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

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 $[0, 1]$?

• 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 Mar 5 '15 at 20:16
• @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.
– whuber
Mar 5 '15 at 20:53
• Doesn't a negative log-likelihood corresponds to a positive likelihood less than one? Jul 25 '17 at 15:57
• @whuber How can that be? Likelihood can be greater than 1, but probability can't! Mar 13 '18 at 16:32
• @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 $1$.
– whuber
Mar 13 '18 at 16:54

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.]

• what do you mean that it is defined up to a multiplicative constant? The wikipedia article doesn't explain it well either. Sep 4 '19 at 0:11
• And the current wikipedia article explains it better why the likelihood can be calculated with the probability density rather than the probability itself. en.wikipedia.org/wiki/… Sep 4 '19 at 0:24
• 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). Sep 4 '19 at 0:27

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.