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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.

  • $\begingroup$ Try to type out your questions and add the self-study tag. $\endgroup$ May 20, 2020 at 14:38
  • $\begingroup$ This similar question might help: stats.stackexchange.com/q/386092/119261. $\endgroup$ May 20, 2020 at 14:44
  • $\begingroup$ 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 $\endgroup$ May 20, 2020 at 18:22
  • $\begingroup$ 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. $\endgroup$
    – BruceET
    May 22, 2020 at 2:12

1 Answer 1


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
[1] 0.410 0.410 0.153 0.025 0.002
[1] 0.008 0.076 0.264 0.412 0.240
[1] 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")

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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)
[1] 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.

enter image description here

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.]

  • $\begingroup$ 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). $\endgroup$ May 24, 2020 at 16:28

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