Center of mass of random points in a general disc

Question

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 $k=1,2, \cdots m.$

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.

Answer 1

The claim is in general not true. Here is a specific case where it fails.

Let $c = 4, R = 1, l = 2$. For any point $z \in \mathbb{C}$ with $|z - c| \leq R$ we can write $z$ as $$z = c + re^{i\theta} = 4 + re^{i\theta}$$ with $r \in [0,1], \theta \in \mathbb{R}$. Squaring we have $$ z^2 = (4+re^{i\theta})^2 = 16 + r^2e^{2i\theta} + 8re^{i\theta} $$ Now let's look at the real part of $z^2$. We have the bound $$\text{Re}(z^2) = \text{Re}(16 + r^2e^{2i\theta} + 8re^{i\theta}) \geq 16 - r^2 - 8 \geq 7$$ where we have used that $\text{Re}(e^{2i\theta}), \text{Re}(e^{i\theta}) \geq -1$ and $r^2 \in [0,1]$.

Note that this implies for every $m$ with probability 1 that \begin{align} \text{Re} \left ( \frac{1}{m}\sum_{i=1}^m Z_i^2 \right ) = \frac{1}{m} \sum_{i=1}^m \text{Re}(Z_i^2) \geq \frac{1}{m} \sum_{i=1}^m 7 = 7 \end{align} Therefore the center of mass cannot approach $c = 4$ because it always has real part greater than 7.

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I believe that in the special case where $c = 0$ the result is true by a symmetry argument.

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Inspecting your calculations, I think what you would want to show is that

$$\lim_{m \rightarrow \infty} \mathbb{E} \left [ \left | \frac{1}{m}\sum_{i=1}^m Z_i^\ell - c \right | \right ] = 0.$$

Normally one would expect a result like $$ \lim_{m \rightarrow \infty} \mathbb{E} \left [ \left | \frac{1}{m}\sum_{i=1}^m Z_i^\ell - \mathbb{E}[Z^\ell] \right | \right ] = 0. $$ so you would want to establish that $\mathbb{E}[Z^\ell] \approx c$ in some appropriate sense, and it does not appear that you have done this (and it also really isn't the case anyway as the counter example showed).