# Finding the maximum likelihood estimator (Theoretical statistics)

Let $$X_1, X_2, ..X_n$$ represent a random sample from this pdf: $$f(x|\theta)= \frac{3x^2}{\theta^3}, 0\leq x\leq \theta$$ (with 0 elsewhere)

Could someone explain how I would find the maximum likelihood estimator (MLE) of this pdf? I know I have to get the likelihood function L, then the log likelihood function, then differentiate, but I am getting lost in the calculations. Thanks!

• Please show us your workings. What I'd expect to have is that the log likelihood function is a decreasing function of $\theta$. Note that part of the likelihood function is $\log \mathbb{1}_{x \in [0,\theta]^n}$, so the maximum value must lie in $\theta \geq \max_{1 \leq i \leq n}(x_i)$. – fGDu94 Feb 5 '20 at 20:08
• Look at the similar question: math.stackexchange.com/q/3527189 – NCh Feb 5 '20 at 20:24
• There are tons of questions here dealing with problems where support depends on parameter. Please search the site and you will find that differentiation is not the way to go. – StubbornAtom Feb 5 '20 at 20:36
• – StubbornAtom Feb 6 '20 at 19:49

$$\mathcal{L}(\theta; X_1,\dots, X_n) = f(X_1)\cdot\ldots\cdot f(X_n)$$ $$= \begin{cases}\frac{3X_1^2}{\theta^3}, & 0\leq X_1\leq \theta \cr 0, & \text{elsewhere}\end{cases}\;\times\;\ldots\;\times\; \begin{cases}\frac{3X_n^2}{\theta^3}, & 0\leq X_n\leq \theta \cr 0, & \text{elsewhere}\end{cases}$$ $$=\begin{cases}\frac{3^n X_1^2\cdot\ldots\cdot X_n^2}{\theta^{3n}}, & 0\leq X_1,\ldots,X_n\leq \theta \cr 0, & \text{elsewhere}\end{cases}$$ $$=\begin{cases}\frac{3^n X_1^2\cdot\ldots\cdot X_n^2}{\theta^{3n}}, & \max(X_1,\ldots,X_n)\leq \theta \cr 0, & \text{elsewhere}\end{cases}$$ $$=\begin{cases}\frac{3^n X_1^2\cdot\ldots\cdot X_n^2}{\theta^{3n}}, & \theta\geq \max(X_1,\ldots,X_n) \cr 0, & \theta< \max(X_1,\ldots,X_n) \end{cases}$$
Since $$\frac{3^nX_1^2\cdot\ldots\cdot X_n^2}{\theta^{3n}}$$ decrease as $$\theta$$ increase, the highest value of $$\mathcal{L}(\theta; X_1,\dots, X_n)$$ is attained at the smallest value of $$\theta$$ satisfying the inequality $$\theta\geq \max(X_1,\ldots,X_n)$$. So, MLE is $$\hat\theta = \max(X_1,\ldots,X_n)$$.