## Friday, 9 November 2012

### Uniformly Distributed Random Points Inside a Circle (2)

So much for posting the answer in a few days! Sorry, I have been Very Very busy.

So here is the answer. Lets assume we have the Cartesian coordinate system with ($$x,y$$) representing the points. Then joint probability distribution function (PDF) of our random points inside the circle (i.e. the joint distribution of random variable $$X$$ and random variable $$Y$$ representing the $$x$$ and $$y$$ coordinate of the random point) is given by:

$$f_{X,Y}(x,y)= \begin{cases} \frac{1}{A}=\frac{1}{\pi R_c^2} & x^2+y^2 \leq R_c^2 \\ 0 & \text{otherwise} \end{cases},$$
where $$R_c$$ is the radius of the circle.

Now lets consider the polar coordinate system where $$r=g_1(x,y)= \sqrt{x^2+y^2}$$ and $$\theta=g_2(x,y)=\arctan (y/x)$$. Also, from inverting the functions $$g_1$$ and $$g_2$$ we have $$x=h_1(r,\theta)=r\cos \theta$$ and $$y=h_2(r,\theta)=r\sin\theta$$. Now calculating the Jacobian of the transformation we have:

$$J(x,y)= \det \begin{bmatrix} \frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} & \frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}} \\ \frac{\partial \theta}{\partial x}=\frac{-y}{x^2+y^2} & \frac{\partial \theta}{\partial y}=\frac{x}{x^2+y^2} \end{bmatrix}= \frac{1}{\sqrt{x^2+y^2}}=\frac{1}{r}.$$
Also we have
$$|J(r,\theta)|=\frac{1}{|J(x,y)|}$$

The joint PDF of random variables $$R$$ and $$\Theta$$ can then be calculated using the joint PDF of $$X$$ and $$Y$$ as:

$$f_{R,\Theta}(r,\theta)= f_{X,Y}\left( h_1(r,\theta),h_2(r,\theta) \right) = \begin{cases} \frac{r}{\pi R_c^2} & 0 \leq \theta \leq 2\pi, 0 \leq r \leq R_c \\ 0 & \text{otherwise} \end{cases}.$$

We can rewrite the non zero part as:
$$f_{R,\Theta}(r,\theta) = \frac{1}{2\pi} \times \frac{2r}{R_c^2} = f_\Theta(\theta)f_R(r)$$
where
$$f_\Theta(\theta) = \frac{1}{2\pi}$$
and
$$f_R(r)=\frac{2r}{R_c^2}.$$
Therefore, $$\Theta$$ is uniformly distributed between $$0$$ and $$2\pi$$. The random variable $$R$$ can be generated by first calculating its cumulative distribution function as
$$F_R(r) = \int_0^r \frac{2\alpha}{R_c^2}d\alpha = \frac{r^2}{R_c^2},$$
and then using a uniformly distributed random variable $$U$$ over the interval $$[0,1]$$ to get
$$U = \frac{R^2}{R_c^2} \implies R=R_c \sqrt{U}.$$

The following MATLAB code generates 1000 random numbers inside a circle with radius $$20$$ centered at $$-30$$, $$-40$$.

%**********************************************
n = 10000;
Rc = 20;
Xc = -30;
Yc = -40;

theta = rand(1,n)*(2*pi);
r = Rc*sqrt(rand(1,n));
x = Xc + r.*cos(theta);
y = Yc + r.*sin(theta);

plot(x,y,'.'); axis square;

%**********************************************


If you don't understand how I did the PDF transformation see
Chapter 6 (6.2.3 to be exact) or
Chapter 8 (8.3 to be exact).

I hope this helps someone out there.

1. Thank you for the informative post. It really helps me out of what I am currently stuck with. One typo I found if I am right is that R_{c} on the right hand side of last expression should not be squared.

1. Thank you. I fixed the typo.

2. Maybe you did not expect your post to help someone after almost one year .. it did for me : )

BTW another typo -- one of the "theta"'s should have been "\theta" ..

Thanks a lot for this helpful post!

1. I am glad it helped. I fixed the typo.

4. i dont know if it's just me but the maths latex writing isn't coming out properly for me

5. just got it sorry ...really helped