limits of marginal probability of polar coordinate probability

Question

I'm self-studying probability and came across a problem of finding the marginal probability $f_\theta$ of a pdf in polar coordinate $f_{r,\theta}$. To find the $f_r$, I know we do $\int^{2\pi}_{0} f_{r,\theta} d\theta$, but how about $f_\theta$? Is it from 0 to 1? Or 0 to infinity?

EDIT: the original problem is find $f_\theta = \int_{0}^{?} \frac{r}{2\pi\sigma^2}exp(\frac{-r^2}{2\sigma^2})dr$

Answer 1

From the looks of it, $r$ is Rayleigh distributed and $\phi$ is uniform over $[0,2\pi]$. The $(r,\phi)$ pair corresponds to the magnitude and the phase of the vector $(x,y)$, where $x$ and $y$ are independent standard Normal variables. In that case, since the support of $x$ and $y$ is $(-\infty, +\infty)$, then the support of $r$ is $[0, +\infty)$.

Answer 2

Suppose $(X,Y)\sim f_{XY}$ where $f_{XY}:\mathbb{R}^2 \longrightarrow (0,\infty)$ is any pdf supported on all of $\mathbb{R}^2$.

Set $X=R\cos(\Theta)$ and $Y=R\sin(\Theta)$.

Then $(R,\Theta)\sim f_{R \Theta}$ where $f_{R \Theta}$ is the density supported on $[0,\infty)\times [0,2\pi)$ given by $$f_{R \Theta}(r,\theta)=f_{XY}\Big(r\cos(\theta),r \sin(\theta)\Big)\Bigg|\frac{\partial\big(r\cos(\theta),r \sin(\theta)\big)}{\partial (r,\theta)}\Bigg|=rf_{XY}\Big(r\cos(\theta),r\sin(\theta)\Big)$$ In order to obtain the marginal density $f_{\Theta}$ you simply "integrate away" the $r$ component and get $$f_{\Theta}(\theta)=\int_0^{\infty}f_{R\Theta}(r,\theta)dr=\int_0^{\infty}rf_{XY}\Big(r\cos(\theta),r\sin(\theta)\Big)dr$$ In your example $f_{XY}(x,y)=\frac{1}{2\pi \sigma^2}e^{-\frac{x^2+y^2}{2 \sigma^2}}$ so $f_{R \Theta}(r,\theta)=\frac{r}{2\pi \sigma^2}e^{-\frac{r^2}{2 \sigma^2}}$ and for $\theta \in [0,2 \pi)$ $$f_{\Theta}(\theta)=\int_0^{\infty}\frac{r}{2\pi \sigma^2}e^{-\frac{r^2}{2 \sigma^2}}dr=\frac{1}{2\pi}$$ i.e. $\Theta \sim \mathcal{U}[0,2\pi)$. If we replaced the upper bound of $\infty$ with any positive constant $\alpha$ notice that $$\int_0^{\alpha}\frac{r}{2\pi \sigma^2}e^{-\frac{r^2}{2\sigma^2}}dr=\frac{1}{2\pi}\Bigg(1-e^{-\frac{\alpha}{2\sigma^2}}\Bigg)$$ Integrating the right hand side from $\theta=0$ to $\theta= 2\pi$ would yield $1-e^{-\frac{\alpha}{2\sigma^2}}\neq 1$ so $f_{\Theta}(\theta)$ wouldn't even be a density. The upper bound has to be $\infty$.

Answer 3

The probability density function $f_{r,\theta}$ is normalized, that is $ \int_0^{2\pi} \int_0^\infty f_{r,\theta} rdr d\theta = 1 $.

We can solve the $f_\theta$ by calculating

$f_\theta = \int_0^\infty \frac{r}{2\pi\sigma^2} exp(- \frac{r^2}{2\sigma^2}) dr $

$ = \frac{1}{2\pi} \int_0^\infty \frac{1}{2\sigma^2} exp( - \frac{r^2}{2\sigma^2}) dr^2 $

$ = -\frac{1}{2\pi} \int_0^\infty exp(- \frac{r^2}{2\sigma^2}) d (-\frac{r^2}{2\sigma^2}) $

$ = -\frac{1}{2\pi} exp(- \frac{r^2}{2\sigma^2}) |_0^\infty $

$ = 0-(-\frac{1}{2\pi}) $

$ = \frac{1}{2\pi}$

Hence,

$f_\theta = \frac{1}{2\pi}$.

It is clear that the integral of probability density function $f_\theta$ over the domain $ \theta \in [0,2\pi] $should be 1, such that

$\int_0^{2\pi} f_\theta d\theta = \frac{\theta}{2\pi} |_0^{2\pi} = 1 - 0 = 1 $

Answer 4

Clearly the domain of $\Theta$ is $[0;2\pi]$ and the domain of $R$ is $[0;+\infty)$

The problem arises from a vector $(X,Y)$ where $X\perp \!\!\!\!\!\! \perp Y$, are identically Centered Gaussian distributed

$$X \sim N(0;\sigma^2)$$

$$Y \sim N(0;\sigma^2)$$

Thus the joint density is

$$f_{XY}(x,y)=f_X(x)\cdot f_Y(y)=\frac{1}{2\pi\sigma^2}e^{-(x^2+y^2)/(2\sigma^2)}$$

Starting from this vector, and passing in polar coordinates you get

$$ \bbox[5px,border:2px solid black] { f_{R;\Theta}(r;\theta)=\frac{1}{2\pi}\cdot \frac{r}{\sigma^2}e^{-r^2/(2\sigma^2)} \qquad (1) } $$

Analyzing this joint density it is evident that it can be factorized in the following way

$$ \bbox[5px,border:2px solid black] { f_{R;\Theta}(r;\theta)=\underbrace{\frac{1}{2\pi}\cdot\mathbb{1}_{[0;2\pi]}(\theta)}_{\text{uniform density}}\times\underbrace{\frac{r}{\sigma^2}e^{-r^2/(2\sigma^2)}\cdot\mathbb{1}_{[0;+\infty)}(r)}_{\text{rayleigh density}}=f_{\Theta}(\theta)\times f_R(r) \qquad (2) } $$

Now observe that:

1. The uniform density is the density of the angle, uniform over all the circle (the circumference of a circle centered in the Origin)

$$\Theta\sim U[0;2\pi]$$

2. The Rayleigh density is the density of the radius, with support $[0;+\infty)$

$$R\sim Rayl(\sigma)$$

3. Angle and Radius are independent rv's

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The following drawing is useful to understand the problem

As you can see, the radius that is $R=\sqrt{X^2+Y^2}$ can takes any real value from zero to $\infty$ while the angle can takes any real value in the interval $[0;2\pi]$

Your original problem starts from the joint density Angle - Radius, independents rv's. Thus to find $f_{\Theta}(\theta)$ you can integrate the joint density in

$$f_{\Theta}(\theta)=\int_0^{\infty}f_{R\Theta}(r,\theta)dr$$

Or simply factorize the joint density (1) as I showed in (2).

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Note for the O.P.: when writing a density

$$f_R(r)$$

the subscrtipt letter indicates the rv thus it must be a capital letter. The letter in the brackets indicates the value that the random variable assumes thus it is a small letter.