0

I am trying to find the MLE for $\{X_i\}_{i=1}^{n}$ that have the following pdf

$$\frac{3x^2}{\theta^3}I(0\le x \le \theta)$$

I tried solving it directly by taking the log of the likelihood and then through the derivative solve for $\theta$, but the indicator function gives me issues. Intuitively, to maximize the mle I need a very small $\theta$ by the restriction $X_{(1)}$ is the MLE. Is this correct or is there a way to solve this directly?

nvm
  • 1,296

1 Answers1

1

You are correct in saying that you need the smallest $\theta$, but for the given problem, the smallest $\theta$ is going to be $X_{(n)}$ as opposed to $X_{(1)}$ since you want $\theta$ to be greater than all the $X_i$.

More concretely, the likelihood here is given by (assuming iid) \begin{equation*} \mathcal{L}(\theta) = \prod_{i=1}^n \frac{3x_i^2}{\theta^3}I(0\leq x_i \leq \theta) = \frac{3^n \prod_{i=1}^n x_i^2}{\theta^{3n}} I (0\leq x_{(n)} \leq \theta). \end{equation*} This is maximized when $\theta = x_{(n)}$.

EuxhenH
  • 1,516
  • But doesn't that imply that $X_{(1)}\le \theta$ which shows that $X_{(1)}$ maximizes the MLE – nvm Feb 08 '20 at 03:23
  • 1
    If $\theta < X_{(n)}$ then $I(0\leq x_{(n)} \leq \theta) = 0$, so the likelihood is $0$. Remember you want $\theta$ to be greater than all $X_i$. If $\theta$ is smaller than one of them, then you couldn't have possibly generated this data. – EuxhenH Feb 08 '20 at 03:24
  • That Makes Sense. Thank You – nvm Feb 08 '20 at 03:27
  • @JeanMarie By definition, the likelihood is a function of $\theta$. – EuxhenH Apr 22 '20 at 23:11
  • You are perfectly right ; I thought I had erased this ridiculous remark. I was thinking to the estimator $\max(x_1,x_2...x_n)$ one can derive from this computation. In fact, I had another remark : the given pdf is that of the RV $max(X,Y,Z)$ (order statistics) of three iid RVs $\approx \ Unif([0,\theta])$. – Jean Marie Apr 22 '20 at 23:17