A sample of size two is taken from the distribution $ f(x;\alpha)=\frac{2}{\alpha^2}(\alpha-x)I_{(0,\alpha)}(x)$. We need to find the maximum likelihood estimator for $\alpha$.
The likelihood function $$L(x_1,x_2;\alpha)=\frac{2}{\alpha^2}(\alpha-x_1)\frac{2}{\alpha^2}(\alpha-x_2)I_{(0,\alpha)}(x_1,x_1)$$
i.e. $$L(x_1,x_2;\alpha)=\frac{4}{\alpha^4}(\alpha-x_1)(\alpha-x_2)I_{(0,y_2)}(y_1)I_{(y_1,\alpha)}(y_2)$$
where $y_1=\min(x_1,x_2)$ and $y_2=\max(x_1,x_2)$.
Since we have a higher power of $\alpha$ in the denominator, first I thought of taking $\hat\alpha=y_1$ but that would make $L=0$, similarly for $\hat \alpha=y_2$.
Considering another method - Maximizing the numerator and minimizing the denominator. Maximizing numerator gives $\hat \alpha=\bar{x}$ and minimizing denominator gives $\hat \alpha=y_1$.
What should we do here?
EDIT: I tried another way, first we'll be able to maximize the likelihood function if the indicator functions are $1$, taking their values as $1$ and maximizing the function gives $$\hat \alpha=\frac{3\sum x_i}{4} \pm \frac{\sqrt{9\sum x_i^2-14x_1x_2}}{4}$$
I also plotted $L(x_1,x_2;\alpha)=\frac{4}{\alpha^4}(\alpha-x_1)(\alpha-x_2)I_{(0,y_2)}(y_1)I_{(y_1,\alpha)}(y_2)$ for fixed values $(x_1,x_2)$ and observed that at $\hat \alpha=\frac{3\sum x_i}{4} + \frac{\sqrt{9\sum x_i^2-14x_1x_2}}{4}$ the function indeed attains a maximum, but what about the other root?
Also the question asks whether this statistic is sufficient or not. How do I tackle that one?
