# Neyman pearson on discrete distribution  I have found the ratio of h1 to h0 and the ratio is increasing .So we should reject H0 for large values of x.How should i find the critical region for such type of questions.It seems to me the rejection region is P(x>k)=α.How i determine k .Thank you.

• Try to type out your questions and add the self-study tag. May 20, 2020 at 14:38
• This similar question might help: stats.stackexchange.com/q/386092/119261. May 20, 2020 at 14:44
• It mights sounds silly . But what is the significance of finding liklihood ratio of null and alternative hypothesis.Why should't i directly reject null hypothesis for all values of P(x<k)=α May 20, 2020 at 18:22
• In answer to your previous Comment, you were asked to make $\alpha = 0.1.$ // Finally got a version of my answer that I think/hope will be helpful. May 22, 2020 at 2:12

The problem in your Question is somewhat quirky. Possibly that's on purpose to show you that likelihood ratio tests do not always provide satisfactory results--especially when the distributions for $$H_0$$ and $$H_1$$ are too nearly alike. I will return to this problem after looking at a likelihood ratio test that works more satisfactorily.

Consider a test of $$H_0$$ vs $$H_1$$ where discrete distributions on integers $$0$$ through $$4$$ are as shown below, followed by the likelihood of the null distribution divided by the alternative distribution.

pdf.0; pdf.1; LR
 0.410 0.410 0.153 0.025 0.002
 0.008 0.076 0.264 0.412 0.240
 51.250  5.395  0.580  0.061  0.008


We want a test with significance level not greater than $$0.1 = 10\%.$$ If we reject for small values of the likelihood ratio, then we can reject for LR < .1, which means rejecting for $$X \ge 3.$$

Then $$\alpha = P(\mathrm{Rej}|H_0) = P(X \ge 3|H_0)\\ = 0.025 + 0.002 = 0.025.$$ [If we tried to increase the rejection region to include $$X=3,$$ then we would have $$\alpha > 0.1.]$$

It follows that the power of this test is $$\pi = P(\mathrm{Rej}|H_1) = P(X \ge 3|H_1)\\ =0.412 + 0.240 = 0.662.$$

The figure below shows the two PDFs with $$H_0$$ in blue and $$H_1$$ in brown. The rejection region is to the right of the vertical dotted line. In the rejection region the blue bars (for $$H_0)$$ are relatively short and the red bars (for $$H_1)$$ are relatively tall. The distributions are sufficiently different that a reasonably useful likelihood ratio test can be found.

plot(x-.02, pdf.0, type="h", lwd=2, col="blue")
lines(x+.02, pdf.1, type="h", lwd=2, col="brown")
abline(h=0, col="green2")
abline(v = 2.5, lty="dotted") Now, we turn to the problem in your Question. Under $$H_0$$ the PDF is given by f.0 and under $$H_1$$ by f.1 for values $$X = 1,2,3,4,5.$$

f.0 = c( .1, .1, .1, .1, .6)
f.1 = c(.05,.06,.08,.09,.72)
like.rat = f.0/f.1;  round(like.rat, 2)
 2.00 1.67 1.25 1.11 0.83


If we reject for small values of the likelihood ratio then we reject when $$X = 4,$$ so the significance level is $$P(X=4|f_0) = 0.1$$ (as requested). We cannot put the $$X$$-value with the smallest likelihood ratio into the rejection region because that would give $$\alpha > 0.1.$$

Accordingly, the power is $$P(X=4|f_1) = 0.09,$$ disappointingly small for practical use.

A useful likelihood ratio test is not possible because the distributions described by $$f_0$$ and $$f_1$$ are not much different from one another, as shown in the figure below. Notes: (1) The distributions in the first example are roughly modeled after $$\mathsf{Binom}(4, 0.2)$$ and $$\mathsf{Binom}(4, 0.7).$$ Rounded to three places and altered to sum to exactly $$1.$$

(2) If you know about randomized tests, then you could reject with probability $$1/6$$ only for $$X = 5$$ (perhaps with the roll of a fair die) to get $$\alpha=0.1.$$ In that case, the power would be $$0.72/6 = 0.18,$$ which is better than $$0.09$$ with the non-randomized test above, but still not large enough to be considered useful.

References for randomized tests: Mood, Graybill, Boes: Intro. to the Theory of Statistics, 3e (1974), p404; Bain & Englehardt, Intro. to Probability and Math. Statistics, 2e (1992), p411. [Productive searching online is difficult because of conflation with 'randomization' tests, a different topic.]

• Isn't the usual definition of likelihood ratio $\frac{f_1}{f_0}$ instead of $\frac{f_0}{f_1}$ the way you have it? One advantage in this problem is that the usual ratio takes on values $0.5, 0.6, 0.8, 0.9, 1.2$ without needing to invoke round(like.rat,2). May 24, 2020 at 16:28