What would be the distribution (p.d.f.) of the following ratio?

$$z = \frac{x_{1}}{|x_{1}|^2 + |x_{2}|^2 + ... + |x_{M}|^2}$$

where $x_{i} \sim \mathcal{CN}(0,a), \forall i$ and $a > 1$. As can be seen, the denominator follows a Chi-square distribution with $2M$ degrees of freedom as $x_{i}$ are i.i.d. R.V.s.

Remark 1: I've run some simulations in Matlab, as shown in the pictures below for $M$ = 10, and the resulting distribution has a bell-shaped histogram. Could it be that the resulting distribution follows one of these distributions: Gaussian/Cauchy/Student's-t?

Remark 2: This is a link to the Matlab/Octave script used to plot the pictures below. Matlab/Octave simulation of the histogram of z

Histogram of the Real part of z

Histogram of the Imaginary part of z

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    $\begingroup$ $z$ itself converges to $0$. You probably mean $Z=Mz$. Since $\sum_{i=1}^M |x_i|^2/M\to 2$, you get that $Z$ converges to a (constant multiple of) a standard complex Gaussian. If you mean to ask something else, please update. $\endgroup$ Jan 7 '18 at 15:27
  • $\begingroup$ @oferzeitouni, thanks for your comment. Your are right, $z$ indeed is missing the $M$ term as you suggested. If you check my simulation script, you will see that $M$ is part of the $z$ ratio. Do you think now with $M$ it is easier to find the p.d.f of this ratio? $\endgroup$ Jan 7 '18 at 15:52
  • $\begingroup$ Asymptotically it converges to Gaussian, as I indicated. I do not know about corrections that eventually vanish in the limit. $\endgroup$ Jan 7 '18 at 16:24
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    $\begingroup$ @FelipeAugustodeFigueiredo : I suggest you restore the original definition of $z$ -- but, just in the captions to your pictures, replace z by Mz. $\endgroup$ Jan 7 '18 at 17:31
  • $\begingroup$ @FelipeAugustodeFigueiredo : Of course, the mere rescaling by a constant factor (say $k>0$) does not make it easier or harder to find the pdf: the pdf of $kY$ is $\frac1k\,f_Y(\frac yk)$, where $f_Y(y)$ is the pdf of a random variable $Y$. $\endgroup$ Jan 7 '18 at 17:43

Let $n:=M$. Since you say "the denominator follows a Chi-square distribution with $2M$ degrees of freedom", it appears that you assume the $x_i$'s to be iid standard complex Gaussian; anyway, if you need $a>1$, then you can do a simple rescaling.

Thus, let us assume indeed that the $x_i$'s are iid standard complex Gaussian. It is then clear that the distribution of $z$ in $\mathbb C=\mathbb R^2$ is rotation invariant. So, it is enough to find the distribution of $|z|^2$, which is the same as that of \begin{equation*} R:=\frac X{(X+Y)^2}, \end{equation*} where $X$ and $Y$ are independent random variables such that $X\sim\text{Gamma}(1,2)$ and $Y\sim\text{Gamma}(n-1,2)$.

Considering now the transformation from $(X,Y)$ to $(R,S)$ with $S:=X$ (say) and using the corresponding formula -- see e.g. formula (41) in Transformations, we obtain the joint density of $(R,S)$, say $f_{R,S}$. Then the density of $R$ is given by the formula \begin{equation*} f_R(r)=\int_0^\infty f_{R,S}(r,s)\,ds. \end{equation*} Using in this integral the substitution $v=\sqrt{rs}$, we find \begin{equation} f_R(r)=\frac1{(n-2)!2^n r^{n+1}}\,\int_0^1 v^n(1-v)^{n-2}\exp\Big\{-\frac v{2r}\Big\}\,dv \tag{1} \end{equation} for $r>0$, with $f_R(r)=0$ for $r\le0$.

Using the binomial expansion for $(1-v)^{n-2}$ and then integrating by parts, we can express the latter integral as a finite sum (with the number of summands depending on $n$) of elementary functions. For instance, for $n=3$ and $r>0$ \begin{equation*} f_R(r)=\exp\Big\{-\frac1{2r}\Big\}\left(\frac{1}{2 r^2}+96 r+\frac{6}{r}+36\right)-96 r+12; \end{equation*} here is the graph of this function:

enter image description here

However, by the dominated convergence theorem, the integral in (1) converges to $c_n:=\int_0^1 v^n(1-v)^{n-2}\,dv=n!(n-2)!/(2n-1)!$ as $r\to\infty$, and so, $f_R(r)\sim \frac{c_n}{(n-2)!2^n r^{n+1}}$ as $r\to\infty$. So, the tail of the distribution of $|z|^2$ is power-like, which shows that "the resulting distribution [of $z$]" is not Gaussian.

Addition in response to the modification of the OP's original question: There are many bell-shaped distributions, in addition to the family of Student $t_d$ distributions with $d$ degrees of freedom -- which includes the Cauchy distribution (for $d=1$) and, as a limit case (for $d=\infty$), the standard normal distribution. Let $U$ denote the real part of $z$: $U:=\Re z$. You seem to be asking if the distribution of $U$ is $t_d$ for some $d\in(0,\infty]$. The answer to this is no, even if you allow rescaling of the distribution.

Indeed, from formula (1), using the Fubini theorem, it is easy to find all the moments of $|z|^2$: \begin{equation*} E|z|^{2m}=ER^m=\frac{m!(n-m-1)!}{2^m(n+m-1)!} \end{equation*} for $m=1,\dots,n-1$. So, the kurtosis of $|z|$ is \begin{equation} \frac{E|z|^4}{(E|z|^2)^2}=\frac{2n(n-1)}{(n+1)(n-2)}, \tag{2} \end{equation} for $n>2$. On the other hand, if the distribution of $U=\Re z$ were $t_d$ for some $d\in(0,\infty]$, then $|z|^2$ would equal $T_1^2+T_2^2$ in distribution, where $T,T_1,T_2$ are iid $t_d$, and so, the kurtosis of $|z|$ would be \begin{equation} \frac{E(T_1^2+T_2^2)^2}{(ET_1^2+ET_2^2)^2}=\frac12+\frac12\,\frac{ET^4}{(ET^2)^2}=\frac{2d-5}{d-4}, \tag{3} \end{equation} for $d>4$. On the other hand, the pdf $f_T(t)$ of $t_d$ is $\sim C_d t^{-d-1}$ (as $t\to\infty$), where $C_d$ does not depend on $t$. So, the pdf $f_{T^2}(u)$ is $\sim \frac12\,C_d u^{-d/2-1}$ (as $u\to\infty$). So, $f_{T_1^2+T_2^2}(u)\sim C_d u^{-d/2-1}$ (as $u\to\infty$). As mentioned above, $f_{|z|^2}(r)=f_R(r)\sim \frac{c_n}{(n-2)!2^n r^{n+1}}$ as $r\to\infty$. So, if the distribution of $U=\Re z$ were $t_d$ for some $d\in(0,\infty]$, then the exponent $-d/2-1$ would be equal $-(n+1)$, so that $d=2n$ -- but then the kurtoses in (2) and (3) would be unequal, for any $n>2$. Moreover, since the kurtosis is not affected by rescaling, the distribution of $U=\Re z$ cannot be $t_d$ for any $d\in(0,\infty]$ -- even if the distribution is rescaled by some positive real constant factor.

Addition in response to a comment by the OP: To find the pdf of $U=\Re z$, one can again use the referenced transformation technique. Indeed, without loss of generality, $U=\sqrt R\,C$, where $C:=\cos\Theta$ and $\Theta$ is a r.v. independent of $R$ and uniformly distributed in the interval $[0,2\pi]$ or, by symmetry, in $[0,\pi]$, so that for the pdf $f_C$ of $C$ one has $f_C(c)=\frac1\pi\,/\sqrt{1-c^2}$ for $-1<c<1$. Considering now the transformation from $(R,C)$ to $(U,W)$, where $U=\sqrt R\,C$ and $W=R$, we have the following rather complicated expression for the pdf of $U$: \begin{equation*} f_U(u)=\int_{u^2}^\infty \frac1{\pi\sqrt w}f_R(w)\frac{dw}{\sqrt{1-u^2/w}} =\frac1\pi\,\int_{u^2}^\infty \frac{f_R(w)\,dw}{\sqrt{w-u^2}} \end{equation*} for all real $u$, where $f_R$ is given by (1). Using the Fubini theorem and Mathematica for the calculation of the iterated double integral, we get \begin{equation*} f_U(u)=\exp\Big\{{-\frac{1}{4 u^2}}\Big\} \,\frac{\left(8 n u^2-1\right) I_n\left(\frac{1}{4 u^2}\right)+I_{n+1}\left(\frac{1}{4 u^2}\right)}{4 |u|^3}, \end{equation*} where $I_n(z)$ is the modified Bessel function of the first kind. Below is the (bell-shaped) graph of $f_U=f_{\Re z}$ for $n=10$, which looks quite similar to your histogram. I am also including the corresponding Mathematica code:

enter image description here

  • $\begingroup$ I followed your answer and it seems OK, however I've run some simulations in Matlab and the distribution indeed is bell-shaped. It might not be a Gaussian distribution but it has a bell like distribution. Could it be a Cauchy or student's-t distribution? I will try to post the histogram of my simulation and the script I have. BTW, your assumption is correct, x_{i} are i.i.d R.V.s $\endgroup$ Jan 7 '18 at 12:55
  • $\begingroup$ I have just updated the question as I forgot to add the term $M$ to $z$ ratio. This error was pointed out by user "ofer zeitouni". $\endgroup$ Jan 7 '18 at 15:54
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    $\begingroup$ The distribution is not Student's $t$ either; in particular, it is not Cauchy -- even after rescaling. But to prove stuff such as something is not something else may be pretty nasty business. Also, your changing the question a number of times makes it hard to keep up with you. $\endgroup$ Jan 7 '18 at 17:18
  • $\begingroup$ Sorry for the change, but I have only added the constant $M$ as it is in line to what I really want to know, the other things (figures and script) are only to show what I have already said since the original post. $\endgroup$ Jan 7 '18 at 18:45
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    $\begingroup$ The graph of the pdf of $R$ is different from the graph of the pdf of $\Re z$ -- because $R$ is not $\Re z$. Rather, $R$ equals $|z|^2$ in distribution, as I noted. $\endgroup$ Jan 7 '18 at 18:51

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