I've been looking at some statistical distribution work and found a question that I can't solve through standard derivation of an MLE. I've been told to maybe look at an indicator function but I've no idea how to use those for these. The question is to find an MLE for a, when given a random sample $X_1 , X_2 , ... , X_n$ with probability density function $$ f(x) = \frac{53x^{52}}{2a^{53}}$$ for $-a < x < a$.
When I try the standard way of taking the product, log and differential the second differential comes out as a positive, which wouldn't make it a maximum. Any explanation of how to work through this would be much appreciated!
To clarify, the method I was taught was as follows:
$$L(a) = \prod_{i=1}^n\frac{53x^{52}}{2a^{53}} = (\frac{53}{2a^{53}})^n\prod_{i=1}^nX_i^{52}$$ Taking logs
$$Log(L(a)) = nLog(53) -53n(Log(a) + Log(2)) + \sum52Log(X_i)$$
To take the derivative from here wrt to a, set to 0 and then rearrange to find $a$ and call this the mle, but this eliminates $a$ entirely, as I've realised I've done something wrong in my previous work.
Remember the likelihood function is a function of $a.$ The likelihood function on data $x_1,\ldots x_n$ is $$L(a) = \left\{\begin{array}{ll}\left(\frac{53}{2a^{53}}\right)^n(x_1x_2\ldots x_n)^{52} & a>\max_i |x_i|\\0&a \le \max_i |x_i|\end{array}\right.$$ since the data is only possible if all of the $x_i$ are in the interval $(-a,a).$ Note that the first piece of this function is always greater than or equal to zero and that it decreases in $a.$ So to maximize, you want to find the smallest value of $a$ such that you are still on the first branch. This is just given by $a=\max_i|x_i|.$
(Technically, the way I wrote the function above (which is from how you wrote the PDF) there is no minimum value of $a$ on the first piece (and $L(\max_i|x_i|)=0$) which is not the maximum value) but in this case its ok to take the limit cause we could have defined the PDF's support to be $[a,a]$ rather than $(a,a).$)
The lesson here is that sometimes we maximize a function by taking derivatives, but not always. But remember when you learned to find global minima, you had to check the endpoints too, as well as the critical points which are only candidate local extrema. That's what's going on here.