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I got a log-likelihood value of -34.82, so I am not getting whether the answer which I have got is right or not.

Can the likelihood take values outside of the range $[0, 1]$?

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    $\begingroup$ This should not be marked as a duplicate. The likelihood is not the same conceptually as a density and the distinction is important especially for beginners. Only an experienced statistician/econometrician would see the equivalence between this question and the one linked as a supposed duplicate. @whuber $\endgroup$
    – Hirek
    Commented Mar 5, 2015 at 20:16
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    $\begingroup$ @Hirek The likelihood is defined as a probability: that is so elementary, it is reasonable to suppose the connection would be apparent to anyone employing the word "likelihood" in any technical sense. Moreover, we have many threads discussing precisely this same question about log likelihood, so the issue is not whether to close this question as a duplicate, but rather what duplicate would be the most helpful. $\endgroup$
    – whuber
    Commented Mar 5, 2015 at 20:53
  • $\begingroup$ Doesn't a negative log-likelihood corresponds to a positive likelihood less than one? $\endgroup$
    – user20637
    Commented Jul 25, 2017 at 15:57
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    $\begingroup$ @whuber How can that be? Likelihood can be greater than 1, but probability can't! $\endgroup$ Commented Mar 13, 2018 at 16:32
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    $\begingroup$ @BjörnLindqvist For continuous distributions, likelihoods drop the infinitesimal probability elements. Thus, they are only defined up to positive multiples in the first place. For discrete distributions likelihoods are indeed probabilities and therefore must be less than $1$. $\endgroup$
    – whuber
    Commented Mar 13, 2018 at 16:54

2 Answers 2

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Likelihood must be at least 0, and can be greater than 1.

Consider, for example, likelihood for three observations from a uniform on (0,0.1); when non-zero, the density is 10, so the product of the densities would be 1000.

Consequently log-likelihood may be negative, but it may also be positive.

[Indeed, according to some definitions the likelihood is only defined up to a multiplicative constant (e.g. see here), so even if the density were bounded by 1, the likelihood still wouldn't be.]

Clarifications as a result of comments/chat: For a continuous distribution, likelihood is defined in terms of density. Density must be at least $0$ and can exceed $1$; and as a result, likelihood can exceed $1$.

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    $\begingroup$ In response to your first comment, Fisher defined likelihood in such a way that $\mathcal{L}(\theta;\underline{x})=c\cdot \prod_i f_{X;\theta}(x_i)$ for any $c>0$ is a likelihood function (as long as any likelihood comparisons were performed with the same $c$, naturally). $\endgroup$
    – Glen_b
    Commented Sep 4, 2019 at 0:27
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    $\begingroup$ @Kirsten Sure (as is indicated by the link in my answer) ... I'm responding to the direct question "Can the likelihood take values outside of the range [0,1]?" to which the short version of my answer is "it cannot be below 0 but it can exceed 1". If the downvote was yours, could you clarify the problem you perceive? $\endgroup$
    – Glen_b
    Commented Oct 9, 2022 at 4:36
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    $\begingroup$ For a continuous random variable, the probability of the data given the parameters is indeed $0$, but in that case the likelihood is not the probability. Rather for continuous variables it's defined in terms of density. Many posts on site discuss the fact that density and probability are not the same thing. The page you link certainly shows likelihoods greater than 0. $\endgroup$
    – Glen_b
    Commented Oct 9, 2022 at 9:55
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    $\begingroup$ Certainly density can exceed 1, just as I state in my answer. Again, density is not probability: (i) Consider a uniform distribution on (0, 0.1). ... what's the height of the density at x=0.05? (ii) consider a normal, mean 0, s.d. 0.1. What's the height of the density at 0? $\endgroup$
    – Glen_b
    Commented Oct 9, 2022 at 10:50
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    $\begingroup$ Please make an edit, @Glen_b, to your post: anything trivial, if you may, so that the downvote could be removed. For context, see the comment. $\endgroup$ Commented Oct 10, 2022 at 5:57
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The likelihood function is a product of density functions for independent samples. A density function can have non-negative values. The log-likelihood is the logarithm of a likelihood function. If your likelihood function $L\left(x\right)$ has values in $\left(0,1\right)$ for some $x$, then the log-likelihood function $\log L\left(x\right)$ will have values between $\left(-\infty,0\right)$. For $L\left(x\right)\in\left[1,\infty\right)$ the $\log L\left(x\right)\in\left[0,\infty\right)$. So $-34.82$ is a typical value for a log-likelihood function.

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