Given $X_1,\ldots, X_n$ are i.i.d random variables with $\ f(x|\theta) = \frac{3x^2}{\theta^3}\ I_{(0,\theta]}(x)$, $\ \theta > 0$. Find the MLE of $\theta$ and its expected value of the MLE.
My attempt: $L(\theta|x) = \frac{3^n(x_1\ldots x_n)^2}{\theta^{3n}}$ for $x_i\in (0,\theta]$ for all $\ i\ =\ 1,\ldots, n$. Thus, $\frac{d}{d\theta} L(\theta|x) = -3^{n+1}(x_1\ldots x_n)^2n\ \theta^{-3n-1}$. Since $x_1,\ldots, x_n\in (0,\theta]$ and $\theta > 0$, $\frac{d}{d\theta} L(\theta|x) < 0$. Therefore, $L(\theta|x)$ only achieves its global maximum at $\theta = 0$ (unreasonable, since $\ \theta > 0$). We conclude that the MLE of $\theta$ does not exist. But if it does not exist, then the same thing is true for its expected value.
My question: Did I get a wrong conclusion about the MLE of $\ \theta$? If the MLE exists, what is the systematic way to find its expected value? Any help on these question would greatly be appreciated.
The likelihood function is a function of $\theta.$ When $\theta$ is less than any of the $x_i$ it is zero. Thus it is equal to $$\frac{3^n}{\theta^{3n}}(x_1\ldots x_n)^2$$ for $\theta\ge\max(x_i)$ and zero otherwise. Thus its maximum at not zero, but rather at the smallest value of $\theta$ so that $\theta\ge \max(x_i).$ In other words, at $\hat\theta = \max(x_i).$
To find its expected value, you can compute the distribution of $\max(x_i)$ and then integrate. To get the distribution of the max, it's easiest to work with CDFs (and then differentiate to get the PDF).