I am having trouble understanding how to use the indicator function to help find the likelihood.
Let $Y_1, Y_2, ... , Y_n$ be a random sample from a population with density function $$ f (y | \theta ) = \begin{cases} 2\theta^2 y^{-3}, & \theta < y < \infty \\ 0, & \text{elsewhere} \end{cases} $$ Find the MLE for $\theta$.
I have two questions about this problem:
We are given that Y min is sufficient for $\theta$. I am told that this is enough to state that Y min is then the MLE for $\theta$ but my question is why this is the case. I know that the MLE is always a function of the sufficient statistic, but I am having trouble understanding how we know that the MLE is ismply $\theta$ and not some other function of the sufficient statistic (such as Y min -1) that involves $\theta$. Ah I hope that was worded clearly.
My next question is this. Assuming that we are not given that the Y min is sufficient for $\theta$, I am having some trouble deriving the MLE. Here is what I have so far: $$ L (\theta) = \prod \frac{2\theta^2}{y^3} = \frac{2^n\theta^{2n}}{\prod y^3} $$
Taking the log of this we obtain:
l ($\theta$) = nln(2) + 2nln($\theta$) - 3 ln( $\prod$ y)
Taking the derivative with respect to theta and setting equal to zero results in:
0 = 0 + $\frac{2n}{\theta}$ + 0
This is obviously an issue because the result does not include the data, so clearly I am missing something. In past experience this result means that the MLE will be either a min or max statistic, but I don't know which one or how to obtain it.
Edit:
Is it enough to just take the log:
l ($\theta$) = nln(2) + 2nln($\theta$) - 3 ln( $\prod$ y)
And say that the likelihood is maximized when the value of y is smallest? So, the MLE of $\theta$ is the minimum statistic?