MLE - Likelihood function has no maximum The density function is : $$ f(x) = \frac{2x}{\alpha^2}$$
If i'm not mistaken, the likelihood function is : $$L(\alpha) = \frac{2^n\prod x_i}{\alpha^{2n}}$$
The likelihood function has no maximum and I know that the maximum likelihood estimator for $\alpha $ is $max\{x_i\}$ but i don't know why. Can someone explain this to me?
Thanks in advance.
 A: View the density of an $X_i$ as a function of two variables. Say
$$
f(x,\alpha)=\begin{cases}
\dfrac{2x}{\alpha^2}&0\leq x\leq\alpha\\
0&\text{otherwise}
\end{cases}
$$
where $f\colon \mathbb{R_{\geq 0}}\times \mathbb{R_{\geq 0}}\to\mathbb{R}$, so that the joint density (by indepdendence) may be written
$$
f(\mathbf{x},\alpha)=f(x_1,\dotsc,x_n,\alpha)=
\begin{cases}
\dfrac{2^n\prod x_i}{\alpha^{2n}}&0\leq x_{(n)}\leq\alpha\\
0&\text{otherwise}
\end{cases}
$$
where $x_{(n)}$ is the maximum of the $x_i$, and note that $0\leq x_i\leq \alpha$ for all $i$ iff $0\leq x_{(n)}\leq\alpha$. The MLE  $\hat{\alpha}=\hat{\alpha}(\mathbf{x})$ is such that
$$
f(\mathbf{x},\hat{\alpha})=\sup_{\alpha>0}f(\mathbf{x},\alpha)\tag{1}.
$$
if it exists But notice  that for fixed $\mathbf{x}$,
$$
\sup_{\alpha>0}f(\mathbf{x},\alpha)\stackrel{\star}{=}\sup_{\alpha\geq {x_{(n)}}}f(\mathbf{x},\alpha)
=\left(2^n\prod x_i\right)\sup_{\alpha\geq x_{(n)}}\frac{1}{\alpha^{2n}}
=
\left(2^n\prod x_i\right)\frac{1}{x_{(n)}^{2n}}\tag{2}
$$
where equality $\star$ is because $f=0$ when $\alpha<x_{(n)}$. Comparing (1) and (2) we see that
$$
\hat{\alpha}=x_{(n)}.
$$
A: The density function is not $\frac{2x}{\alpha^2}.$ It is $\frac{2x}{\alpha^2}$ for $0<x<\alpha$ and zero otherwise. This means that the joint density for $n$ variables is zero if any of the $x_i$ are greater than $\alpha.$ As a result, the likelihood function is zero for $\alpha < \max_i(X_i).$ 
