5
$\begingroup$

Example 1 of the wikipedia article on the Likelihood Function suggests that given the likelihood function $\mathcal{L}(\theta | X)$, we can find the probability of $\theta$ given $X$, $p(\theta|X)$.

Is this just a confusion of notation? That is, some of the literature likes to write $\mathcal{L}(\theta|X) = p(X|\theta)$ even though the likelihood is $\textit{not}$ a conditional probability distribution, and is perhaps more aptly notated as $p(X;\theta)$.

$\endgroup$
4
  • 1
    $\begingroup$ Example 1 correctly shows that likelihood is a probability of $X$ given a fixed value of parameter $\theta$. For it to be a conditional probability, you'd need to assume $\theta$ to be a random variable and have a distribution, what inevitably leads you to making Bayesian assumptions rather then frequentist. $\endgroup$
    – Tim
    Oct 26, 2017 at 14:00
  • 1
    $\begingroup$ Ah, so the last paragraph makes bayesian assumptions? "This is not the same as saying that the probability that ${\displaystyle p_{\text{H}}=0.5} p_\text{H} = 0.5$, given the observation $HH$, is $0.25$. For that, we could apply Bayes' theorem, which implies that the posterior probability (density) is proportional to the likelihood times the prior probability." $\endgroup$
    – user393454
    Oct 26, 2017 at 14:04
  • 1
    $\begingroup$ Likelihood is not a density on $\Theta$, i.e. $\int_{\Theta} L(\theta | X) d \theta \neq 1$. But it is a density on $X$. $\endgroup$ Oct 26, 2017 at 14:04
  • $\begingroup$ See stats.stackexchange.com/questions/2641/… and stats.stackexchange.com/questions/224037/… $\endgroup$
    – Tim
    Oct 26, 2017 at 14:08

3 Answers 3

4
$\begingroup$

Let ${\displaystyle p_{\text{H}}}$ be the probability that a certain coin lands heads up (H) when tossed. So, the probability of getting two heads in two tosses (HH) is ${\displaystyle p_{\text{H}}^{2}}$. If ${\displaystyle p_{\text{H}}=0.5}$, then the probability of seeing two heads is 0.25:

$${\displaystyle P({\text{HH}}\mid p_{\text{H}}=0.5)=0.25.}$$

With this, we can say that the likelihood of ${\displaystyle p_{\text{H}}=0.5}$, given the observation HH, is 0.25, that is

$${\displaystyle {\mathcal {L}}(p_{\text{H}}=0.5\mid {\text{HH}})=P({\text{HH}}\mid p_{\text{H}}=0.5)=0.25.}$$

This is not the same as saying that the probability that ${\displaystyle p_{\text{H}}=0.5}$, given the observation HH, is $0.25$. For that, we could apply Bayes' theorem, which implies that the posterior probability (density) is proportional to the likelihood times the prior probability. [Wikipedia Example 1 on Likelihood function]

This Wikipedia Example states exactly what it should:

the likelihood (function) $$\mathcal{L}(\theta|x)$$ as a function of $\theta$ indexed by the realised observation $x$, takes an image value at a particular value of the parameter (like $\theta={\displaystyle p_{\text{H}}=0.5}$) that is the value of the sampling distribution (pmf or pdf) at the observed sample for that value of the parameter $$p(x|\theta).$$ The final paragraph is a proper warning that a likelihood value or function is in general not a probability value or density/mass function on the parameter. To turn the likelihood function into a density function, the parameter space needs to be endowed with a probability structure, including a prior distribution/measure, which turns the sampling probability density into a conditional probability density.

The last sentence could always be turned into something clearer, like

For producing a probability statement on a value of the parameter, one needs to consider this parameter as a random variable, which requires a probability measure on the parameter space, called a prior distribution. With this preliminary, one applies Bayes' theorem, defining the posterior probability (density) on the parameter as proportional to the likelihood times the prior probability.

$\endgroup$
4
  • $\begingroup$ I believe your last paragraph addresses my concern: the wikipedia article does not mention that the likelihood function would first need to be "endowed with a probability structure", as you put it, because $p(x|\theta)$ is not necessarily a probability distribution. Correct? $\endgroup$
    – user393454
    Oct 26, 2017 at 14:29
  • $\begingroup$ Is there any reason the article should be left as is, not addressing the fact that one cannot simply apply Bayes' theorem? (Since $p(x|\theta)$ may not be a probability distribution.) $\endgroup$
    – user393454
    Oct 26, 2017 at 14:37
  • $\begingroup$ Yes, I see what you're saying, but the sentence is worded such that the existence of a prior probability is implicitly assumed. I think the paragraph would be clearer with an addendum like: "Note that the application of Bayes' theorem assumes that there exists a prior probability over $\theta$, and thus that $\theta$ is a random variable." Please correct me if you find anything wrong with this sentence. $\endgroup$
    – user393454
    Oct 26, 2017 at 14:49
  • $\begingroup$ Please see my suggestion for a rewording, as I find the "thus that θ is a random variable" open to confusion (and the customary objection that $\theta$ is a fixed number and hence cannot be a random variable, which is missing the point of the Bayesian resolution). $\endgroup$
    – Xi'an
    Oct 26, 2017 at 14:54
0
$\begingroup$

Yes you can. The distribution function you get when you do that is called the posterior distribution function. You need to specify a marginal distribution for the parameter though, which is called the prior distribution.

$\endgroup$
-1
$\begingroup$

I think wiki uses a confusing notation and term. It uses likelihood to distinguish it from posterior probability.

If we write down the full Bayes'r rule, it should be as: $P(p_H=0.5|HH) = \frac{P(HH|p_H=0.5)*P(p_H=0.5)}{p(HH)}$.

In Bayesian inference, $P(HH|p_H=0.5)$ is always referred to as likelihood where as $P(p_H=0.5)$ is referred to as prior and $P(p_H=0.5|HH)$ is referred as posterior probability.

Given the formula above, obviously wiki cannot claim $P(p_H=0.5|HH) = P(HH|p_H=0.5)$ so I guess that is why it uses likelihood here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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