Let $u, v, w, x, y$ be vectors in $\mathbb{R}^m$ where each component is an observation of a random variable $U, V, W, X, Y$. Let $\bar{u}, \bar{v}, \bar{w}, \bar{x}, \bar{y}$ be the average of $u, v, w, x, y$, i.e., $\bar{u}=\frac{1}{m}\sum_{i=1}^m u_i, \bar{v}=\frac{1}{m}\sum_{i=1}^m v_i, \bar{w}=\frac{1}{m}\sum_{i=1}^m w_i, \bar{x}=\frac{1}{m}\sum_{i=1}^m x_i, \bar{y}=\frac{1}{m}\sum_{i=1}^m y_i$ and $S^2_u=\frac{1}{m-1}\sum_{i=1}^m (u_i-\bar{u})^2, S^2_v=\frac{1}{m-1}\sum_{i=1}^m (v_i-\bar{v})^2, S^2_w=\frac{1}{m-1}\sum_{i=1}^m (w_i-\bar{w})^2, S^2_x=\frac{1}{m-1}\sum_{i=1}^m (x_i-\bar{x})^2, S^2_y=\frac{1}{m-1}\sum_{i=1}^m (y_i-\bar{y})^2$. Suppose $U, V, W, X, Y$ are independent random variables. Is it possible to find $p_1, p_2, p_3, p_4, p_5 \geq 0$ such that $\sum_{i=1}^5 p_i=1$ and $$ \min_{\sum_{i=1}^5 p_i=1, p_i \geq 0} S^2_z $$ where $z=p_1u + p_2v + p_3w + p_4x+ p_5y$ and $S^2_z=\frac{1}{m-1}\sum_{i=1}^m (z_i-\bar{z})^2$ and $\bar{z}=\frac{1}{m}\sum_{i=1}^m z_i$.
Question Is it possible to find $p_1, p_2, p_3, p_4, p_5$? Is there a neat way to represent this problem in a matrix form where $A$ contains columns $u, v, w, x, y$? I fill this problem should already be addressed in statistics because it tries to find the best linear combination of 5 random variables that has the least variance, but I have not figured it out.
My try
I can see that the objective function is continuous and the constraint set is closed and bounded so the minimizer exists.
If I understand well you want to minimize $p_1^2\sigma^2_1+\cdots+p_m^2\sigma^2_m$ with respect to $(p_i)$ with the constraint $p_i>0$ and $p_1+\cdots+p_m=1$. It seems to me that the minimum is reached for $$p_i=\frac{c}{\sigma_i^2}.$$ For this enough is to minimize for $x_i>0$ for all $i>0.$
$$x\mapsto \frac{x_1^2\sigma^2_1+\cdots+x_m^2\sigma^2_m}{(x_1+\cdots+x_m)^2}$$
by using the partial derivatives. To see that the stationary point is a minimum you may use
$$x_i^2\frac{\sigma^2_i}{\sigma^2_j}+x_j^2\frac{\sigma^2_j}{\sigma^2_i}\geq 2x_ix_j.$$