Expected value of the sample median from a folded normal distribution

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Suppose $X_1, ldots, X_n sim N(0,sigma^2)$ are iid. Find the expected value of $M$, the median of $vert X_1 vert, ldots vert X_n vert$



What I have so far:



The density of $vert X_i vert$ is given by:
$$f_vert X vert(x) = frac2sqrt2pi sigma^2expleft(-fracx^22sigma^2right)$$



If $n = 2m - 1$, the median $M$ is the $m^th$ order statistic of $vert X_1 vert, ldots, vert X_n vert$. This has density
$$f_vert X vert_(m) (x) = fracn!(m-1)!(n-m)! f_vert X vert(x) (F_vert X vert)^m-1(1 - F_vert X vert)^n-m$$
where $F_vert X vert(x) = F_X(x) - F_X(-x)$ is the cdf of $vert X vert$. So the expected value of the median is
$$mathbb E M = int_0^infty x f_vert X vert_(m) (x) dx$$



Is there a better way to approach this, rather than having to compute this integral?




probability-distributions normal-distribution median




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    up vote
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    Suppose $X_1, ldots, X_n sim N(0,sigma^2)$ are iid. Find the expected value of $M$, the median of $vert X_1 vert, ldots vert X_n vert$



    What I have so far:



    The density of $vert X_i vert$ is given by:
    $$f_vert X vert(x) = frac2sqrt2pi sigma^2expleft(-fracx^22sigma^2right)$$



    If $n = 2m - 1$, the median $M$ is the $m^th$ order statistic of $vert X_1 vert, ldots, vert X_n vert$. This has density
    $$f_vert X vert_(m) (x) = fracn!(m-1)!(n-m)! f_vert X vert(x) (F_vert X vert)^m-1(1 - F_vert X vert)^n-m$$
    where $F_vert X vert(x) = F_X(x) - F_X(-x)$ is the cdf of $vert X vert$. So the expected value of the median is
    $$mathbb E M = int_0^infty x f_vert X vert_(m) (x) dx$$



    Is there a better way to approach this, rather than having to compute this integral?




    probability-distributions normal-distribution median




    share|cite|improve this question





















      up vote
      5
      down vote

      favorite









      up vote
      5
      down vote

      favorite











      Suppose $X_1, ldots, X_n sim N(0,sigma^2)$ are iid. Find the expected value of $M$, the median of $vert X_1 vert, ldots vert X_n vert$



      What I have so far:



      The density of $vert X_i vert$ is given by:
      $$f_vert X vert(x) = frac2sqrt2pi sigma^2expleft(-fracx^22sigma^2right)$$



      If $n = 2m - 1$, the median $M$ is the $m^th$ order statistic of $vert X_1 vert, ldots, vert X_n vert$. This has density
      $$f_vert X vert_(m) (x) = fracn!(m-1)!(n-m)! f_vert X vert(x) (F_vert X vert)^m-1(1 - F_vert X vert)^n-m$$
      where $F_vert X vert(x) = F_X(x) - F_X(-x)$ is the cdf of $vert X vert$. So the expected value of the median is
      $$mathbb E M = int_0^infty x f_vert X vert_(m) (x) dx$$



      Is there a better way to approach this, rather than having to compute this integral?




      probability-distributions normal-distribution median




      share|cite|improve this question











      Suppose $X_1, ldots, X_n sim N(0,sigma^2)$ are iid. Find the expected value of $M$, the median of $vert X_1 vert, ldots vert X_n vert$



      What I have so far:



      The density of $vert X_i vert$ is given by:
      $$f_vert X vert(x) = frac2sqrt2pi sigma^2expleft(-fracx^22sigma^2right)$$



      If $n = 2m - 1$, the median $M$ is the $m^th$ order statistic of $vert X_1 vert, ldots, vert X_n vert$. This has density
      $$f_vert X vert_(m) (x) = fracn!(m-1)!(n-m)! f_vert X vert(x) (F_vert X vert)^m-1(1 - F_vert X vert)^n-m$$
      where $F_vert X vert(x) = F_X(x) - F_X(-x)$ is the cdf of $vert X vert$. So the expected value of the median is
      $$mathbb E M = int_0^infty x f_vert X vert_(m) (x) dx$$



      Is there a better way to approach this, rather than having to compute this integral?





      probability-distributions normal-distribution median





      share|cite|improve this question










      share|cite|improve this question




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