Compute the MLE of variance for $f(x) = 3x^3 /\theta^3$

Question

We are given $X_1, X_2,\dots, X_n$ that are independent and with the same disturbution function $f(x) = 3x^2 /\theta^3$ for $ 0 \leq x \leq \theta$, where $\theta$ is an unknown positive variable. Using the maximum likelyhood method find :

1. An MLE for $\theta$
2. An MLE for the variance of Xi's

My answer to 1 is the well -known method : $L(\theta) = \prod_{i=1}^nf(X_i:\theta) \implies $ $$ L(\theta) = (\frac{3}{\theta^3})^n\prod_{i=1}^{n}x_i^3 $$ Now this has a well-known answer : $\hat \theta = \max_{1\leq i\leq n}x_i$

My problem is question 2 : I haven't got the concept of MLE quite well . I computed $\sigma^2 = Var(Xi) = \frac{3\theta^2}{5}$. When we are trying to find an MLE for $Var(X_i)$ is it like we are trying to find an MLE for $\theta^2$ ? And if we already have an MLE for $\theta$ can't we just say that the MLE for $\theta^2$ is $\hat \theta^2$?
So I would answer something like :
$\hat Var(X_i) =\frac{3}{5}\hat \theta^2 $ but I am not sure at all .

Even if question 1) wasn't present , the only way I know to use this method is to -blindly - take the definition of maximum likelyhood function (which hopefully will be a function of the parameter I want to estimate) $L(\theta) = \prod_{i=1}^nf(X_i:\theta) $ . So I thought I could maybe write $f(x) $ as a function of $3/5 \theta^2$ : $$f(x:\frac{3}{5}\theta^2) = \sqrt{\frac{3}{5}}\frac{3x^3}{\sqrt{\frac{3}{5}\theta^2}^3}$$ Then I would have : $$L(\frac{3}{5}\theta^2) = (\sqrt{\frac{3}{5}}\frac{3}{\sqrt{\frac{3}{5}\theta^2}^3})^n\prod_{i=1}^{n}x_i^3 \implies $$

$$L(\sigma^2) = \Big(\sqrt{\frac{3}{5} }\frac{3}{\sigma^2}\Big)^n\prod_{i=1}^{n}x_i^3 $$ So then the answer would be just $\hat {\sigma^2} = \max\limits_{1 \leq i \leq n}{x_i}$ ?

Answer 1

By the invariance property of MLE, if $\widehat{\theta}$ is a MLE of a parameter $\theta$ and $g: \mathbb{R} \to \mathbb{R}$ is a function, then $g(\widehat{\theta})$ is a MLE of $g(\theta)$. Since you have obtained an expression for the variance in terms of $\theta$, your first answer would be correct.

Your second approach by using first principles again should give the same answer. However, you should be careful to reparametrise from $\theta$ to the variance parameter $V$ more carefully.