0
$\begingroup$

Suppose that $Y_1,Y_2,...,Y_n$ are a sample of size $n$ and IID, each of which have distribution

$f_Y(y)=\frac{2y}{\theta^2}$ with support $0<y<\theta$

Find the MLE (method of maximum likelihood) estimate of $\theta$.

I decided to use the log-likelihood technique, but after taking the derivative I end up with this equation:

$$\frac{-2n}{\theta}\sum_{i=1}^nln(Y_i)=0$$

So this suggests that there is no value of $\theta$ that acts as an MLE estimate... but I'm not sure if there are other approaches or if I've made a mistake. How would you find the MLE estimate of $\theta$?

$\endgroup$
1
  • $\begingroup$ If you are correct so far then the derivative with respect to $\theta$ has one sign, provided of course $\theta$ is at least $\max Y_i$. If that sign is in fact negative then the MLE is just $\max Y_i$. $\endgroup$
    – Ian
    Feb 7, 2018 at 2:07

1 Answer 1

Reset to default
2
$\begingroup$

First, notice that

$$ L := \log \left( \prod_{i=1}^n \frac{2y_i}{\theta} \right) = \log \left( \frac{2^n}{\theta^{2n}} \prod_{i=1}^n y_i \right) = n \log 2 + \sum_{i=1}^n \log (y_i) - 2n \log \theta $$

Taking the first derivative w.r.t. $\theta$, we find

$$ \frac{\partial L}{\partial \theta} = -\frac{2n}{\theta} < 0, \forall \theta > 0 $$

This means that $L$ is a strictly decreasing function of $\theta$, therefore picking the minimum allowed value of $\theta$ maximizes $L$. Observe that

$$ 0 < y_i < \theta, \ \forall i \Rightarrow \theta > \max_i y_i $$

Therefore, $\theta_{\text{MLE}} = \max_i y_i$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .