0

Let $Y_{1}, \ldots, Y_{n}$ be a random sample from from a $U(0,2 \theta)$ distribution. The probability density function for each $Y_{i}(i=1, \ldots, n)$ is given by:

$f_{Y_{i}}(y \mid \theta)=\left\{\begin{array}{ll}\frac{1}{2 \theta}, & 0 \leq y \leq 2 \theta \\ 0, & \text { otherwise }\end{array}\right.$

The likelihood function for $\theta$ given the sample $Y=y$ may be written as: $ L(\theta \mid y)=\prod_{i=1}^{n} \frac{1}{2 \theta} I\left(y_{i} \leq 2 \theta\right)=(2 \theta)^{-n} I\left(y_{n} \leq 2 \theta\right) $ where $y_{n}$ denotes the sample maximum, $I$ is an indicator function.

I need to find the MLE for $\theta$

The text I'm following starts by pointing out that the likelihood function takes strictly positive values for $2 \theta \geq y_{n}$ or else $\theta \geq y_{n} / 2 .$ Specifically: $ L(\theta \mid Y=y)=\theta^{-n}, \quad \theta \geq y_{n / 2} $

If I accept this, which I don't see where it's coming from, then... since the likelihood is decreasing in $\theta,$ I can maximise at $y_{n} / 2 .$ And hence the MLE for $\theta$ would be $\hat{\theta}^{M L E}=Y_{n} / 2$

But then again, this intermediate step is not entirely clear.

RobPratt
  • 45,619
JKM
  • 449
  • 1
    Should be $\prod_{i=1}^n I(y_i\le 2\theta)=I(y_n \color{red}\le 2\theta)$. – StubbornAtom Jul 14 '20 at 19:26
  • Ok this is where the confusion was coming from. Just changed it. The perils of taking your texts at face value. – JKM Jul 14 '20 at 19:31
  • Well the confusion was stemming from typo in the formulation, so your comment for the product of the indicator solved my question. But that thread is certainly helpful and related. – JKM Jul 14 '20 at 19:57

0 Answers0