Suppose $\theta_i$ for $i=1,\dots, n$ are my parameters and $w_i$ for $i=1,\dots, n$ are known constants , $w_i, \theta_i \in(0,1)$, and $\sum_{i=1}^n w_i=1$
$$f(\mathbf{\theta})=\frac{(\sum^n_{i=1}w_i \theta_i)^2}{n\sum_{i=1}^n(w_i\theta_i)^2}$$
If it is not convex, can it be minimised?
I don't think it is convex, but its minimum value is $1/n$, achieved when all the $\theta_i=0$ except for one of them. Your function is the reciprocal of a convex function of the vector $v=(w_1\theta_1,\dots,w_n\theta_n)/\sum_{i=1}^n w_i\theta_i,$ which is a point on the probability simplex. That is, your $1/f(\theta)=n\sum v_i^2$, where $\sum v_i^2$ is well known to be convex. Convex functions defined on convex domains (such as the simplex) attain their maxima at extreme points, a.k.a. vertices of the simplex, which have all $v_i=0$ except for one of them, which is 1.