Distribution of distance from center of sample group We have a bivariate normal process where $X, Y \sim N(0, \sigma)$, with no covariance.
(For convenience we can assert that $\sigma = 1$, or that we have a good estimate for its value.)
What is the distribution of the random variable $R(n) =  \sqrt{\overline{x_i}^2 + \overline{y_i}^2}$ — i.e., the Euclidean distance between the sample center of a n points and the true center (at the origin)?
Note that as defined:


*

*$R(n) \ge 0$

*$E[R(n)] \to 0$ monotonically as $n \to \infty$

*The Rayleigh distribution gives us $R(1) = \sigma \sqrt{\pi/2} \approx \sigma 1.25$


Furthermore, based on a Monte Carlo simulation for $n \in [2, 25]$ with $\sigma = 1$:


*

*Variance decreases monotonically as n increases

*Skewness appears constant across n at 0.63

*Kurtosis appears constant across n at about 0.24


(This question is a simplified version of this slightly more complicated one that seems to have gotten derailed in complications.)
 A: The sample means are zero mean normal r.v.'s , 
$$\bar X \sim N(0,\sigma^2/n),\;\; \bar Y \sim N(0,\sigma^2/n)$$
Then we have the r.v.'s
$$Z_x = \left(\frac{\bar X}{\sigma/\sqrt n}\right)^2 = \frac n{\sigma^2}\bar X^2 \sim \chi^2(1),\;\;Z_y = \left(\frac{\bar Y}{\sigma/\sqrt n}\right)^2 =\frac n{\sigma^2}\bar Y^2\sim \chi^2(1),$$
Therefore, the r.v.
$$W = Z_x + Z_y =\frac n{\sigma^2}\left(\bar X^2+\bar Y^2\right)\sim \chi^2(2)$$
By the properties of a chi-square random variable, we have
$$W_n=\frac {\sigma^2}nW \sim \text {Gamma}(k=1, \theta = 2\sigma^2/n) = \text{Exp}(2\sigma^2/n)$$
i.e.
$$f_{W_n}(w_n) = \frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} w_n\Big\}$$
Define $R_n = \sqrt {W_n}$. By the change-of-variable formula we have
$$W_n = R_n^2 \Rightarrow \frac {dW_n}{dR_n} = 2R_n$$ and so
$$f_{R_n}(r_n) = 2r_n\frac {n}{2\sigma^2}\cdot \exp\Big \{-\frac {n}{2\sigma^2} r_n^2\Big\} = \frac {r_n}{\alpha^2} \exp\Big \{-\frac {r_n^2}{2\alpha^2} \Big\},\;\;\alpha \equiv \sigma/\sqrt n$$
So $R_n$ also follows a Rayleigh distribution with parameter $\alpha$. We have
$$\begin{align}
&E(R_n) = (\sigma/\sqrt n)\sqrt {\pi/2} \Rightarrow \lim_{n\rightarrow \infty} E(R_n) =0\\
&\operatorname {Var}(R_n) = \frac {4-\pi}{2}(\sigma/\sqrt n)^2\Rightarrow \lim_{n\rightarrow \infty} \operatorname {Var}(R_n) =0\\
&\text{skewness}   =\frac{2\sqrt{\pi}(\pi - 3)}{(4-\pi)^{3/2}} =0.6311\\
&\text {kurtosis}  =-\frac{6\pi^2 - 24\pi +16}{(4-\pi)^2} =0.2450\\
\end{align}$$
as the Monte Carlo simulation has given. Skewness and Kurtosis do not indeed depend on the parameter.
