Statistics: finding an MLE for theta

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I have to obtain a likelihood-ratio test for $H_0: \theta =0 $ vs $H_1:\theta \neq 0$.

$\theta \in [-1,1]$

I also think the size of the random sample is $n=1$, it says "given an observation of $X$". It is written in another language so this was my attempt at translating the problem.

The pdf of our random variable is:

$$f_\theta (x) = (2\theta x + 1 - \theta) I_{(0,1)}(x)$$

In order to design the test, I know that I have to obtain the MLE (maximum likelihood estimator) for $\theta$, but I just cannot do it, and it seems very trivial.

Any help will be very appreciated

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Now the text is complete and it is possible to show a solution.

First of all let's have a look at the density

$$f(x;\theta)=\theta(2x-1)+1$$

$x \in (0;1)$

$\theta \in [-1;1]$

For $\theta=0$ f is uniform For the rest, f is a straight line so, the MLE can be only $\pm1$, depending on the sign of the coefficient $(2x-1)$

Thus, the MLE for $\theta$ is the following

$$ \hat{\theta} = \begin{cases} 1, & \text{if $x>\frac{1}{2}$} \\ -1, & \text{if $x<\frac{1}{2}$} \end{cases}$$

for x=0 any value of $\theta$ is MLE

Now we can set the generalized likelyhood ratio

$$ \lambda(x) = \begin{cases} \frac{1}{2x}, & \text{if $x>\frac{1}{2}$} \\ \frac{1}{2-2x}, & \text{if $x<\frac{1}{2}$} \end{cases}$$

we finished.

Knowing the distribution of X, that is uniform in (0;1) under $H_0$, it is easy to find the critical region after setting a size for $\alpha$ for the test