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  First, we note $E ( \eta ) = \sqrt { n } E ( | \rho | )$ . Then $\begin{array} { r } { E ( | \rho | ) = \int _ { 0 } ^ { 1 } d \rho \rho \frac { 2 } { \sqrt { \pi } } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } ( 1 - \rho ^ { 2 } ) ^ { \frac { n - 3 } { 2 } } = } \end{array}$ $\frac { 2 } { \sqrt { \pi } } \ast \frac { 1 } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) }$ (substitute $u \ : = \ : \rho ^ { 2 }$ and use $\Gamma ( x + 1 ) = x \Gamma ( x )$ ). Lemma
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- two gives the $n \infty$ limit. $\begin{array} { r l r } { \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } \ \approx \ \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \sqrt { \frac { n } { 2 } } \left[ 1 + \frac { 0 . 5 * 0 . 5 } { 2 ( n / 2 ) } \right] \ = } \end{array}$ q 2π 1 + 54 ∗ 1n  + O(1/n2)
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  $$
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  { V a r } ( \eta ) = \frac { 1 } { n } \left( 1 - \frac { 1 } { \pi } \right) + { \cal O } ( 1 / n ^ { 2 } )
 
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  First, we note $E ( \eta ) = \sqrt { n } E ( | \rho | )$ . Then $\begin{array} { r } { E ( | \rho | ) = \int _ { 0 } ^ { 1 } d \rho \rho \frac { 2 } { \sqrt { \pi } } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } ( 1 - \rho ^ { 2 } ) ^ { \frac { n - 3 } { 2 } } = } \end{array}$ $\frac { 2 } { \sqrt { \pi } } \ast \frac { 1 } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) }$ (substitute $u \ : = \ : \rho ^ { 2 }$ and use $\Gamma ( x + 1 ) = x \Gamma ( x )$ ). Lemma
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+ two gives the $n \infty$ limit. $\begin{array} { r l r } { \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } \ \approx \ \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \sqrt { \frac { n } { 2 } } \left[ 1 + \frac { 0 . 5 * 0 . 5 } { 2 ( n / 2 ) } \right] \ = } \end{array}$ q 2π 1 + 54 ∗ 1n  + O(1/n2)
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  $$
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  { V a r } ( \eta ) = \frac { 1 } { n } \left( 1 - \frac { 1 } { \pi } \right) + { \cal O } ( 1 / n ^ { 2 } )