Consider the disc $|z-k| \leq r.$For the case of a unit disc ,the probability desnity of the distance from origin of a randomly and uniformly chosen point in the disc is given by \begin{equation} f(u) = \begin{cases} 2u, & \text{if } \quad 0 \leq u \leq 1 \\ 0, & \text{otherwise} \end{cases} \end{equation} However ,it is not obvious to me as to how we can find the same in case of the general disc mentioned .
Any help\hints\suggestions would be greatly appreciated.
PSBy origin i mean the origin of the coordinate system i.e. $(0,0)$ so I want the distribution of a randomly and uniformly chosen point in the ball from $(0,0)$ not from the center of the circle.
My apologies, this is sort of an answer in the making;it would have been too long for a comment Kindly somebody point out mistakes ,if any any and the possibility of proving the following assertion via a direct approach. I want to prove that for a fixed positive integer $l,$ the center of mass of the $lth$ powers of $m$ randomly and uniformly chosen complex numbers from the disc $|z-c| \leq R,R>0 \quad \text{and} \quad c \in \mathbf{C} $ approaches $c,$ the center of the disc.This is intuitively expected but i want to prove in direct way.Here is what I have tried so far.
Let $ Z_1,Z_2, \cdots Z_m$ be randomly and uniformly chosen point in the disc $ |z-c| \leq R$ .My goal is to find the the bounds of $ \left |\frac{ \sum_1^m Z_k}{m} \right|$ and then take the limit as $m$ becomes arbitrarily large.We expect from symmetry conditions that the limit approaches $|c|$ as $m \to \infty.$
Under the above conditions we see we can write \begin{equation} Z_k=c+\sqrt{U_k}e^{\iota \theta_k} \end{equation} where $U_k $ is circularly uniform over $(0,R)$ $\theta_k's$ are uniform in the interval $(0,2 \pi)$ } and $k=1,2, \cdots m.$The square root is used because the probability density function (pdf) for the radius must be proportional to the area of the circle
We note \begin{align*} Z_k^l &= (c + r_k e^{i \theta_k})^l \\ &= \sum_{j=0}^{l} \binom{l}{j} c^{l-j} r_k^j e^{i (j\theta_k)}, \quad k=1,2, \cdots, m \end{align*} Therfore, \begin{equation*} \frac{\sum_1^m Z^k}{m}=\frac{1}{m} \sum_{k=1}^{m} \sum_{j=0}^{l} \binom{j}{l} c^{n-j}r_k^je^{\iota (j\theta_k)} \end{equation*} Using triangular inequality and taking expectations : \begin{equation} \mathbf{E}\left ( \left| \frac{\sum_1^m Z^k}{m}\right|\right) \leq \frac{1}{m}\sum_{k=1}^{m} \sum_{j=0}^{l} \binom{l}{j} |c|^{n-j} \mathbf{E}(r_k^j) \mathbf{E} (|e^{\iota (j\theta_k)}|) \end{equation} Now we note that \begin{equation} \mathbf{E} (|e^{\iota (j\theta_k)}|)=1 \end{equation} and \begin{equation} \mathbf{E}(r_k^j)= \frac{1}{R^2}\int_{0}^{R} 2x.x^{ \frac{j}{2}} dx = \frac{4}{R^2} \cdot \frac{R^{j+4}}{j+4} \end{equation} Hence , \begin{align*} \mathbf{E}\left ( \left| \frac{\sum_1^m Z^k}{m}\right|\right) & \leq \frac{1}{m}\sum_{j=0}^{l} \binom{l}{j} |c|^{n-j} \sum_{k=1}^{m} \mathbf{E}(r_k^j)\\ & \leq 4 \sum_{j=0}^{l} \binom{l}{j} |c|^{n-j} \frac{R^{j+4}}{j+2} \end{align*} If my calculations are correct ,the RHS of the last inequality must approach $|c|$ as $m \to \infty.$Unfortunately I do not know how I Can prove that from here.I would be highly obliged for any help/hints/suggestions.