Return to Question

This is a follow-up question to Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such explore types of convergence which are not encapsulated by the Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Now clearly \begin{eqnarray} &&\int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta =\\ && \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) \right) d\delta + O(k)\\ &&= \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} - \frac{(-\imath k)^\delta}{\delta} \right) d\delta + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \left.\left( Ei(-\delta) - Ei(-A \delta)\right)\right|_0^\infty+ \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \log(-A) + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k) \end{eqnarray} where $A=-\log(-\imath k)= \imath \pi/2 - \log(k)$. Now we have checked numericaly that the integral in the middle above decays monotonically when $k\rightarrow 0$. Since now $\log(-A) \rightarrow \log(\log(1/k))$ when $k\rightarrow 0$ the claim is established.

Define $c_n:=\sqrt{n\log(\log(n))}$. Indeed we have: \begin{eqnarray} \log\left( \kappa_{S_n}(k)\right) &=& -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &=& \frac{1+4 \log(2)-4 \log(2-2 \log(k)+\log(n) + \log(\log(\log(n))))}{8\log(\log(n))} k^2 + O(\frac{k^3}{\sqrt{n \cdot \log(\log(n))}}) \end{eqnarray} where in the second line I simply used the Series[] command in Mathematica to obtain the expansion . Now we can clearly see that: \begin{equation} \lim_{n\rightarrow \infty} \log\left( \kappa_{S_n}(k)\right) = -\frac{k^2}{2} \end{equation} as it should be.

This is a follow-up question to Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such explore types of convergence which are not encapsulated by the Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f^{-1}(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Now clearly \begin{eqnarray} &&\int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta =\\ && \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) \right) d\delta + O(k)\\ &&= \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} - \frac{(-\imath k)^\delta}{\delta} \right) d\delta + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \left.\left( Ei(-\delta) - Ei(-A \delta)\right)\right|_0^\infty+ \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \log(-A) + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k) \end{eqnarray} where $A=-\log(-\imath k)= \imath \pi/2 - \log(k)$. Now we have checked numericaly that the integral in the middle above decays monotonically when $k\rightarrow 0$. Since now $\log(-A) = \log(-\imath \pi/2 + \log(k))= \log(-\imath \pi/2-\log(1/k)) \rightarrow \log(-\log(1/k)) = -\imath \pi/2 + \log(\log(1/k)) \rightarrow \log(\log(1/k))$ when $k\rightarrow 0$ the claim is established.

Define $c_n:=\sqrt{n\log(\log(n))}$. Indeed we have: \begin{eqnarray} &&\log\left( \kappa_{S_n}(k)\right) =\\ && -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ e^{\imath \frac{k}{c_n}}(1+\imath \frac{k}{2 c_n}) - \frac{1}{2} \frac{k^2}{c_n^2} \log(1+\log(c_n)-\log(k))\right]\\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ 1+\imath \frac{3}{2} \frac{k}{c_n} - \frac{k^2}{c_n^2} + O(\frac{k^3}{c_n^3}) - \frac{1}{2} \frac{k^2}{c_n^2} \log(1+\log(c_n)-\log(k)) \right]\\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + \imath k \frac{3}{2} \frac{n}{c_n} + \left(\frac{1}{8} - \frac{1}{2} \log(1+\log(c_n)-\log(k))\right) \frac{k^2}{c_n^2} n + O(\frac{k^3}{c_n^3})\\ &&= \left(\frac{1}{8} - \frac{1}{2} \log(1+\log(c_n)-\log(k))\right) \frac{k^2}{c_n^2} n + O(\frac{k^3}{c_n^3}) \end{eqnarray} Now for the statistic to be properly normalized we have to have: \begin{equation} \lim_{n\rightarrow \infty} \frac{n}{c_n^2} \log(\log(c_n)) = 1 \end{equation} which is indeed the case as one can readily check by plugging the definition of $c_n$ into the lhs.

This is a follow-up question to Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such to formulate a counterexample to Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation}

Unfortunately for the time being I am unable to prove or disprove that claim above. However if we assume that the above is true then at least we can make sure that our test statistic is properly normalized.

This is a follow-up question to Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such explore types of convergence which are not encapsulated by the Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation} Indeed by using the integral representation of the Gamma function along with integration by parts we quickly establish the following identity: \begin{eqnarray} (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) = \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) + \sum\limits_{n=1}^\infty \frac{(\imath k)^n}{n!}\cdot \frac{e^{n-\delta}}{\delta-n} \end{eqnarray}

Now clearly \begin{eqnarray} &&\int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta =\\ && \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) \right) d\delta + O(k)\\ &&= \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} - \frac{(-\imath k)^\delta}{\delta} \right) d\delta + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \left.\left( Ei(-\delta) - Ei(-A \delta)\right)\right|_0^\infty+ \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \log(-A) + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k) \end{eqnarray} where $A=-\log(-\imath k)= \imath \pi/2 - \log(k)$. Now we have checked numericaly that the integral in the middle above decays monotonically when $k\rightarrow 0$. Since now $\log(-A) \rightarrow \log(\log(1/k))$ when $k\rightarrow 0$ the claim is established.

Now we check that our test statistic is properly normalized.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation} Unfortunately for the time being I am unable to prove or disprove that claim above. However if we assume that the above is true then at least we can make sure that our test statistic is properly normalized. Define $c_n:=\sqrt{n\log(\log(n))}$. Indeed we have: \begin{eqnarray} \log\left( \kappa_{S_n}(k)\right) &=& -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &=& \frac{1+4 \log(2)-4 \log(2-2 \log(k)+\log(n) + \log(\log(\log(n))))}{8\log(\log(n))} k^2 + O(\frac{k^3}{\sqrt{n \cdot \log(\log(n))}}) \end{eqnarray} where in the second line I simply used the Series[] command in Mathematica to obtain the expansion . Now we can clearly see that: \begin{equation} \lim_{n\rightarrow \infty} \log\left( \kappa_{S_n}(k)\right) = -\frac{k^2}{2} \end{equation} as it should be.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation}

Unfortunately for the time being I am unable to prove or disprove that claim above. However if we assume that the above is true then at least we can make sure that our test statistic is properly normalized.

Define $c_n:=\sqrt{n\log(\log(n))}$. Indeed we have: \begin{eqnarray} \log\left( \kappa_{S_n}(k)\right) &=& -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &=& \frac{1+4 \log(2)-4 \log(2-2 \log(k)+\log(n) + \log(\log(\log(n))))}{8\log(\log(n))} k^2 + O(\frac{k^3}{\sqrt{n \cdot \log(\log(n))}}) \end{eqnarray} where in the second line I simply used the Series[] command in Mathematica to obtain the expansion . Now we can clearly see that: \begin{equation} \lim_{n\rightarrow \infty} \log\left( \kappa_{S_n}(k)\right) = -\frac{k^2}{2} \end{equation} as it should be.