Question #104554
Examine whether a best critical region exists for testing the null hypothesis H0: θ=θ0 against the alternative hypothesis H1:θ>θ0 for the parameter θ of the distribution:


f(X, θ)=(1+θ)/(x+2)^2 1≤x<∞
1
Expert's answer
2020-03-06T12:08:09-0500

Suppose x1,,xnx_1,\ldots,x_n is a random sample from a population. We will find the test with the best critical region, that is, find the uniformly most powerful test, with a sample size nn and a significance level α\alpha to test the simple H0:θ=θ0H_0: \theta=\theta_0 against the composite H1:θ>θ0H_1: \theta>\theta_0.

Let α=P(C;θ0)\alpha=P(C;\theta_0).

For each simple H1:θ=θaH_1: \theta=\theta_a, say, the ratio of the likelihood functions is:

L(θ0)L(θa)=1+θ0(x1+2)21+θ0(xn+2)21+θa(x1+2)21+θa(xn+2)2k,k=const.(1+θ01+θa)nk.1+θ01+θak1/n.\frac{L(\theta_0)}{L(\theta_a)}=\frac{\frac{1+\theta_0}{(x_1+2)^2}\ldots\frac{1+\theta_0}{(x_n+2)^2}}{\frac{1+\theta_a}{(x_1+2)^2}\ldots\frac{1+\theta_a}{(x_n+2)^2}}\leq k, k=const.\\ (\frac{1+\theta_0}{1+\theta_a})^n\leq k.\\ \frac{1+\theta_0}{1+\theta_a}\leq k^{1/n}.

Here we use the Neyman Pearson Lemma.

The left side of the last inequality does not depend on x1,,xnx_1,\ldots,x_n. So each critical region is "the best". The best critical region does not exist.


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