How to find MLE when samples depend on the estimated parameter Can you show me what I'm doing wrong here?  This is the homework problem:
Consider a random sample $Y_1, \ldots , Y_n$ from the pdf $f_Y(y;\theta) = 2y\theta^2$ where $0\le y \le \frac{1}{\theta}$.  Find the maximum likelihood estimator for $\theta$.
The likelihood function for this pdf is $\prod_{i=1}^{n} f_Y(y;\theta) = \prod_{i=1}^{n} 2y_{i}\theta^2 = 2^n \theta^{2n} \prod_{i=1}^{n} y_i$.
So to me, it looks like I can maximize the likelihood function by choosing $y_{max}$ for the MLE.  But then what do I do about the condition $0\le y \le \frac{1}{\theta}$?  
I tried looking up how to do this but the only example I can find is the one about the uniform distribution, but that one is much more straightforward because the condition on y is just $0 \le y \le \theta$ and not $\frac{1}{\theta}$.  To maximize $\frac{1}{\theta}$ I would choose my MLE to be $y_{min}$ but that wouldn't maximize the likelihood function.
Any help would be GREATLY appreciated.  
 A: I suggest you draw the likelihood as a function of $\theta$, without forgetting that $1/\theta$ must be greater than any observation (i.e. what are the bounds on $\theta$?). 
Keep in mind that everything but $\theta^{2n}$ in the likelihood is a constant, and so you can write it as $c.\theta^{2n}$; so just draw $\cal{L}/c$ over the domain of $\theta$. You may find it more convenient to deal with the log-likelihood, or you may not.
It should help you clarify what you're doing.
If you're still stuck, consider thinking in terms of $\psi = 1/\theta$ and then go back to doing it in terms of $\theta$.
(Alternatively - look back at that uniform example. What would the MLE of $\theta$ be if the data were uniform on $[0,1/\theta]$? Can you see how to do the original problem now?)
A: In this kind of problem, it helps a lot to write explicitly the indicators in the definitions of the densities. You have $Y_1,\dots,Y_n$, conditionally IID given $\Theta=\theta$, such that
$$
  f_{Y_i\mid\Theta}(y_i\mid\theta) = 2y_i\theta^2\, I_{[0,\,1/\theta]}(y_i) \, .
$$
Since $0\leq y_i\leq 1/\theta$ if and only if $0\leq \theta\leq 1/y_i$, the indicator can be rewritten as
$$
  I_{[0,\,1/\theta]}(y_i) = I_{[0,\,1/y_i]}(\theta) \, .
$$
Introducing the notations $Y=(Y_1,\dots,Y_n)$ and $y=(y_1,\dots,y_n)$, the likelihood is
$$
  L_y(\theta) = f_{Y\mid\Theta}(y\mid\theta) = \prod_{i=1}^n f_{Y_i\mid\Theta}(y_i\mid\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} \left( \prod_{i=1}^n I_{[0,\,1/y_i]}(\theta) \right) \, .
$$
But a product of indicators is the indicator of the intersection (why?), so
$$
  L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) \, ,
$$
and (why?)
$$
  I_{\cap_{i=1}^n [0,\,1/y_i]}(\theta) = I_{[0,\,1/y_{(n)}]}(\theta) \, ,
$$
in which $y_{(n)}=\max \{y_1,\dots,y_n\}$. Hence,
$$
  L_y(\theta) = 2^n \left(\prod_{i=1}^n y_i\right) \theta^{2n} I_{[0,\,1/y_{(n)}]}(\theta) \, ,
$$
and this is maximized (why? draw the graph for some fixed sample $y$) when $\theta=1/y_{(n)}$, yielding the maximum likelihood estimator
$$
  \hat{\Theta}_{ML}=\frac{1}{Y_{(n)}} \, .
$$
