# 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 $\theta \geq \max_{1 \leq i \leq n}(x_i)$. Feb 5, 2020 at 20:08
• Look at the similar question: math.stackexchange.com/q/3527189
– NCh
Feb 5, 2020 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. Feb 5, 2020 at 20:36
• Feb 6, 2020 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 $$\hat\theta = \max(X_1,\ldots,X_n)$$.