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I have the following problem:

Consider two probability density functions on $[0,1]: f_0(x) = 1$, and $f_1(x) = 2x$. We want to test such hypotheses: $H_0: X \sim f_0(x)$ versus the alternative $X \sim f_1(x)$, with significance level $\alpha = 0.1$, how large can the power possibly be?

My approach:

Let $\{\Gamma_0, \Gamma_1\}$ be a partition of $[0,1]$ into sets and the decision rule be that hypothesis $H_i$ is true if the observation $X$ belongs to $\Gamma_i$. Then, the probability of type 1 error is $$\alpha = \int_{\Gamma_1} f_0(x)\,\mathrm dx = 0.1$$ while the power of the test is $$power = 1 - \beta = \int_{\Gamma_1} f_1(x)\,\mathrm dx.$$ Now,since $f_0(x)$ is the uniform density, we know that the total "length" of the set $\Gamma_1$ is $0.1$

Then looking at the graph of density $f_1(x)$ i have realized, tha in order to maximize the power of the test i should take the region from 0.9 to 1.

So the maximal power of the test is :

$$power = 1 - \beta = \int_{0.9}^{1} 2x\,\mathrm dx. = 0.19$$

Is my solution correct? And are there some other approaches for solving this exercise? I have thought about Likelihood-ratio test but i don't know how to apply it to my problem.

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Your answer looks correct to me.

The likelihood ratio test would have the rejection rule for the form reject if $\frac{L_0}{L_1} \leq c$, where $L_0$ and $L_1$ are the likelihood under the null and alternative.

So in your problem the rejection rule becomes: "reject if $\frac{1}{2x} \leq c$".

i.e. reject if $x \geq k$ (for some $k$)

Now the $k$ required to get $\alpha = 0.1$ is $k=0.9$.

So in this case the LRT is the same as your test.

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