{to} U$ where $T_n,mu,U in mathbb R^d$. Then, under mild regularity assumptions on $f:mathbb R^d to mathbb R$, we have
    begin{align*}
    sqrt{n} big(f(T_n) - f(mu)big) stackrel{d}{to} J_mu U
    end{align*}
    where $J_mu$ is the Jacobian matrix of $f$ evaluated at $mu$, that is $ [J_mu]_{ij} = partial f_i / partial x_j ; big|_{x = mu}$.



    In this example, let $bar Y_n = Y_n / n$ and $bar Z_n = Z_n/n$. Then, by the multivariate CLT (or arguing by independence from univariate CLT), we have
    begin{align*}
    sqrt{n} Big[
    begin{pmatrix}
    bar Y_n \ bar Z_n
    end{pmatrix} -
    begin{pmatrix}
    1 \ 1
    end{pmatrix}
    Big]
    stackrel{d}{to} U :=
    begin{pmatrix}
    U_1 \ U_2
    end{pmatrix} sim N(0,I_2).
    end{align*}
    Let $f(y,z) = y/(y+z)$. The Jacobian can be identified with a row vector $$Big[frac{z}{(y+z)^2},; frac{-y}{(y+z)^2}Big].$$ Since $X_n = f(bar Y_n, bar Z_n)$, we have by the multivariate delta method
    begin{align*}
    sqrt{n} big( X_n - f((1,1))big) stackrel{d}{to}
    begin{bmatrix}
    1/4 & -1/4
    end{bmatrix}
    begin{bmatrix}
    U_1 \ U_2
    end{bmatrix} = frac{U_1-U_2}{4} sim N(0, frac18).
    end{align*}
    Rescaling, we have $sqrt{8n} (X_n - 1/2) stackrel{d}{to} N(0,1)$, the desired result.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      One can use the multivariate version of the delta method:



      Assume that $a_n to infty$ and $a_n(T_n - mu) stackrel{d}{to} U$ where $T_n,mu,U in mathbb R^d$. Then, under mild regularity assumptions on $f:mathbb R^d to mathbb R$, we have
      begin{align*}
      sqrt{n} big(f(T_n) - f(mu)big) stackrel{d}{to} J_mu U
      end{align*}
      where $J_mu$ is the Jacobian matrix of $f$ evaluated at $mu$, that is $ [J_mu]_{ij} = partial f_i / partial x_j ; big|_{x = mu}$.



      In this example, let $bar Y_n = Y_n / n$ and $bar Z_n = Z_n/n$. Then, by the multivariate CLT (or arguing by independence from univariate CLT), we have
      begin{align*}
      sqrt{n} Big[
      begin{pmatrix}
      bar Y_n \ bar Z_n
      end{pmatrix} -
      begin{pmatrix}
      1 \ 1
      end{pmatrix}
      Big]
      stackrel{d}{to} U :=
      begin{pmatrix}
      U_1 \ U_2
      end{pmatrix} sim N(0,I_2).
      end{align*}
      Let $f(y,z) = y/(y+z)$. The Jacobian can be identified with a row vector $$Big[frac{z}{(y+z)^2},; frac{-y}{(y+z)^2}Big].$$ Since $X_n = f(bar Y_n, bar Z_n)$, we have by the multivariate delta method
      begin{align*}
      sqrt{n} big( X_n - f((1,1))big) stackrel{d}{to}
      begin{bmatrix}
      1/4 & -1/4
      end{bmatrix}
      begin{bmatrix}
      U_1 \ U_2
      end{bmatrix} = frac{U_1-U_2}{4} sim N(0, frac18).
      end{align*}
      Rescaling, we have $sqrt{8n} (X_n - 1/2) stackrel{d}{to} N(0,1)$, the desired result.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        One can use the multivariate version of the delta method:



        Assume that $a_n to infty$ and $a_n(T_n - mu) stackrel{d}{to} U$ where $T_n,mu,U in mathbb R^d$. Then, under mild regularity assumptions on $f:mathbb R^d to mathbb R$, we have
        begin{align*}
        sqrt{n} big(f(T_n) - f(mu)big) stackrel{d}{to} J_mu U
        end{align*}
        where $J_mu$ is the Jacobian matrix of $f$ evaluated at $mu$, that is $ [J_mu]_{ij} = partial f_i / partial x_j ; big|_{x = mu}$.



        In this example, let $bar Y_n = Y_n / n$ and $bar Z_n = Z_n/n$. Then, by the multivariate CLT (or arguing by independence from univariate CLT), we have
        begin{align*}
        sqrt{n} Big[
        begin{pmatrix}
        bar Y_n \ bar Z_n
        end{pmatrix} -
        begin{pmatrix}
        1 \ 1
        end{pmatrix}
        Big]
        stackrel{d}{to} U :=
        begin{pmatrix}
        U_1 \ U_2
        end{pmatrix} sim N(0,I_2).
        end{align*}
        Let $f(y,z) = y/(y+z)$. The Jacobian can be identified with a row vector $$Big[frac{z}{(y+z)^2},; frac{-y}{(y+z)^2}Big].$$ Since $X_n = f(bar Y_n, bar Z_n)$, we have by the multivariate delta method
        begin{align*}
        sqrt{n} big( X_n - f((1,1))big) stackrel{d}{to}
        begin{bmatrix}
        1/4 & -1/4
        end{bmatrix}
        begin{bmatrix}
        U_1 \ U_2
        end{bmatrix} = frac{U_1-U_2}{4} sim N(0, frac18).
        end{align*}
        Rescaling, we have $sqrt{8n} (X_n - 1/2) stackrel{d}{to} N(0,1)$, the desired result.






        share|cite|improve this answer









        $endgroup$



        One can use the multivariate version of the delta method:



        Assume that $a_n to infty$ and $a_n(T_n - mu) stackrel{d}{to} U$ where $T_n,mu,U in mathbb R^d$. Then, under mild regularity assumptions on $f:mathbb R^d to mathbb R$, we have
        begin{align*}
        sqrt{n} big(f(T_n) - f(mu)big) stackrel{d}{to} J_mu U
        end{align*}
        where $J_mu$ is the Jacobian matrix of $f$ evaluated at $mu$, that is $ [J_mu]_{ij} = partial f_i / partial x_j ; big|_{x = mu}$.



        In this example, let $bar Y_n = Y_n / n$ and $bar Z_n = Z_n/n$. Then, by the multivariate CLT (or arguing by independence from univariate CLT), we have
        begin{align*}
        sqrt{n} Big[
        begin{pmatrix}
        bar Y_n \ bar Z_n
        end{pmatrix} -
        begin{pmatrix}
        1 \ 1
        end{pmatrix}
        Big]
        stackrel{d}{to} U :=
        begin{pmatrix}
        U_1 \ U_2
        end{pmatrix} sim N(0,I_2).
        end{align*}
        Let $f(y,z) = y/(y+z)$. The Jacobian can be identified with a row vector $$Big[frac{z}{(y+z)^2},; frac{-y}{(y+z)^2}Big].$$ Since $X_n = f(bar Y_n, bar Z_n)$, we have by the multivariate delta method
        begin{align*}
        sqrt{n} big( X_n - f((1,1))big) stackrel{d}{to}
        begin{bmatrix}
        1/4 & -1/4
        end{bmatrix}
        begin{bmatrix}
        U_1 \ U_2
        end{bmatrix} = frac{U_1-U_2}{4} sim N(0, frac18).
        end{align*}
        Rescaling, we have $sqrt{8n} (X_n - 1/2) stackrel{d}{to} N(0,1)$, the desired result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 2:55









        passerby51passerby51

        2,1411018




        2,1411018






























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            Are they similar matrix

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            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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