Finding a maximum likelihood estimate of a parameter when regularity conditions don't hold I am currently trying to find the maximum likelihood estimate (or determine that it doesn't exist) of the parameter $a$ for a random sample of the variables $X_1, X_2 ...X_n$ with pdf $$f(x;a)=
\begin{cases}
 ax^{-2}&\text{if}\, 0 < a \leq x < \infty \\
 0&\text{otherwise}
\end{cases}$$
I've determined that the likelihood $L(a;x)$ is $$L(a;x)=
\begin{cases}
 a^n\prod_{i=1}^{n} x_i^{-2}&\text{if}\, \min(x_i) \geq a \\
 0&\text{otherwise}
\end{cases}$$
I know that this case doesn't satisfy the regularity conditions for using the derivative to maximize the likelihood, so I'm not sure where to go with this problem. 
 A: The answer by jbowman covers what you need to know to obtain the MLE, and I have nothing to add to that derivation.  However, it is worth noting that in cases like this, where you have a sampling density that is monotonic in the parameter, it is usual for the maximum-likelihood estimator (MLE) to be bounded by the true parameter value and so you obtain a biased estimator.  In this particular case you have $\hat{a} \geqslant a$ so the MLE will be biased upward.  In cases like this it is useful to derive the distribution of the MLE and have a look at its expected value, to quantify the bias.  You might also consider adjusting the estimator to correct the bias.  
Letting $\hat{A} = \hat{A}(X_1, ..., X_n) = \min X_i$ be the estimator, and assuming for our derivation that $n>2$, we have the distribution function:
$$\begin{equation} \begin{aligned}
F_{\hat{A}}(r) \equiv \mathbb{P}(\hat{A} \leqslant r) = \mathbb{P}(\min X_i \leqslant r) 
&= 1- \mathbb{P}(\min X_i > r) \\[6pt]
&= 1-\mathbb{P}(X_1 > r) \cdot ... \cdot \mathbb{P}(X_n > r) \\[6pt]
&= \begin{cases} 
      1-a^n r^{-n} & \text{ for } r \geqslant a, \\[6pt]
      0 & \text{ for } r < a, \end{cases}
\end{aligned} \end{equation}$$
and corresponding density function:
$$\begin{equation} \begin{aligned}
f_{\hat{A}}(r) = \begin{cases} 
      n a^n r^{-n-1} & \text{ for } r \geqslant a, \\[6pt]
      0 & \text{ for } r < a. \end{cases}
\end{aligned} \end{equation}$$
This gives the expected value:
$$\begin{equation} \begin{aligned}
\mathbb{E}(\hat{A}) = \int \limits_a^\infty r f_{\hat{A}}(r) dr &= \int \limits_a^\infty n a^n r^{-n} dr \\[8pt]
&= \Bigg[ \frac{n}{n-1} \cdot a^n r^{-(n-1)} \Bigg]_{r=a}^{r \rightarrow \infty} \\[8pt]
&= \frac{n}{n-1} \cdot a.
\end{aligned} \end{equation}$$

Correcting for bias: We can see from the above result that the MLE is biased upward, but it can easily be "corrected" by using the corresponding estimator:
$$A^* \equiv \frac{n-1}{n} \hat{A} = \frac{n-1}{n} \min X_i.$$
With some additional algebra it can be shown that the bias-adjusted MLE has moments:
$$\mathbb{E}(A^*) = a \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \mathbb{V}(A^*) = \frac{a^2}{n(n-2)}.$$
This latter estimator is unbiased and consistent.  It is probably a preferable estimator to the MLE.
A: You are correct that the problem doesn't satisfy the regularity conditions, but you can still find the MLE by looking at the derivative.  Let's work with the log likelihood:
$$l(a;x) = n\log(a) -2\Sigma\log(x_i)$$
Clearly this is monotonically increasing in $a$ (this is all the looking at the derivative we're going to do.)  So... our MLE $\hat{a}$ will be as large as possible.  How large is that?  We know $a \leq \min(x_i)$, so the largest $a$ can be is $\min(x_i)$.  It follows that $\hat{a} = \min(x_i)$.
