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