Most powerful test - hypothesis

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sample of size n = 1 from p.d.f.

$f(x| \theta) = 1 +\theta^2 (0.5-x)$ if $0<x<1, 0 $ o/w

where $-1 ≤ \theta ≤ 1$

Derive the MP test for testing $$H_0 : \theta = 0$$ $$H_A : \theta = \theta_1$$ at level significance $\alpha$

Derive the power of this test at $\theta = \theta_1$

So we have been given this question but only covered it briefly so was wondering if someone could help please.

My thoughts are as follows: The first part I'm not too sure how to do, it is a bit challenging but i have thoughts of using likelihood ratio test?

The second part we just use the definition maybe?

I am pretty new to this topic but would like to get better + if someone could help it would be really helpful, thank you :)

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Using Neyman Pearson's lemma you get

$$\frac{L(\theta_0|\mathbf{x})}{L(\theta_1|\mathbf{x})}\leq c$$

which leads immediately to the following critial region

$$x\leq c^*$$

In fact, simply substituting, you get

$$\frac{1}{1+\theta_1^2(0.5-x)}\leq c$$

that is very easy to be solved w.r.t. $x$ finding

$$x\leq c^*$$

In the constant $c^*$ all the expressions not depending on $x$ are included...but it is not a problem as $c^*$ for you is only a point at which evaluate an integral...

thus you can reject $H_0$ if

$$\mathbb{P}[X\leq c^*|\theta=0]=\alpha$$

that is

$$\int_0^{c^*} dx=\alpha$$

$$c^*=\alpha$$

to get the power, simply using the definition you get

$$\mathbb{P}[X\leq \alpha|\theta=\theta_1]=\gamma$$

that is

$$\int_0^{\alpha}[1+\theta^2(0.5-x)]dx=\left(1+\frac{\theta^2}{2}\right)\alpha-\frac{\theta^2}{2}\alpha^2$$

As you can see (fixing a certain $\alpha$) the power increases as $|\theta|$ increases...

this is the graphic of your power function w.r.t. $\theta$

(for this graphic I set $\alpha=5\%$)

enter image description here