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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!

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  • $\begingroup$ Please add the self-study tag, read its tag-wiki and modify your question to follow the guidelines on asking such questions. In particular, you'll need to clearly identify what you've done to solve the problem yourself, and indicate the specific help you need at the point you struck difficulty. $\endgroup$ – Glen_b -Reinstate Monica Nov 14 '15 at 4:51
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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 $\hat{\theta} < \theta_0 \cdot \alpha^{\frac{1}{n}}$.

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  • $\begingroup$ So where would you reject $H_0$? $\endgroup$ – jchaykow Nov 14 '15 at 20:39
  • $\begingroup$ I modified my answer, hope it helps you. $\endgroup$ – Silvestris Nov 19 '15 at 19:20

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