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.

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.

ReplyDeleteThank you. I fixed the typo.

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

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

Thanks a lot for this helpful post!

I am glad it helped. I fixed the typo.

DeleteHelpful . really :)

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

ReplyDeletejust got it sorry ...really helped

ReplyDelete