I am reading the book "Patter recognition" by Cristopher Bishop. At Chater 2.3.6 "Bayesian inference for the Gaussian", there is written

The likelihood function, that is, the probability of the observed data given $\mu$, viewed as a function of $\mu$, is given by

$$p(\mathbf{X}|\mu) = \prod_{n=1}^{N} p(x_n|\mu)$$

Again, we emphasize that the likelihood function $p(\mathbf{X}|\mu)$ is not a probability distribution over $\mu$ and is not normalized

I can understand that the likehood is not a probability distribution over $\mu$, that is, it's not the probability of obtaining $\mu$, but rather the probability of obtaining the observation given $\mu$. But why it is normalized? Is it not true that $\int_{-\infty}^{+\infty} p(\mathbf{X}|\mu) =1$?

  • $\begingroup$ Note that the likelihood is a function of $\mu$; the data are fixed. $\endgroup$
    – Glen_b
    Apr 27, 2019 at 14:39
  • $\begingroup$ What are you integrating with respect to? If it’s $dx$ it is $=1$, if it’s $d\mu$ it’s $\ne1$. $\endgroup$
    – Don Slowik
    Apr 29, 2019 at 15:25

1 Answer 1


Likelihood is the probability density over X given $\mu$, so it can be normalized as $$\int_xp(X|\mu)dx$$ However, since it's not the probability, and it will not be treated as probability it doesn't need to be normalized. So, the normalization constants can be dropped without losing anything, and thus they often are omitted.

  • $\begingroup$ So to normalize it I have to divide by $$\int_xp(X|\mu)dx$$? $\endgroup$ Apr 27, 2019 at 0:45
  • $\begingroup$ You could by it’s not necessary for mle $\endgroup$
    – Aksakal
    Apr 27, 2019 at 1:36
  • $\begingroup$ I think "Likelihood is the probability density over X given μ" is misleading. The function of x that you called "likelihood" is a PDF but not a likelihood function. $\endgroup$
    – Nownuri
    Sep 1, 2019 at 15:58

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