Finding MLE of $\alpha$ when a sample is drawn from $f(x) = \frac{2}{\alpha^2}(\alpha-x)I_{(0,\alpha)}(x)$ So I had this question on a test and still can't do much about it, it's like this:
"Given a random sample with a size of 2, from a population with density:
$$f(x) = \frac{2}{\alpha^2}(\alpha-x)I_{(0,\alpha)}(x)$$
We also observed that $x_1 = 1.2$ and $x_2 = 4.4$. Find a MLE for $\alpha$."
So, calculating the likelihood function:
$$
L(x,\alpha)=\prod_{1}^{2}\frac{2}{\alpha^2}(\alpha-x_i)I_{(0,\alpha)}(x_i)
$$
$$
L(x,\alpha) = \frac{4}{\alpha^4}(\alpha-x_1)(\alpha-x_2)I_{(\alpha \geq \max(x_1,x_2))}(\alpha)
$$
Given that $\frac{4}{\alpha^4}(\alpha-x_1)(\alpha-x_2)$ is not really a crescent or decrescent function for all $\alpha$, I'm really not sure what to do next!
Checking for $\alpha \geq 4.4$ only, I got that the point the function stops increasing and starts decreasing again is at $\approx 6.861$ (replacing the values in the function, and using desmos to graph it), so I thought this would be the MLE, but my lecturer said that the right answer would be the $\max(x_1,x_2)$, so I'm really lost.
 A: I take it $I_{0, \alpha}(x)$ is the indicator function that takes 1 if $x \in (0, \alpha)$ and $0$ otherwise.
$$L(x,\alpha)=\prod_{1}^{2}\frac{2}{\alpha^2}(\alpha-x_i)I_{(0,\alpha)}(x_i)$$
I take the log because: (1) maximizing the likelihood is the same as maximizing the log of the likelihood since log is a monotonic increasing transformation (2) log makes multiplication sum and that's nicer.
$$\log L(x,\alpha)=\sum_i \left[ \log 2 - 2 \log \alpha + \log (\alpha -
 x_i) + \log I_{(0,\alpha)}(x_i)\right] $$
Our maximum log likelihood problem is
\begin{equation}
 \begin{array}{*2{>{\displaystyle}r}}
 \mbox{maximize (over $\alpha$)} & \sum_i \left[ \log 2 - 2 \log \alpha + \log (\alpha -
 x_i) + \log I_{(0,\alpha)}(x_i)\right] \end{array}
\end{equation}
We immediately can see that if $x_i > \alpha$ for any $x_i$ that we have a $\log 0 = -\infty$ for our objective (which kinda is bad if you're trying to maximize). So we know we want $\alpha \geq \max(x_1, x_2)$. An equivalent optimizaiton problem is.
\begin{equation}
 \begin{array}{*2{>{\displaystyle}r}}
 \mbox{maximize (over $\alpha$)} & \sum_i \left[ - 2 \log \alpha + \log (\alpha -
 x_i) \right] \\
 \mbox{subject to} & \alpha \geq \max(x_1, x_2)
 \end{array}
\end{equation}
This isn't a thrilling maximization problem because the objective isn't inherently concave. The first order conditions won't be sufficient for a maximum. The first order (necessary but not sufficient) condition though (ignoring the constraint) is:
$$ - \frac{4}{\alpha} + \frac{1}{\alpha - 1.2} + \frac{1}{\alpha - 4.4} = 0  $$
A bunch of algebra leads to the quadratic equation:
$$ -2a^2 + \frac{84}{5}a - \frac{528}{25} = 0$$
You get roots $1.5392$ and $6.8608$, and $6.860826939130014$ looks like a maximum. Graphing the original likelihood function:

