Problem definition
Consider the following bivariate random variable \begin{equation*} \overline{z}\triangleq \frac{1}{4} \sum_{j=1}^4 z_j \end{equation*} where $z_1, \dots, z_4$ are uniformly distributed over the edges of a square centered in the origin and with edge of unitary length. More precisely, \begin{equation*} \begin{aligned} z_1 &= \left[\begin{array}{c} x_1 \\ -0.5 \end{array}\right] \qquad x_1 \sim \mathcal{U}([-0.5,0.5])\\ z_2 &= \left[\begin{array}{c} +0.5 \\ y_2 \end{array}\right] \qquad y_2 \sim \mathcal{U}([-0.5,0.5])\\ z_3 &= \left[\begin{array}{c} x_3 \\ +0.5 \end{array}\right] \qquad x_3 \sim \mathcal{U}([-0.5,0.5])\\ z_4 &= \left[\begin{array}{c} -0.5 \\ y_4 \end{array}\right] \qquad y_4 \sim \mathcal{U}([-0.5,0.5]) \end{aligned} \end{equation*} my question is the following: what is the distribution of $\overline{z}$?
My attempt
Clearly, $\overline{z}$ is a point that falls inside the square. I'm wondering if considering $\overline{z}$ as uniformly distributed inside the square is, at least, a reasonable approximation. I can be wrong, but, by symmetry, it seems to me that each possible point inside the square are equally likely. To disprove my belief, we can check the expectation and the covariance of $\overline{z}$.
Computing the expectation of $\overline{z}$ is trivial, because by linearity it turns out immediately that $\mathbb{E}[\overline{z}]=0_{2\times1}$. This fact is in line with my hope that $\overline{z}$ is (at least, roughly) uniformly distributed over the square.
Then, for the second step, we can check the covariance. If $\overline{z}$ is uniform, then it must be $\textrm{Cov}[\overline{z}]=I_2/12$, where $I_2$ is the $2\times2$ identity matrix. A possible, but tedius, way to compute the covariance is given by the formula \begin{equation*} \textrm{Cov}[\overline{z}]= \mathbb{E}[\overline{z}\overline{z}'] \end{equation*} where it is noted that $\overline{z}$ is zero-mean and $'$ denotes the transpose operator. By linearity, it follows \begin{equation*} \textrm{Cov}[\overline{z}]= \frac{1}{16}\sum_{i,j=1}^4\mathbb{E}[z_{ij}] \end{equation*} where $z_{ij}\triangleq z_i z_j'$. At this point, before digging into the computations, I would like to have an external opinion.
Your random variable comes out to be $(\frac{x1+x2}{4},\frac{y1+y2}{4})$.
So the probability distribution of the X-coordinate of the random variable is the convolution of the probability distributions of $x_1$ and $x_2$ (scaled by $\frac{1}{4}$ at the end). This turns out to be the autoconvolution of a uniform probability distribution from $-0.5$ to $0.5$. This autoconvolution will give you a triangular probability distribution function. We can say the same for the Y-coordinate.
Here's a simulated result:
And this is the final probability distribution: