# Most powerful test. Help using Neyman Pearson

$$X_1,X_2,...$$ independent continuous random variables with p.d.f

$$f(x) = \theta x^{\theta-1}$$ if $$0 otherwise for $$\theta > 0$$

sample size = 1

Use Neyman-Pearson Lemma to drive MP test for the hypothesis $$H_0: \theta = 4$$ $$H_1: \theta = 6$$ at level of significance $$\alpha$$

Derive the power of the above test at $$\theta = 6$$

So my thoughts are we use the NP lemma, so this would go $$\frac{L(\theta_0|\mathbf{x})}{L(\theta_1|\mathbf{x})}\leq c$$

this would be the same as $$\frac{4x^3}{6x^5} < c$$

simplifies to $$\frac{2}{3x^2} < c$$ which after this point I'm not too sure how to complete if i'm being honest

But im thinking along the lines of

$$P(X> \sqrt \frac{2}{3c} | \theta_0 = 4) =\alpha$$ and for the power part the same thing but use $$\theta = 6$$

If someone could walk me through this please, it'll be really helpful

Thank you

• The likelihoods should involve $x_1, \ldots, x_n$ (each likelihood will be the product of densities). Apr 29 '21 at 15:28
• Huh? You have two null hypotheses Apr 29 '21 at 15:28

## 2 Answers

This exercise is very simple because you have a single observation thus your random sample is $$X_1$$

Your solution is almost correct as you arrived at

$$\frac{4x^3}{6x^5}\leq c$$

that is the same as

$$x>k$$

Thus simply applying the definition you get (fixing a certain $$\alpha$$)

$$\mathbb{P}[X>k|\theta=4]=\int_{k}^1 4x^3dx=\alpha \rightarrow k=\sqrt[4]{1-\alpha}$$

thus you reject the null hypotesis iff you single observation $$x_1>\sqrt[4]{1-\alpha}$$

to derive the power at $$\theta=6$$ always using the definition you get

$$\gamma=\mathbb{P}[X>\sqrt[4]{1-\alpha}|\theta=6]=\int_{\sqrt[4]{1-\alpha}}^1 6x^5dx=1-(1-\alpha)^{3/2}$$

• I did not see that was sample size was $n=1$! Good on you. Apr 29 '21 at 19:53
• can i ask why $\frac{4x^3}{6x^5} ≤ c$ is the same as $x > k$?
– Amy
Apr 30 '21 at 10:28
• @Amy : Yes, of course! You have that $$\frac{4x^3}{6x^5} \leq c$$ to be solved w.r.t. $x$. Easy find $$|x|>\sqrt{\frac{2}{3c}}$$ but being $x$ always positive you have $$x>k$$ where $k$ is the expression you found. There is no need to explicitate this constant because for your purposes this is only a point at which evaluate an integral... Apr 30 '21 at 10:34
• Thank you so much that makes a lot of sense :)
– Amy
Apr 30 '21 at 10:35

As was mentioned in the comment sections, $$L$$ depends on the entire observed simple random sample. $$L(\theta|x_1,\ldots,x_n)=\theta^n (x_1 \times \ldots \times x_n)^{\theta -1}$$ This gives us the following likelihood ratio: $$\frac{L(\theta_0|x_1,\ldots,x_n)}{L(\theta_1|x_1,\ldots,x_n)}=\Big(\frac{\theta_0}{\theta_1}\Big)^n(x_1 \times \ldots \times x_n)^{\theta_0 - \theta_1}$$ Assume $$\theta=\theta_0$$ is true, and let $$X_1,\ldots,X_n \sim f$$ be iid. Put $$Q=-\Big[\ln(X_1)+\ldots + \ln(X_n)\Big]$$ You should check that, when $$n$$ is large, $$Q$$ is approximately $$N\Big(\frac{n}{\theta_0},\frac{n}{\theta_0^2}\Big)$$. Since $$\theta_1>\theta_0$$, $$P\Bigg(\frac{L(\theta_0|X_1,\ldots, X_n)}{L(\theta_1|X_1,\ldots, X_n)} where $$z^*=\frac{\theta_0 \ln\big[c(\theta_1/\theta_0)^n\big]}{(\theta_1 - \theta_0)\sqrt{n}}-\sqrt{n}$$. If we want the aforementioned probability to equal $$\alpha$$, we can take $$z^*=z_{1-\alpha}$$ and solve this for $$c$$, yielding the following: $$c=(\theta_0/\theta_1)^n\exp\Bigg\{\frac{(z_{1-\alpha}+\sqrt{n})(\theta_1 - \theta_0)\sqrt{n}}{\theta_0}\Bigg\}$$

Remark: I neglected to see that you only had a single observation. This is how you'd proceed with you have a large sample size.