# what the distribution of test statistic X as −1≤α≤1 and $f_\alpha(x) = 2\alpha x+1 - \alpha$ w/ $H_0: \alpha= 0$ and $H_a: \alpha > 0$? [closed]

For a real −1≤α≤1 , define $$f_α(x)=2αx+1−α$$. It is easy to see that fα is nonnegative and integrates to 1, namely is a distribution, over [0,1] . Consider the null hypothesis that α=0 , namely $$f_α$$ is uniform, and the alternative hypothesis that α>0 . Given a single sample, 0.8, from $$f_α$$, find the p -value.

With the question, I just assume that $$\alpha$$ is uniform distribution and try to solve it from there but failed. any suggestion to figure out the distribution?

• What do you mean by distribution of $\alpha$? If your sample is $X$, then that is your test statistic. And you know the distribution of $X$ under null hypothesis. Aug 1, 2020 at 6:31
• It's distribution of X not $\alpha$. Aug 25, 2020 at 23:31

You can use the Neyman-Pearson lemma to determine the most powerful test to apply.

First, we want to move into testing two simple hypotheses:

$$H_0: \quad \alpha=0\\ H_1: \quad \alpha=\hat{\alpha}$$

where $$\hat{\alpha} \in \{ \alpha : \alpha > 0 \}$$. Please bear in mind that $$\hat{\alpha}$$ is a fixed value, so we have the full distribution defined. We just don't know the value yet.

If $$\alpha \neq 0$$, the derivative $$\partial_\alpha f_\alpha = 2x-1$$. Then, we have that $$f_\alpha$$ is monotonically decreasing for $$x<1/2$$, and monotonically increasing for $$x>1/2$$.
For $$x=0.8$$, we can set the alternative hypothesis as $$H_1: \, \alpha=1$$, since it corresponds to the MLE.

The Neyman-Pearson lemma allows us to define the most powerful test:

$$\Lambda(x)=\frac{\mathcal{L}(x|H_0)}{\mathcal{L}(x|H_1)}=\frac{1}{2x} \leq k \iff \\ x \geq \frac{1}{2k} = k^*$$

The latter represents the rejection region for $$H_0$$.
This implies that the smallest $$x$$ for a p-value = 0.05 is

$$P(x > k^*)=\int_{k^*}^1 \mathcal{L}(x|H_0) \, dx = \int_{k^*}^1 \, dx = 1 - k^* = 0.05 \implies \\ x > 1 - 0.05 = 0.95$$

In order to get the p-value corresponding to the observed $$x=0.8$$, you simply recalculate the integral:

$$P(x > 0.8) = \int_{0.8}^1 dx = 0.2$$

P.S. Thanks to @whuber for the enlightening comments.

• The question can be solved by null distribution. $H_0: f_\alpha i\text{ is uniform distribution} \qquad H_a: f_\alpha \text{ not uniform distribution}$. When $\alpha=0.8$, the p-value = 1 - 0.8 = 0.2. Aug 25, 2020 at 23:35
• You can test against the null that f is uniform. This doesn't exclude the likelihood ratio test. I tried to show you how that is derived. Aug 26, 2020 at 9:33