Consider a random sample $X_1,…,X_n$ from a distribution with PDF $$f(x;\alpha,\beta)=\frac{\alpha}{\beta^\alpha}x^{\alpha-1}, 0\leq x\leq\beta$$ and parameters $\alpha,\beta>0$. Write down the likelihood and maximize it for any fixed $\alpha$ with respect to $\beta$. Write down the resulting profile likelihood.
Here is what I have so far, but the result of my computation leaves me uncertain:
Observe that $l(\alpha,\beta;x)=n\ln{\alpha}-n\alpha\ln{\beta}+(\alpha-1)\sum_{x=1}^n\ln{x_i}$.
It follows that $\frac{\partial{l}}{\partial{\beta}}=\frac{-n\alpha}{\beta}=0$.
Thus, for any fixed $\alpha$, $\hat{\beta}=0$ and $l^*(\alpha)=l(\alpha,\hat{\beta)}$ (but $ln{0}$ is undefined?).
Best Answer
Let's calculate profile likelihood for $\beta$
$$L(\beta)\propto \frac{1}{\beta^{n \alpha}}\cdot\mathbb{1}_{[x_{(n)};+\infty)}(\beta)$$
It is evident that L is strictly decreasing thus
$$\hat{\beta}=x_{(n)}=max(\mathbf{x})$$
If you want to get the estimator of $\alpha$ too ( not requested) you can get the constraint log-likelihood
$$l(\alpha)=n \log\alpha- n \alpha\log x_{(n)}+(\alpha-1)\Sigma_i x_i$$
... and you can easy conclude