If $Y_1,Y_2,...,Y_n$ ~ $U[0, \theta]$, and we want to test
$$H_0: \theta=\theta_0$$ $$H_1: \theta < \theta_0$$
what would be the likelihood ratio test for a given $\alpha$.
What I know so far:
We know that the MLE of $\theta$ is $Y_\text{MAX}$.
And $F_{Y_\text{MAX}}(y)=P(Y\leq y)=P(\text{all }Y_i \leq y)$
So I think my likelihood function is $(\frac{y}{\theta})^n$
But I don't understand where the supremum comes in and how to find it.
Thanks!
If I understood you question correctly here is my answer
$$\hat{\theta} = max\{Y_1,\ldots,Y_n\}$$ $$max\{Y_1,\ldots,Y_n\} \leq y \Leftrightarrow Y_1 \leq y,\ldots,Y_n\leq y$$
$$P(\hat{\theta} \leq y)= P(max\{Y_1,\ldots,Y_n\} \leq y) = P(Y_1 \leq y,\ldots, Y_n\leq y)$$ Since $Y_i$ are iid we have $$=P(Y_1 \leq y) \cdot \ldots \cdot P(Y_n \leq y) = \left( \frac{y}{\theta}\right)^n$$
In your case $H_1: \theta < \theta_0$ so the critical area looks like $C = \{ Y: \hat{\theta} \leq q\}$ and $P_{H_0}(C) = \alpha$.
$$\alpha = \left(\frac{q}{\theta_0}\right)^n \Rightarrow q = \theta_0 \cdot \alpha^{\frac{1}{n}}$$.
So you reject $H_0$ when $\hat{\theta} < \theta_0 \cdot \alpha^{\frac{1}{n}}$.
