Timeline for Is the exact value of any likelihood meaningless?
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Jan 4, 2023 at 6:55 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 13, 2022 at 21:44 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 13, 2022 at 21:34 | comment | added | Michael Lew | The first paragraph is wrong wrong wrong. Likelihood ratios can be used as measures of evidence but an individual likelihood cannot. The "hypotheses" that can be assessed by a likelihood ratio are parameter values within the statistical model, not the usual things brought to mind by the word 'hypothesis '. Likelihoods are entirely model-dependent and and so a likelihood of, say, 0.0023 from a coin tossing experiment with the usual statistical model is not the same as a likelihood of 0.0023 from an experiment where the data are on a continuous scale. That is not a difference between kg and lb. | |
Jun 13, 2022 at 15:30 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 13, 2022 at 8:05 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 13, 2022 at 0:33 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 12, 2022 at 23:28 | comment | added | Henry | Yes it does, but the same issue arises there. If the parameter there is say the probability $\theta$ that the next result is the same as the previous result, then obviously you have to look at the results in order. The probability of seeing those particular sequences is very small for any value of that parameter, simply because the sequences are very long. But even then, you would need to decide whether the likelihood with the first sequence is $\theta^{95}(1-\theta)^{204}$ or ${299 \choose 95}\theta^{95}(1-\theta)^{204}$ while I would say it is proportional to each of those | |
Jun 12, 2022 at 12:11 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 12, 2022 at 2:08 | history | edited | dipetkov | CC BY-SA 4.0 |
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Jun 12, 2022 at 1:08 | comment | added | Henry | Not at all, and I did not say that - it is obviously important in finding a binomial probability and is not meaningless. I am saying it can be used or not in the calculation of a likelihood (so long as this is done consistently) and in that sense does not affect the likelihood of a parameter taking a particular value. Hence my original comment | |
Jun 12, 2022 at 0:32 | comment | added | Henry | $\theta^2(1-\theta)$ is the probability of Heads then Heads then Tails in that order. ${3\choose 2}\theta^2(1-\theta)$ is the probability of $2$ Heads and $1$ Tails in any order. It would be peculiar if your assessment of the likelihood of a particular value of $\theta$ having observed $HHT$ depends on whether you use the full observation or a sufficient statistic but, since likelihood is relative, you do need to be consistent in your choice | |
Jun 12, 2022 at 0:17 | comment | added | Henry | You seem to be suggesting that if one model thinks order matters and the other that order does not matter then there is an issue over the likelihood of the latter. I am saying that since the calculations you do for the likelihoods are only meaningful up to proportionality, you can scale them to be on a comparable basis | |
Jun 12, 2022 at 0:17 | comment | added | Henry | In your $\theta_0 = 0.1$ and $\theta_1 = 0.12$ case and my $HHT$ example, it still does not matter whether you say the likelihood is $\theta^2(1-\theta)$ or ${3\choose 2}\theta^2(1-\theta)$ as you would get a ratio of $1.408$ either way. | |
Jun 11, 2022 at 21:44 | comment | added | Henry | But clearly you can rescale both likelihoods by the same amount without affecting the Likelihood Ratio or Bayes Factor. Suppose you toss a biased coin three times and want to consider the models of the probability of heads as $\theta_0=\frac15$ or $\theta_1=\frac45$. If you see $HHT$ it does not matter whether you say the likelihood of $\theta$ is then $\theta^2(1-\theta)$ or say ${3\choose 2}\theta^2(1-\theta)$ as you will get a ratio of $4$ in both cases. | |
Jun 11, 2022 at 17:19 | history | answered | dipetkov | CC BY-SA 4.0 |