Showing $max(X_i)$ is the consistent estimator for pdf $f(x;θ) = 2x / θ^2$ Suppose $X_1,...,X_n$ are $iid$ with pdf $f(x;θ) = 2x / θ^2, 0 < x ≤ θ$, I need to find the MLE for the median of the distribution and show that it is a consistent estimator.
I don't understand why the solution set the function $F_X(M)=M^2/\sigma^2 = 1/2$ ... how do we know $1/2$ is the median? And how can I show that it is a consistent estimator. 
 A: The median is defined as the point where:
$$\int^M p(x)dx = 0.5$$
(it is the halfway point - half the samples are expected to lie beneath it, and half above).
So, we're looking for a median M which satisfies:
$$\int_0^M \frac{2x}{\theta^2}dx = 0.5$$
$$\Rightarrow \frac{2}{\theta^2} * \frac{M^2}{2} = 0.5$$
$$ \Rightarrow M = \theta/\sqrt2 $$
... which is your answer.
Edit: I'm assuming that you're asking why the estimator M stated above is consistent - this is weird in practice, because if you have the PDF, you know the median surely.
The estimator here doesn't depend on the sample $x$, hence it is a constant (it doesn't have a sampling distribution). It doesn't depend on the sample size and it's unbiased. So it's consistent.
A: For continuous distributions, median is defined to be that value of $x$ for which the distribution function $F(x)$ equals $1/2$. So $1/2$ is not median of the distribution, you would have to solve for $x$ to find the median.
We have the population pdf $$f_{\theta}(x)=\frac{2x}{\theta^2}\mathbf1_{0<x<\theta}$$
The distribution function is of the form 
\begin{align}
F_{\theta}(x)&=\int_0^x f_{\theta}(t)\,dt
\\&=\begin{cases}0&,\text{ if }x<0\\\left(\frac{x}{\theta}\right)^2&,\text{ if }0<x<\theta\\1&,\text{ if }x>\theta\end{cases}
\end{align}
So, for $0<x<\theta$,
$$F_{\theta}(x)=\frac{1}{2}\implies x=\frac{\theta}{\sqrt{2}}=g(\theta)\quad,\,\text{say}$$
By invariance property, MLE of $g(\theta)$ is $g(\hat\theta)$ where $\hat\theta$ is the MLE of $\theta$.
Indeed, for a sample of size $n$, $$\hat\theta=\max_{1\le i\le n} X_i$$
As noted in a comment by @Glen_b, here we are talking about consistency of a (point) estimator that estimates some function of the parameter of interest. In this case, you are probably asking about checking consistency of the estimator $\hat\theta$ in estimating $\theta$.
Assuming that is the case, if we can show that $\hat\theta$ is a consistent estimator of $\theta$, then by the continuous mapping theorem $g(\hat\theta)$ will be a consistent estimator of $g(\theta)$. 
To show that $\hat\theta$ is consistent, you can work directly with the definition of consistency. That is, use the pdf of $\hat\theta$ to show that for some positive $\epsilon$, 
$$\lim_{n\to\infty}P_{\theta}\left[\left|\hat\theta-\theta\right|<\epsilon\right]=1$$
Or you can use a sufficient condition of consistency to show that as $n\to \infty$, $$E(\hat\theta)\to \theta\qquad\text{ and }\qquad \operatorname{Var}(\hat\theta)\to 0$$
As this is only a sufficient condition, using it to show consistency might not work always.
(Maximum likelihood estimators are often consistent estimators of the unknown parameter provided some regularity conditions are met. Here, one such regularity condition does not hold; notably the support of the distribution depends on the parameter. So we are resorting to the definitions to prove consistency.)
