Suppose $x_i\sim N(0,1)_\mathbb{C}$ is a i.i.d. random variable from a complex distribution and

$$P=2^n|x_1| |x_2|...|x_n|$$

Rewriting $x_i=r_ie^{i\theta}$; that is,

$$P=2^n r_1 r_2...r_n,~r\geq 0.$$

Now define $L:= Ln(P)$,


(i) Question:I'd like to know if there is a pdf for $L$.

For $r_i$ there is the Rayleigh distribution, which arises in the case of random complex numbers whose real and imaginary components are i.i.d.with equal variance and zero mean (in this case suppose $u,v\sim N(0,\frac{1}{2})$ and $x=u+iv$, where $x_i\sim N(0,1)_\mathbb{C}$), i.e. , by a change of variable to polar coordinate in the pdf of the complex gaussian distribution.

  • $\begingroup$ Of course $L$ has a distribution, so I presume you are asking for some kind of formula. It looks like a sum of Gumbel distributions to me, which means a formula will be quite complicated, but apparently there exist good approximations. What kind of an answer are you looking for, then? $\endgroup$
    – whuber
    Commented Feb 20, 2020 at 19:55
  • 1
    $\begingroup$ If there is a close pdf for $L$, but if not, an approximation looks good to me. I will be looking into your references and try to find something. Thanks. $\endgroup$ Commented Feb 20, 2020 at 20:04

1 Answer 1


Letting $Y_i \equiv - 2 \ln R_i$ and noting that $R_i \sim \text{Rayleigh}(1)$ you have the CDF:

$$\begin{equation} \begin{aligned} F_Y(y) \equiv \mathbb{P}(Y_i \leqslant y) &= \mathbb{P}(-Y_i/2 \geqslant -y/2) \\[6pt] &= \mathbb{P}(e^{-Y_i/2} \geqslant e^{-y/2}) \\[6pt] &= \mathbb{P}(R_i \geqslant e^{- y/2}) \\[6pt] &= \exp \Big( -\frac{e^{- y}}{2} \Big). \\[6pt] \end{aligned} \end{equation}$$

You therefore obtain the density function:

$$\begin{equation} \begin{aligned} f_Y(y) = \frac{d F_Y}{dy} (y) &= \frac{1}{2} \cdot \exp \Big( -y -\frac{e^{- y}}{2} \Big) \\[6pt] &= \text{Gumbel}(y| \mu = \ln 2, \beta = 1). \\[6pt] \end{aligned} \end{equation}$$

The mean and variance of this distribution are $\mathbb{E}(Y_i) = \ln 2 + \gamma$ and $\mathbb{V}(Y_i) = \pi^2/6$, where $\gamma$ denotes the Euler-Mascheroni constant. Your random variable $L_n$ can be written as an affine function of the sample mean of the random variables $Y_1,...,Y_n$ as follows:

$$L_n = n \Big( \ln 2 - \frac{1}{2} \cdot \bar{Y}_n \Big).$$

It follows that this random variable has mean:

$$\begin{equation} \begin{aligned} \mathbb{E}(L_n) &= n \Big( \ln 2 - \frac{1}{2} \cdot \mathbb{E}(\bar{Y}_n) \Big) \\[6pt] &= n \Big( \ln 2 - \frac{1}{2} \cdot (\ln 2 + \gamma) \Big) \\[6pt] &= \frac{\ln 2 - \gamma}{2} \cdot n, \\[6pt] &\approx 0.05796576 \cdot n \\[6pt] \end{aligned} \end{equation}$$

and variance:

$$\begin{equation} \begin{aligned} \mathbb{V}(L_n) &= \frac{n^2}{4} \cdot \mathbb{V}(\bar{Y}_n) \quad \quad \quad \quad \quad \ \ \\[6pt] &= \frac{n^2}{4} \cdot \frac{\pi^2}{6n} \\[6pt] &= \frac{\pi^2}{24} \cdot n \\[6pt] &\approx 0.4112335 \cdot n. \\[6pt] \end{aligned} \end{equation}$$

As whuber points out in the comments, the exact distribution of $L$ is a convolution of Gumbel distributions (see e.g., Nadarajah 2006). This exact distribution is complicated, but we can apply the classical central limit theorem to obtain an approximation using the normal distribution. For large values of $n$ we have:

$$L_n \overset{\text{Approx}}{\sim} \text{N} \Big( \text{Mean} = \frac{\ln 2 - \gamma}{2} \cdot n, \text{Var} = \frac{\pi^2}{24} \cdot n \Big).$$

  • $\begingroup$ This is a very pretty answer, thanks! $\endgroup$ Commented Feb 24, 2020 at 14:51

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