We have likelihood function

$$ L\left(\mathbf{\Theta} \middle| x_1, \ldots, x_n \right) = \prod^n_{i=1} f\left( x_i \middle| \mathbf{\Theta} \right) $$

and a score function

$$ V\left(\mathbf{\Theta} \middle| x_1, \ldots, x_n \right) = \nabla_\Theta \sum^n_{i=1} \ln{f\left(x_i, \mathbf{\Theta} \right)}, $$

where $x_i$ are observations and $\mathbf{\Theta}$ is the vector of parameters.

If we consider the maximum likelihood estimator $\mathbf{\Theta^{\ast}}$, it's the extreme of $L$, so

$$ V\left(\mathbf{\Theta^{\ast}}\right) = 0, $$

but we need $L\left(\mathbf{\Theta^{\ast}}\middle| x_1, \ldots, x_n \right) \neq 0$.

So, is this inequality always true or is there a situation, where the likelihood with the MLE could be equal to zero?


If you have a very inadequate model such that at least one discretely (or continuously) distributed observation has zero probability (or probability density) for any parameter value $\theta\in\Theta$, that is, you essentially observe something that is impossible under that model, then yes, your maximum likelihood would be zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.