8

I would like to find the PDF of the real random variable $$z=\frac{AB-CD}{A^2+B^2+C^2+D^2},$$ where $A$, $B$, $C$, and $D$ are independent Gaussian random variables with mean $0$ and variance $1$. Alternatively, the problem can be seen as finding the PDF of $$z=\frac{\Re\{AB^*\}}{\vert A\vert^2+\vert B\vert^2}=\frac{1}{2}\left(\frac{\vert A+B\vert^2}{\vert A\vert^2+\vert B\vert^2}-1\right),$$ where $A$ and $B$ are independent circularly symmetric complex Gaussian random variables with mean $0$ and variance $1$.

After running a Monte Carlo simulation in Matlab, I believe $z$ is uniformly distributed in $[-1/2, 1/2]$, but I cannot prove it. Here is the Matlab code I used:

N = 1e6;
vars = randn(N,4);
z = nan(N,1);
for n = 1:N
    z(n) = (vars(n,1)*vars(n,2)-vars(n,3)*vars(n,4))/norm(vars(n,:))^2;
end

figure; cdfplot(z);
figure; histogram(z);

Any help will be much appreciated!

Matteo Nerini
  • 83

4 Answers4

7

As $-D\sim D$, we have

$$Z=\frac{AB-CD}{A^2+B^2+C^2+D^2} \sim W=\frac{AB+CD}{A^2+B^2+C^2+D^2}.$$

It is enough to show that $W+\frac12\sim \mathcal U (0,1)$. We can write

$$W+\frac12=\frac{(A+B)^2+(C+D)^2}{(A+B)^2+(C+D)^2+(A-B)^2+(C-D)^2}$$

where all $A+B$, $A-B$, $C+D$, $C-B$ are independent and identically distributed, following $\mathcal N(0,2)$, using the characterization of the sum of and difference of two normal distributions [1]. Hence, by noting that the distribution of the sum of two squared independent standard normal random variables is $\text{Gamma}\left (1,\frac{1}{2} \right)$, we have

$$W+\frac12 \sim \frac{T_1}{T_1+T_2} \sim \text{Beta}(1,1) \sim \mathcal U(0,1)$$

where $$\small T_1=\frac{(A+B)^2+(C+D)^2}{2}$$ $$ \small T_2= \frac{(A-B)^2+(C-D)^2}{2}$$ are independent and have $\text{Gamma}\left (1,\frac{1}{2} \right)$ distribution, and where the relation between Gamma and Beta distributions [2] is used.

Amir
  • 11,124
4

The following relies on 3 basic results regarding $\chi^2$ distribution and quadratic forms:

If $x\sim \mathrm{MVN}(0,I)$, then for symmetric matrix $A$, $x'Ax$ is $\chi^2$ distributed iff $A$ is idempotent, and the d.f. is the rank of $A$.

If $x\sim \mathrm{MVN}(0,I)$, then $x'Ax$ and $x'Bx$ are independent iff $AB=0$, for symmetric matrices $A$ and $B$.

If $x\sim \mathrm{Gamma}(\alpha, \theta)$ independent of $y\sim \mathrm{Gamma}(\beta, \theta)$, then $\frac{x}{x+y}$ is Beta$(\alpha,\beta)$ distributed.

Let $x\sim \mathrm{MVN}_4(0,I)$, and $$M=\begin{bmatrix} 0 & \frac12 & 0 & 0 \\ \frac12 & 0 & 0 & 0 \\ 0 & 0 & 0 &-\frac12 \\ 0 & 0 & -\frac12 & 0 \end{bmatrix}.$$ Then the quantity of interest is $$z=\frac{x'Mx}{x'x},$$ i.e., a ratio of two quadratic forms. The denominator is clearly $\chi^2_4$ distributed. But the numerator is not a $\chi^2$ distributed variable because $M$ is not idempotent. We can add $\frac12$ of the denominator to the numerator to make the numerator also a $\chi^2$. Namely, $$ z+\frac12 = \frac{x'Mx+\frac12 x'x}{x'x}=\frac{x'Nx}{x'x} $$ where $$N=\begin{bmatrix} \frac12 & \frac12 & 0 & 0 \\ \frac12 & \frac12 & 0 & 0 \\ 0 & 0 & \frac12 &-\frac12 \\ 0 & 0 & -\frac12 & \frac12 \\\end{bmatrix},$$which is now idempotent with rank 2. So the numerator is $\chi^2_2$ distributed.

But $$ z+\frac12 =\frac{x'Nx}{x'Nx+x'(I-N)x}. $$ Note that $I-N$ is also idempotent with rank 2 and $(I-N)N=0$. Thus the denominator is a sum of two independent $\chi^2_2$ variables, one of them being the numerator.

It follows that $z+\frac12$ has a $\text{Beta}\left(\frac22, \frac22\right)$ distribution, i.e., $\text{Uniform}(0,1)$, by standard Gamma vs Beta relationship and $\chi^2_\nu \stackrel{D}= \text{Gamma}\left(\frac{\nu}{2},2\right)$.

Zack Fisher
  • 2,481
2

Let $X$ and $Y$ be independent circularly symmetric gaussian random variables with mean $0$ and unit variance. In this paper, it is proved that the law of $Z = Y/X$ has the density of the form

$$ f_Z(z) = \frac{1}{\pi(1 + |z|^2)^2}, \qquad z \in \mathbb{C}. $$

Now, let $X = A + iD$ and $Y = C + iB$ so that $A, B, C, D$ are IID standard normal variables and

$$ V := \frac{AB - CD}{A^2 + B^2 + C^2 + D^2} = \frac{\operatorname{Im}(\overline{X}Y)}{|X|^2 + |Y|^2} = \frac{\operatorname{Im}(Y/X)}{1 + |Y/X|^2}. $$

Then the moment-generating function of $V$ is given by

\begin{align*} \mathbb{E}[e^{\xi V}] &= \int_{0}^{\infty} \mathrm{d}r \int_{-\pi}^{\pi} \mathrm{d}\theta \, \frac{r}{\pi(1 + r^2)^2} \exp\biggl( \xi \frac{r \sin \theta}{1 + r^2} \biggr) \\ &= \frac{1}{4\pi} \int_{0}^{\pi} \mathrm{d}\psi \int_{-\pi}^{\pi} \mathrm{d}\theta \, (\sin \psi)\exp\biggl( \frac{\xi}{2}\sin\theta\sin\psi \biggr) \tag{$r=\tan(\psi/2)$} \\ &= \frac{1}{4\pi} \sum_{n=0}^{\infty} \frac{(\xi/2)^n}{n!} \int_{0}^{\pi} \mathrm{d}\psi \int_{-\pi}^{\pi} \mathrm{d}\theta \, (\sin\theta)^n (\sin\psi)^{n+1}. \end{align*}

By using the symmetry of the sine function and the beta function identity (or by integration by parts combined with mathematical induction), it is not hard to check that

$$ \int_{0}^{\pi} \mathrm{d}\psi \int_{-\pi}^{\pi} \mathrm{d}\theta \, (\sin\theta)^n (\sin\psi)^{n+1} = \begin{cases} \frac{4\pi}{n+1}, & \text{$n$ is even}, \\ 0, & \text{$n$ is odd}. \end{cases}$$

Therefore

\begin{align*} \mathbb{E}[e^{\xi V}] = \sum_{k=0}^{\infty} \frac{(\xi/2)^{2k}}{(2k+1)!} = \frac{\sin(\xi/2)}{\xi/2} = \int_{-\frac{1}{2}}^{\frac{1}{2}} \mathrm{d}v \, e^{\xi v}. \end{align*}

Therefore it follows that

$$ f_V(v) = \mathbf{1}_{[-\frac{1}{2}, \frac{1}{2}]}(v), $$

or equivalently, the law of $V$ is the uniform distribution over $[-\frac{1}{2}, \frac{1}{2}]$.

Sangchul Lee
  • 181,930
0

Okay, so there are a few key facts of note. Here, I assume the following:

  • $\newcommand{\N}{\operatorname{Normal}} X \sim \N(\mu_X,\sigma_Y^2)$
  • $Y \sim \N(\mu_Y,\sigma_Y^2)$
  • $N_i \sim \N(0,1)$ are independent

Then:

  • $\chi^2(k)$ is the sum of $k$-many squared independent standard normals, i.e. $\sum_1^k N_i^2 \sim \chi^2(k)$.

  • Consequently, $\displaystyle X^2/\sigma_X^2 \sim \chi^2(1)$

  • Meanwhile, $X+Y \sim \N(\mu_X + \mu_Y, \sigma_X^2+\sigma_Y^2)$, and for any constant $\alpha$ we have $\alpha X \sim \N(\alpha \mu, \alpha^2 \sigma_X^2)$.

  • Also, $X/Y \sim \operatorname{Cauchy}(0,1)$ if $X,Y$ are i.i.d. standard normals too. (The usual shift-and-division, i.e. $$X \mapsto \frac{X-\mu_X}{\sigma_X^2}$$ will achieve this.)

  • Also of note is the polarization identity, $$ XY = \frac{(X+Y)^2 - (X-Y)^2}{4} $$

This list of facts should be more than enough to derive the distribution, so I'll leave the details to you.

PrincessEev
  • 50,606