Distribusi

Apakah ada ekspresi kanonik atau analitik untuk distribusi probabilitas untuk variabel acak kompleks sirkular-simetris $Z$ :

Z = e^{j θ},

$Z = e^{j\theta},$ dimana

θ \sim U (0, 2 π)

$\theta \sim \mathcal U(0, 2\pi)$ ?

Catatan samping:

Diketahui bahwa bagian nyata dan imajiner, yaitu:

ℜ (Z) = \cos θ ℑ (Z) = \sin θ

$\Re(Z) = \cos \theta \\ \Im(Z) = \sin \theta$ memiliki kepadatan marginal yang diberikan oleh :

f_{ℜ (Z)} (z) = f_{ℑ (Z)} (z) = \frac{1}{π \sqrt{1 - z^{2}}}, - 1 < z < 1,

$f_{\Re(Z)}(z) = f_{\Im(Z)}(z) = \frac{1}{\pi\sqrt{1-z^2}}, \quad -1 < z < 1,$ tetapi karena mereka tidak independen, menghitung PDF bersama mereka tidak trivial.

EDIT: $Z$ berbeda dari normal kompleks di sini, amplitudo $|Z|$ adalah deterministik dan identik 1, sedangkan jika $Z$ normal kompleks, $|Z|$ akan didistribusikan Rayleigh.

complex-random-variable

— Robert L.
sumber

Bagaimana hal ini berbeda dari kompleks sirkular-simetris normal ?

— Maxtron

@ Maxtron amplitudo di sini adalah 1, sedangkan kompleks normal memiliki amplitudo yang itu sendiri adalah variabel acak dengan distribusi Rayleigh.

— Robert L.

@OlliNiemitalo punya jawaban untuk hal yang sama seperti yang saya ingat ...

— Fat32

@ Fat32 Ya ini berdering. Pertanyaan lain adalah: Apa distribusi itu?

— Olli Niemitalo

@OlliNiemitalo Ya itu pertanyaannya!

— Fat32

Jawaban:

Karena bagian nyata dan imajiner sangat tergantung satu sama lain (jika Anda memiliki nilai satu, Anda tahu nilai yang lain persis), sepertinya Anda bisa menerapkan pdf marginal dari bagian nyata $r$ , diberi nilai bagian imajiner $i$ :

f_{r i} (r, i) = f_{r | i} (r | i) f_{i} (i)

$f_{ri}(r, i) = f_{r | i}(r\ |\ i) f_i(i)$

Anda mencatat pdf dari bagian nyata dan imajiner secara individual:

f_{r} (z) = f_{i} (z) = \frac{1}{π \sqrt{1 - z^{2}}}

$f_r(z) = f_i(z) = \frac{1}{\pi\sqrt{1-z^2}}$

Itu meninggalkan pdf marginal $f_{r | i}(r\ |\ i)$ . Ingatlah bahwa untuk realisasi variabel acak yang diberikan $Z$ , dua komponen terkait secara deterministik:

r^{2} + i^{2} = \cos^{2} (θ) + \sin^{2} (θ) = 1

$r^2 + i^2 = \cos^2(\theta) + \sin^2(\theta) = 1$

Dengan adanya hubungan ini, kita dapat menyelesaikannya $r$ dengan kondisi $i$ :

r^{2} = 1 - i^{2}

$r^2 = 1 - i^2$

r = \pm \sqrt{1 - i^{2}}

$r = \pm \sqrt{1-i^2}$

Oleh karena itu, pdf marginal dari $r$ diberi nilai $i$ adalah sepasang impuls:

f_{r | i} (r | i) = \frac{1}{2} δ (r - \sqrt{1 - i^{2}}) + \frac{1}{2} δ (r + \sqrt{1 - i^{2}})

$f_{r | i}(r\ |\ i) = \frac{1}{2}\delta\left(r - \sqrt{1 - i^2}\right) + \frac{1}{2}\delta\left(r + \sqrt{1 - i^2}\right)$

Menyatukan ini akan menghasilkan:

f_{r i} (r, i) = \frac{δ (r - \sqrt{1 - i^{2}}) + δ (r + \sqrt{1 - i^{2}})}{2 π \sqrt{1 - i^{2}}}

$f_{ri}(r, i) = \frac{\delta\left(r - \sqrt{1 - i^2}\right) + \delta\left(r + \sqrt{1 - i^2}\right)}{2\pi\sqrt{1-i^2}}$

Memikirkan hal ini secara geometris, untuk setiap garis horizontal $i = i_0$ (untuk $i_0 \in [-1, 1]$ ) dalam $ri$ pesawat, hanya ada dua titik $r_0 = \pm \sqrt{1 - i_0^2}$ itu bukan nol, dan pdf memiliki ketinggian tak terbatas pada titik-titik tersebut. Seperti yang kita harapkan, titik-titik persimpangan (yaitu titik di mana pdf adalah nol) adalah di mana garis horizontal bersilangan dengan lingkaran satuan!

Ini berarti bahwa pdf bersama bernilai nol, kecuali di sepanjang lingkaran unit, di mana ia mengambil ketinggian tak terbatas. Itu sejalan dengan intuisi, sebagai definisi dari variabel acak $Z$ memastikan bahwa hanya dapat mengambil nilai yang ada di lingkaran unit.

Tidak ada yang istimewa tentang cara spesifik saya menguraikan ini; Anda juga dapat memindahkan masalah dan melihat garis vertikal di $ri$ bidang bentuk $r = r_0$ dan Anda akan menemukan hubungan yang sama karena kopling dekat dari dua variabel acak.

Saya percaya formulasi ini setara dengan yang ada di jawaban AlexTP , tetapi turunannya mungkin lebih intuitif.

— Jason R
sumber

apakah ada salah ketik di

r = 1 - i^{2}

$r = 1- i^2$ ; mungkin maksudmu

r^{2} = 1 - i^{2}

$r^2=1-i^2$ ? Juga dapat kami tunjukkan bahwa pdf gabungan terintegrasi ke 1 dalam bidang xy (atau eqv ri). Lagi-lagi juga pdf bersama

f_{r i} (r, i)

$f_{ri}(r,i)$ is not circularly symmetric? (the denominator?). Cannot we simply use the logic that given

z = e^{j θ} = \cos (θ) + j \sin (θ) = x + j y

$z = e^{j \theta} = \cos(\theta) + j \sin(\theta) = x + jy$ with

θ

$\theta$ uniform in [

0, 2 π

$0,2\pi$ ] and the constraint

x^{2} + y^{2} = 1

$x^2+y^2=1$ simply offers a circularly symmetric joint pdf of the type

f_{x y} (x, y) = K δ_{2} (x^{2} + y^{2} - 1)

$f_{xy}(x,y) = K \delta_2(x^2+y^2-1)$ or

f_{R, θ} = K δ_{2} (R - 1)

$f_{R,\theta} = K \delta_2(R-1)$ ; a ring impulse on x-y plane? where K is possibly

1 / 2 π

$1/2\pi$ ?

— Fat32

I corrected the typos you referred to, thanks.

— Jason R

@Fat32: I made some fixes that led me to what sounds like what you were getting at in your comment above. AlexTP's answer still is probably more intuitively pleasing though.

— Jason R

There is a typo in copying

f_{r | i} (r | i)

$f_{r|i}(r|i)$ to

f_{r, i} (r, i)

$f_{r,i}(r,i)$ . Dan saya perhatikan itu

f_{r, i} (0, 1) = \infty

$f_{r,i}(0,1) = \infty$ berbeda dengan

f_{r, i} (1, 0) = \frac{1}{2 π}

$f_{r,i}(1,0) = \frac{1}{2\pi}$ . Saya pikir mereka harus memiliki nilai yang sama, bukan?

— AlexTP

semuanya, mungkin simulasi cepat akan bermanfaat untuk menunjukkan poin Anda :)

— AlexTP

Hindari perhitungan yang rumit, biarkan $X$ dan $Y$ menjadi variabel acak normal standar , variabel acak Anda $Z$ memiliki distribusi yang sama antara $V$

V ≜ (\frac{X}{\sqrt{X^{2} + Y^{2}}}, \frac{Y}{\sqrt{X^{2} + Y^{2}}})

$V \triangleq \left(\frac{X}{\sqrt{X^2+Y^2}},\frac{Y}{\sqrt{X^2+Y^2}} \right)$ (easy to see

‖ V ‖ = 1

$\lVert V\rVert=1$ and the angle of

V

$V$ is equivalent to the angle of a circularly symmetric Normal hence uniform).

This kind of $V$ is one of the constructions of a point uniformly distributed on circle (which can be generalized to $(n-1)$ -sphere, see Sphere Point Picking and for example this answer).

Thus the PDF of $Z$ is simply the reciprocal of the unit circle's circumference. For $Z_\rho = \rho e^{j\Theta}$ with fixed $\rho$ and uniform $\Theta$ ,

in polar coordinates (where infinitesimal area is $r dr d\theta$ ),

f_{R, Θ} (r, θ) = \frac{1}{2 π} δ (r - ρ)

$f_{R,\Theta}(r,\theta)=\frac{1}{2\pi} \delta(r - \rho)$

— AlexTP
sumber

i think this is correct.

— robert bristow-johnson

The intuition seems to be so simple behind the proof. All you need to do is pick points from the surface of a unit sphere :)

— Maxtron

Can't we just remove the

| | z | |

$||z||$ in the denominator? Otherwise, how would you evaluate the expression for

| | z | | \to 0

$||z||\rightarrow 0$ ? Of course, we would like

f_{Z} (0) = 0

$f_Z(0)=0$ to hold.

— Matt L.

@MattL. you are right, it is wrong having

‖ z ‖

$\lVert z \rVert$ in the denominator, not due to the evaluation at limits but because it is simply wrong. I have fixed the final result.

— AlexTP

@AlexTP: I still think that there shouldn't be any variable in the denominator. Have a look at my answer and please tell me if/where I go wrong.

— Matt L.

Based on the existing answers, which opened my eyes for what's going on here, I would like to present yet another very simple expression for the solution, which is just slightly different from the one in AlexTP's answer (and which turned out to be equivalent to the one given in Jason R's answer, as shown below in the EDIT-part).

[EDIT: now that AlexTP has edited his answer, our expressions for the PDF are identical; so all three answers finally agree with each other].

Let the complex random variable $Z=X+jY$ be defined as

\begin{matrix} (1) & Z = ρ e^{j θ} \end{matrix}

$Z=\rho e^{j\theta}\tag{1}$

where the radius $\rho$ is deterministic and given, whereas the angle $\theta$ is random and uniformly distributed on $[0,2\pi)$ . I state without further proof that $Z$ is circularly symmetrical, from which it follows that its probability density function (PDF) must satisfy

\begin{matrix} (2) & f_{Z} (z) = f_{Z} (x + j y) = f_{Z} (r), with r = \sqrt{x^{2} + y^{2}} \end{matrix}

$f_Z(z)=f_Z(x+jy)=f_Z(r),\qquad\textrm{with}\quad r=\sqrt{x^2+y^2}\tag{2}$

i.e., it can be written as a function of the radius (magnitude) $r$ .

Since the PDF must be zero everywhere except for $r=\rho$ , and since it must integrate to unity (when integrated over the 2-dimensional plane), the only possible PDF is

$\begin{matrix} (3) & f_{Z} (r) = \frac{1}{2 π} δ (r - ρ) \end{matrix}$ $f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)\tag{3}$

It can be shown that $(3)$ leads to the correct marginal densities for the random variables $X$ and $Y$ .

EDIT:

After some very useful discussion in the comments it appears that we've managed to agree on one solution to the problem. I will show in the following that the unassuming formula $(3)$ is actually equivalent to the more involved looking formula in Jason R's answer. Note that I use $r$ for the magnitude (radius) of the complex RV $Z$ , whereas in Jason's answer $r$ denotes the real part of $Z$ . I will use $x$ and $y$ for the real and imaginary parts, respectively. Here we go:

\begin{matrix} (4) & f_{Z} (r) = \frac{1}{2 π} δ (r - ρ) = \frac{1}{2 π} δ (\sqrt{x^{2} + y^{2}} - ρ) \end{matrix}

$f_Z(r)=\frac{1}{2\pi}\delta(r-\rho)=\frac{1}{2\pi}\delta\left(\sqrt{x^2+y^2}-\rho\right)\tag{4}$

We know that $\delta\big(g(x)\big)$ is given by

\begin{matrix} (5) & δ (g (x)) = \sum_{i} \frac{δ (x - x_{i})}{| g^{'} (x_{i}) |} \end{matrix}

$\delta\big(g(x)\big)=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}\tag{5}$

where $x_i$ are the (simple) roots of $g(x)$ . We have

\begin{matrix} (6) & g (x) = \sqrt{x^{2} + y^{2}} - ρ and g^{'} (x) = \frac{x}{\sqrt{x^{2} + y^{2}}} = \frac{x}{r} \end{matrix}

$g(x)=\sqrt{x^2+y^2}-\rho\quad\textrm{and}\quad g'(x)=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}\tag{6}$

The two roots $x_i$ are

\begin{matrix} (7) & x_{1, 2} = \pm \sqrt{ρ^{2} - y^{2}} \end{matrix}

$x_{1,2}=\pm\sqrt{\rho^2-y^2}\tag{7}$

Consequently,

\begin{matrix} (8) & | g^{'} (x_{1}) | = | g^{'} (x_{2}) | = \frac{\sqrt{ρ^{2} - y^{2}}}{ρ} = \sqrt{1 - {(\frac{y}{ρ})}^{2}} \end{matrix}

$|g'(x_1)|=|g'(x_2)|=\frac{\sqrt{\rho^2-y^2}}{\rho}=\sqrt{1-\left(\frac{y}{\rho}\right)^2}\tag{8}$

With $(5)$ - $(8)$ , Eq. $(4)$ can be written as

\begin{matrix} (9) & f_{X, Y} (x, y) = \frac{1}{2 π \sqrt{1 - {(\frac{y}{ρ})}^{2}}} [δ (x - \sqrt{ρ^{2} - y^{2}}) + δ (x + \sqrt{ρ^{2} - y^{2}})] \end{matrix}

$f_{X,Y}(x,y)=\frac{1}{2\pi\sqrt{1-\left(\frac{y}{\rho}\right)^2}}\left[\delta\left(x-\sqrt{\rho^2-y^2}\right)+\delta\left(x+\sqrt{\rho^2-y^2}\right)\right]\tag{9}$

For $\rho=1$ , Eq. $(9)$ is identical to the expression given in Jason R's answer.

I think we can now agree that Eq. $(3)$ is a correct (and very simple) expression for the PDF of the complex RV $Z=\rho e^{j\theta}$ with deterministic $\rho$ and uniformly distributed $\theta$ .

— Matt L.
sumber

I think the determinant of Jacobian transform should be

r

$r$ then

\int_{θ = 0}^{2 π} \int_{r = 0}^{+ \infty} f_{Z} (r) r d r d θ \neq 1

$\int_{\theta=0}^{2\pi} \int_{r=0}^{+\infty} f_Z(r) r dr d\theta \neq 1$

— AlexTP

@AlexTP: Take as an extreme example

ρ = 0

$\rho=0$ , which makes the RV

Z

$Z$ deterministic, it's always zero. It's PDF should then be a Dirac impulse at

r = z = y = 0

$r=z=y=0$ , which is what I get with the formula I suggested. With the formula in your answer we get an undefined term in that case.

— Matt L.

it's true, but I think the value of PDF can be infinity and only the CDF should be bounded. In other words, the Dirac impulse matters only inside integrals.

— AlexTP

Yes, but if you integrate your pdf (for

ρ = 0

$\rho=0$ ), what do you get (and how would you do it)?

— Matt L.

I would do like this

lim_{ρ \to 0^{+}} \int_{θ} \int_{r = 0}^{+ \infty} \frac{1}{2 π ρ} δ (r - ρ) r d r d θ = lim_{ρ \to 0^{+}} \int_{θ} \frac{1}{2 π ρ} ρ d θ = \int_{θ} \frac{1}{2 π} d θ = 1

$\lim_{\rho \to 0^+}\int_\theta \int_{r=0}^{+\infty}\frac{1}{2\pi\rho}\delta(r-\rho) r dr d\theta = \lim_{\rho \to 0^+} \int_\theta \frac{1}{2\pi\rho} \rho d\theta = \int_\theta \frac{1}{2\pi} d\theta = 1$ The integral is finite while the PDF is not thanks to the Jacobian determinant. My point is that you need that Jacobian determinant

r

$r$ whenever you integrate the PDF in polar coordinate. Tell me if you find it unconvinced.

— AlexTP