Finding MLE of $\alpha$ when a sample is drawn from $f(x) = \frac{2}{\alpha^2}(\alpha-x)I_{(0,\alpha)}(x)$

Question

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.

Answer 1

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: