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

$f(x) = \theta x^{\theta-1}$ if $0<x<1 , 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

  • $\begingroup$ The likelihoods should involve $x_1, \ldots, x_n$ (each likelihood will be the product of densities). $\endgroup$
    – angryavian
    Apr 29, 2021 at 15:28
  • $\begingroup$ Huh? You have two null hypotheses $\endgroup$ Apr 29, 2021 at 15:28

2 Answers 2


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


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

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

  • $\begingroup$ I did not see that was sample size was $n=1$! Good on you. $\endgroup$
    – Matthew H.
    Apr 29, 2021 at 19:53
  • $\begingroup$ can i ask why $\frac{4x^3}{6x^5} ≤ c$ is the same as $x > k$? $\endgroup$
    – Amy
    Apr 30, 2021 at 10:28
  • $\begingroup$ @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... $\endgroup$
    – tommik
    Apr 30, 2021 at 10:34
  • $\begingroup$ Thank you so much that makes a lot of sense :) $\endgroup$
    – Amy
    Apr 30, 2021 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 $\theta_1>\theta_0$, $$P\Bigg(\frac{L(\theta_0|X_1,\ldots, X_n)}{L(\theta_1|X_1,\ldots, X_n)}<c\Bigg)=P\Bigg(Q<\frac{\ln\big[c(\theta_1/\theta_0)^n\big]}{\theta_1-\theta_0}\Bigg)\approx P(Z<z^*)$$ 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.


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