Skip to main content
deleted 47 characters in body
Source Link
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and an explicit asymptotic variance $\tilde\sigma/\sqrt n$$\tilde\sigma^2/n$; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given. More specifically, $\tilde\sigma=\sqrt {{\tilde\sigma}^2}$ and \begin{equation} \tilde\sigma^2=EL(V)^2=E(2\xi-\xi^2+\mu^T\Sigma^{-1}\mu)^2, \end{equation} where \begin{equation} V:=(Y-EY,(Y-EY)(Y-EY)^T-I),\quad Y:=\Sigma^{-1/2}z_1, \end{equation} \begin{equation} L(x_1,x_2):=2x_1^T\,EY-EY^T\,x_2\,EY, \end{equation} \begin{equation} \xi:=EY^T\,(Y-EY)=\mu^T\Sigma^{-1}(z_1-\mu). \end{equation}

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and an explicit asymptotic variance $\tilde\sigma/\sqrt n$; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given. More specifically, $\tilde\sigma=\sqrt {{\tilde\sigma}^2}$ and \begin{equation} \tilde\sigma^2=EL(V)^2=E(2\xi-\xi^2+\mu^T\Sigma^{-1}\mu)^2, \end{equation} where \begin{equation} V:=(Y-EY,(Y-EY)(Y-EY)^T-I),\quad Y:=\Sigma^{-1/2}z_1, \end{equation} \begin{equation} L(x_1,x_2):=2x_1^T\,EY-EY^T\,x_2\,EY, \end{equation} \begin{equation} \xi:=EY^T\,(Y-EY)=\mu^T\Sigma^{-1}(z_1-\mu). \end{equation}

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and an explicit asymptotic variance $\tilde\sigma^2/n$; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given. More specifically, \begin{equation} \tilde\sigma^2=EL(V)^2=E(2\xi-\xi^2+\mu^T\Sigma^{-1}\mu)^2, \end{equation} where \begin{equation} V:=(Y-EY,(Y-EY)(Y-EY)^T-I),\quad Y:=\Sigma^{-1/2}z_1, \end{equation} \begin{equation} L(x_1,x_2):=2x_1^T\,EY-EY^T\,x_2\,EY, \end{equation} \begin{equation} \xi:=EY^T\,(Y-EY)=\mu^T\Sigma^{-1}(z_1-\mu). \end{equation}

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.

added 446 characters in body
Source Link
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and a certainan explicit variance;asymptotic variance $\tilde\sigma/\sqrt n$; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given. More specifically, $\tilde\sigma=\sqrt {{\tilde\sigma}^2}$ and \begin{equation} \tilde\sigma^2=EL(V)^2=E(2\xi-\xi^2+\mu^T\Sigma^{-1}\mu)^2, \end{equation} where \begin{equation} V:=(Y-EY,(Y-EY)(Y-EY)^T-I),\quad Y:=\Sigma^{-1/2}z_1, \end{equation} \begin{equation} L(x_1,x_2):=2x_1^T\,EY-EY^T\,x_2\,EY, \end{equation} \begin{equation} \xi:=EY^T\,(Y-EY)=\mu^T\Sigma^{-1}(z_1-\mu). \end{equation}

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and a certain explicit variance; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given.

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and an explicit asymptotic variance $\tilde\sigma/\sqrt n$; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given. More specifically, $\tilde\sigma=\sqrt {{\tilde\sigma}^2}$ and \begin{equation} \tilde\sigma^2=EL(V)^2=E(2\xi-\xi^2+\mu^T\Sigma^{-1}\mu)^2, \end{equation} where \begin{equation} V:=(Y-EY,(Y-EY)(Y-EY)^T-I),\quad Y:=\Sigma^{-1/2}z_1, \end{equation} \begin{equation} L(x_1,x_2):=2x_1^T\,EY-EY^T\,x_2\,EY, \end{equation} \begin{equation} \xi:=EY^T\,(Y-EY)=\mu^T\Sigma^{-1}(z_1-\mu). \end{equation}

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.

Source Link
Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229

If $\mu\ne0$, then the distribution of $R_n$ is asymptotically normal with asymptotic mean $\mu^T\Sigma^{-1}\mu$ and a certain explicit variance; see e.g. Theorem 3.9, page 1018, where a bound on the rate of convergence is also given.

If $\mu=0$, then the distribution of properly normalized $R_n$ is asymptotically chi-squared; see e.g. Theorem 3, page 48.