Find the MLE of $p$ where $f(y;p)=2p^2y^{-3}$.
Attempt:
Method: find the likelihood function, differentiate with respect to $p$ then set to zero and solve for $p$.
$L(p;y)=\prod\limits_{i=1}^n [2p^2y^{-3}_i] =\prod\limits_{i=1}^n[2] \prod\limits_{i=1}^n[p^2] \prod\limits_{i=1}^n [y^{-3}_i] = 2^n p^{2n} \prod\limits_{i=1}^n [y^{-3}_i]$
Differentiating the log likelihood gives $2n/p$ (i think) and the $y_i$ all disappear. So what happens when you set this to zero? Because if $0=2n/p$ then surely $0=2n$ and $n=0$.
Have I misstated $L(p;y)$? Or is something going wrong with my differentiation? (Or both?)
Both the PDF and the likelihood in your post are incorrect, due to the fact that you forget to include indicator functions.
In fact, the PDF is $$f(y;p)=2p^2y^{-3}\mathbf 1_{y\geqslant p}$$ hence, the likelihood of some i.i.d. sample $\mathbf y=(y_i)_{1\leqslant i\leqslant n}$ for the PDF $f(\ ;p)$ is $$L(p;\mathbf y)=2^np^{2n}\left(\prod_iy_i^{-3}\right)\mathbf 1_{m(\mathbf y)\geqslant p}$$ where $$m(\mathbf y)=\min_{1\leqslant i\leqslant n} y_i$$ One sees readily that $$L(p;\mathbf y)=c(\mathbf y)p^{2n}\mathbf 1_{m(\mathbf y)\geqslant p}$$ for some positive constant $c(\mathbf y)$ independent of $p$, hence $L(\ ;\mathbf y)$ is maximum at $$\hat p=m(\mathbf y)$$ No differentiation is involved, rather a precise understanding of the situation.