Suppose $x_i\sim N(0,1)$ is a i.i.d. random variable from a normal distribution and $X=|x_1| |x_2|...|x_n|$ . Now define $R:= Ln(X)$,


Then, I'd like to know if there is a distribution for $R$ or a way to calculate the mean and the variance of $R$ with another well known distribution.

  • 1
    $\begingroup$ Since Normal variables can be negative, do you want to define e.g. $\ln(-x)=\ln x+\pi i$ for $x>0$, or change the problem to $R=\sum_i\ln|X_i|$? $\endgroup$
    – J.G.
    Feb 18, 2020 at 16:58
  • $\begingroup$ Sorry about that, indeed it is the absolute value of $x$. $\endgroup$ Feb 18, 2020 at 17:01
  • $\begingroup$ Do you need necessarily another known distribution? Since the $x_i$'s are i.i.d., $\mathbb{E}[R]=n \mathbb{E}[\log |x_1|]$, $\operatorname{Var}[R]=n \operatorname{Var}[\log |x_1|]$. Then, this boils down to compute a couple integrals (not necessarily nice, admittedly). $\endgroup$
    – Clement C.
    Feb 18, 2020 at 17:07
  • $\begingroup$ So, to calculate $ \mathbb{E}[\log |x_1|]$, what distribution I can use? $\endgroup$ Feb 18, 2020 at 17:20
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    $\begingroup$ Use distribution of $x_1$ which is given. $\endgroup$
    – NCh
    Feb 18, 2020 at 17:24

1 Answer 1


The reader's familiarity is assumed with the Euler-Mascheroni constant and the values at $s=\tfrac12$ of the Gamma, digamma and trigamma functions. Warning: you should double-check all my arithmetic, but this code suggests I'm right.

Let $I_k:=\int_{\Bbb R}\frac{1}{\sqrt{2\pi}}\exp(-\tfrac12x^2)\ln^k|x|dx$, so $R$ has mean $nI_1$ and $n(I_2-I_1^2)$. But$$I_k=\sqrt{\frac{2}{\pi}}\int_0^\infty\exp(-\tfrac12x^2)\ln^kxdx=\frac{1}{\sqrt{\pi}}\int_0^\infty y^{-1/2}\exp(-y)\ln^k(\sqrt{2}y^{1/2})dy.$$Next define$$J_l:=\int_0^\infty y^{-1/2}\exp(-y)\ln^lydy=\Gamma^{(l)}(\tfrac12),$$so special values of the gamma, digamma and trigamma functions give$$J_0=\sqrt{\pi},\,\frac{J_1}{J_0}=-\gamma-\ln4,\,\frac{J_2}{J_0}=(\gamma+\ln4)^2+\frac{\pi^2}{2}.$$Hence$$I_1=\frac{1}{\sqrt{\pi}}\left(\ln\sqrt{2}\Gamma(\tfrac12)+\tfrac12\Gamma^\prime(\tfrac12)\right)=\frac{-\gamma-\ln 2}{2}$$and$$I_2=\frac{1}{\sqrt{\pi}}(\ln^2\sqrt{2}\Gamma(\tfrac12)+\ln\sqrt{2}\Gamma^\prime(\tfrac12)+\frac14\Gamma^{\prime\prime}(\tfrac12))=\frac{\gamma\ln 2}{2}+\frac{\ln^22}{4}+\frac{\gamma^2}{4}+\frac{\pi^2}{8}.$$In particular,$$I_2-I_1^2=\frac{\pi^2}{8}.$$


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