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.
On
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 $\alpha < \max_i(X_i).$
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 $$ \hat{\alpha}=x_{(n)}. $$