Let $X_1,\ldots,X_n$ be i.i.d. random variables from the population with pdf $$ f(x)= \begin{cases} 3\beta^3 x^{-4},\quad &x\geq\beta,\\ 0, &\text{elsewhere}. \end{cases} $$
I need to derive MLE of $\beta$.
I found the likelihood function $L(\beta\mid x)$ to be $$ \ 3^n \beta^{3n}\prod_{i=1}^nx_i^{-4} $$
I have trouble to log-likelihood function, because after that when I set it to be $0$ i can not solve it. I think that this is because support depend on estimator $\beta$. Can anyone show me how to do this problem? Thank you in advance!
Once again an example of the dead-end one gets into when one writes incorrectly the densities, that is, omitting the indicator functions.
Here, the PDF $f_\beta$ is defined by $$ f_\beta(x)=3\cdot\beta^3\cdot x^{-4}\cdot\mathbf 1_{x\geqslant\beta}, $$ hence the likelihood $L_x(\beta)$ of a sample $x=(x_i)_{1\leqslant i\leqslant n}$ is NOT what you write but $$ L_x(\beta)=\prod_{i=1}^nf_\beta(x_i)=3^n\cdot\beta^{3n}\cdot\prod_{i=1}^nx_i^{-4}\cdot\prod_{i=1}^n\mathbf 1_{x_i\geqslant\beta}. $$ The product of indicator functions on the RHS is $\mathbf 1_{m(x)\geqslant\beta}$ where $$ m(x)=\min\{x_i\mid 1\leqslant i\leqslant n\}, $$ hence $$ L_x(\beta)=c(x)\cdot\beta^{3n}\cdot\mathbf 1_{m(x)\geqslant\beta}, $$ where $c(x)$ is independent on $\beta$. To maximize $L_x(\beta)$, one should look for $\beta\leqslant m(x)$ (otherwise $L_x(\beta)=0$) such that $\beta^{3n}$ is maximal. Thus, the MLE of $\beta$ based on $x$ is $$ \hat\beta(x)=m(x). $$