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.



1 Answer 1


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 $\hat{\theta} < \theta_0 \cdot \alpha^{\frac{1}{n}}$.

  • $\begingroup$ So where would you reject $H_0$? $\endgroup$
    – conv3d
    Nov 14, 2015 at 20:39
  • $\begingroup$ I modified my answer, hope it helps you. $\endgroup$
    – Silvestris
    Nov 19, 2015 at 19:20

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