The wikipedia page claims that likelihood and probability are distinct concepts.

In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the observed outcomes.

Can someone give a more down-to-earth description of what this means? In addition, some examples of how "probability" and "likelihood" disagree would be nice.

  • 7
    Great question. I would add "odds" and "chance" in there too :) – Neil McGuigan Sep 14 '10 at 5:28
  • 4
    I think you should take a look at this question stats.stackexchange.com/questions/665/… because Likelihood is for statistic purpose and probability for probability. – robin girard Sep 14 '10 at 6:04
  • 2
    Wow, these are some really good answers. So a big thanks for that! Some point soon, I'll pick one I particularly like as the "accepted" answer (although there are several that I think are equally deserved). – Douglas S. Stones Sep 15 '10 at 1:13
  • Also note that the "likelihood ratio" is actually a "probability ratio" since is a function of the observations. – JohnRos Nov 2 '11 at 10:29
up vote 259 down vote accepted

The answer depends on whether you are dealing with discrete or continuous random variables. So, I will split my answer accordingly. I will assume that you want some technical details and not necessarily an explanation in plain English. If my assumption is not correct please let me know and I will revise my answer.

Discrete Random Variables

Suppose that you have a stochastic process that takes discrete values (e.g., outcomes of tossing a coin 10 times, number of customers who arrive at a store in 10 minutes etc). In such cases, we can calculate the probability of observing a particular set of outcomes by making suitable assumptions about the underlying stochastic process (e.g., probability of coin landing heads is $p$ and that coin tosses are independent).

Denote the observed outcomes by $O$ and the set of parameters that describe the stochastic process as $\theta$. Thus, when we speak of probability we want to calculate $P(O|\theta)$. In other words, given specific values for $\theta$, $P(O|\theta)$ is the probability that we would observe the outcomes represented by $O$.

However, when we model a real life stochastic process, we often do not know $\theta$. We simply observe $O$ and the goal then is to arrive at an estimate for $\theta$ that would be a plausible choice given the observed outcomes $O$. We know that given a value of $\theta$ the probability of observing $O$ is $P(O|\theta)$. Thus, a 'natural' estimation process is to choose that value of $\theta$ that would maximize the probability that we would actually observe $O$. In other words, we find the parameter values $\theta$ that maximize the following function:

$L(\theta|O) = P(O|\theta)$

$L(\theta|O)$ is called the likelihood function. Notice that by definition the likelihood function is conditioned on the observed $O$ and that it is a function of the unknown parameters $\theta$.

Continuous Random Variables

In the continuous case the situation is similar with one important difference. We can no longer talk about the probability that we observed $O$ given $\theta$ because in the continuous case $P(O|\theta) = 0$. Without getting into technicalities, the basic idea is as follows:

Denote the probability density function (pdf) associated with the outcomes $O$ as: $f(O|\theta)$. Thus, in the continuous case we estimate $\theta$ given observed outcomes $O$ by maximizing the following function:

$L(\theta|O) = f(O|\theta)$

In this situation, we cannot technically assert that we are finding the parameter value that maximizes the probability that we observe $O$ as we maximize the PDF associated with the observed outcomes $O$.

  • 26
    The distinction between discrete and continuous variables disappears from the point of view of measure theory. – whuber Sep 14 '10 at 15:48
  • 17
    @whuber yes but an answer using measure theory is not that accessible to everyone. – user28 Sep 14 '10 at 20:09
  • 13
    @Srikant: Agreed. The comment was for the benefit of the OP, who is a mathematician (but perhaps not a statistician) to avoid being misled into thinking there is something fundamental about the distinction. – whuber Sep 14 '10 at 20:36
  • 5
    You can interpret a continuous density the same as the discrete case if $O$ is replaced by $dO$, in the sense that if we ask for $Pr(O\in(O',O'+dO') |\theta)$ (i.e. probability that the data $O$ is contained in an infinintesimal region about $O'$) and the answer is $f(O'|\theta)dO'$ (the $dO'$ makes this clear that we are calculating the area of an infinintesimaly thin "bin" of a histogram). – probabilityislogic Jan 28 '11 at 13:40
  • 5
    I am over 5 years late to the party, but I think that a very crucial follow-up to this answer would be stats.stackexchange.com/questions/31238/… which stresses upon the fact that likelihood function $L(\theta)$ is not a pdf with respect to $\theta$. $L(\theta$) is indeed a pdf of data given the parameter value, but since the since $L$ is a function of $\theta$ alone (with data held as a constant), it is irrelevant that $L(\theta)$ is a pdf of data given $\theta$. – Shobhit Jan 8 '16 at 16:04

This is the kind of question that just about everybody is going to answer and I would expect all the answers to be good. But you're a mathematician, Douglas, so let me offer a mathematical reply.

A statistical model has to connect two distinct conceptual entities: data, which are elements $x$ of some set (such as a vector space), and a possible quantitative model of the data behavior. Models are usually represented by points $\theta$ on a finite dimensional manifold, a manifold with boundary, or a function space (the latter is termed a "non-parametric" problem).

The data $x$ are connected to the possible models $\theta$ by means of a function $\Lambda(x, \theta)$. For any given $\theta$, $\Lambda(x, \theta)$ is intended to be the probability (or probability density) of $x$. For any given $x$, on the other hand, $\Lambda(x, \theta)$ can be viewed as a function of $\theta$ and is usually assumed to have certain nice properties, such as being continuously second differentiable. The intention to view $\Lambda$ in this way and to invoke these assumptions is announced by calling $\Lambda$ the "likelihood."

It's quite like the distinction between variables and parameters in a differential equation: sometimes we want to study the solution (i.e., we focus on the variables as the argument) and sometimes we want to study how the solution varies with the parameters. The main distinction is that in statistics we rarely need to study the simultaneous variation of both sets of arguments; there is no statistical object that naturally corresponds to changing both the data $x$ and the model parameters $\theta$. That's why you hear more about this dichotomy than you would in analogous mathematical settings.

  • 6
    +1, what a cool answer. Analogy with differential equations seems very apropriate. – mpiktas Mar 5 '12 at 20:15
  • 3
    As an economist, although this answer does not relate as closely as the previous to the concepts I've learnt, it was the most informative one in an intuitive sense. Many thanks. – Robson Jan 20 '16 at 14:25
  • Actually, this statement is not really true "there is no statistical object that naturally corresponds to changing both the data x and the model parameters θ.". There is, it's called "smoothing, filtering, and prediction", in linear models its the Kalman filter, in nonlinear models, they have the full nonlinear filters, en.wikipedia.org/wiki/Kushner_equation etc – crow Nov 17 '17 at 18:26
  • Yes, great answer! As lame as this sounds, by choosing $\Lambda\left(x, \theta\right)$ instead of the standard notation of $P\left(x, \theta\right)$, it made it easier for me to see that we're starting off with a joint probability that can be defined as either a likelihood or a conditional probability. Plus, the "certain nice properties" comment helped. Thanks! – Mike Williamson Aug 18 at 19:20
  • 1
    @whuber Yes, I know $\Lambda$ isn't the usual notation. That's exactly why it helped! I stopped thinking that it must have a particular meaning and instead just followed the logic. ;-p – Mike Williamson Aug 19 at 15:14

I'll try and minimise the mathematics in my explanation as there are some good mathematical explanations already.

As Robin Girand points out the difference between probability and likelihood is closely related to the difference between probability and statistics. In a sense probability and statistics concern themselves with problems that are opposite or inverse to one another.

Consider a coin toss. (My answer will be similar to Example 1 on Wikipedia.) If we know the coin is fair ($p=0.5$) a typical probability question is: What is the probability of getting two heads in a row. The answer is $P(HH) = P(H)\times P(H) = 0.5\times0.5 = 0.25$.

A typical statistical question is: Is the coin fair? To answer this we need to ask: To what extent does our sample support the our hypothesis that $P(H) = P(T) = 0.5$?

The first point to note is that the direction of the question has reversed. In probability we start with an assumed parameter ($P(head)$) and estimate the probability of a given sample (two heads in a row). In statistics we start with the observation (two heads in a row) and make INFERENCE about our parameter ($p = P(H) = 1- P(T) = 1 - q$).

Example 1 on Wikipedia shows us that the maximum likelihood estimate of $P(H)$ after 2 heads in a row is $p_{MLE} = 1$. But the data in no way rule out the the true parameter value $p(H) = 0.5$ (let's not concern ourselves with the details at the moment). Indeed only very small values of $p(H)$ and particularly $p(H)=0$ can be reasonably eliminated after $n = 2$ (two throws of the coin). After the third throw comes up tails we can now eliminate the possibility that $P(H) = 1.0$ (i.e. it is not a two-headed coin), but most values in between can be reasonably supported by the data. (An exact binomial 95% confidence interval for $p(H)$ is 0.094 to 0.992.

After 100 coin tosses and (say) 70 heads, we now have a reasonable basis for the suspicion that the coin is not in fact fair. An exact 95% CI on $p(H)$ is now 0.600 to 0.787 and the probability of observing a result as extreme as 70 or more heads (or tails) from 100 tosses given $p(H) = 0.5$ is 0.0000785.

Although I have not explicitly used likelihood calculations this example captures the concept of likelihood: Likelihood is a measure of the extent to which a sample provides support for particular values of a parameter in a parametric model.

  • 3
    Great answer! Especially the three last paragraphs are very useful. How would you extend this to describe the continuous case? – Demetris Sep 2 '14 at 11:53
  • 7
    For me, best answer. I don't mind math at all, but for me math is a tool ruled by what I want (I don't enjoy math for its own sake, but for what it helps me do). Only with this answer do I know the latter. – Mörre Apr 20 '15 at 13:28

I will give you the perspective from the view of Likelihood Theory which originated with Fisher -- and is the basis for the statistical definition in the cited Wikipedia article.

Suppose you have random variates $X$ which arise from a parameterized distribution $F(X; \theta)$, where $\theta$ is the parameter characterizing $F$. Then the probability of $X = x$ would be: $P(X = x) = F(x; \theta)$, with known $\theta$.

More often, you have data $X$ and $\theta$ is unknown. Given the assumed model $F$, the likelihood is defined as the probability of observed data as a function of $\theta$: $L(\theta) = P(\theta; X = x)$. Note that $X$ is known, but $\theta$ is unknown; in fact the motivation for defining the likelihood is to determine the parameter of the distribution.

Although it seems like we have simply re-written the probability function, a key consequence of this is that the likelihood function does not obey the laws of probability (for example, it's not bound to the [0, 1] interval). However, the likelihood function is proportional to the probability of the observed data.

This concept of likelihood actually leads to a different school of thought, "likelihoodists" (distinct from frequentist and bayesian) and you can google to search for all the various historical debates. The cornerstone is the Likelihood Principle which essentially says that we can perform inference directly from the likelihood function (neither Bayesians nor frequentists accept this since it is not probability based inference). These days a lot of what is taught as "frequentist" in schools is actually an amalgam of frequentist and likelihood thinking.

For deeper insight, a nice start and historical reference is Edwards' Likelihood. For a modern take, I'd recommend Richard Royall's wonderful monograph, Statistical Evidence: A Likelihood Paradigm.

  • 3
    Interesting answer, I actually thought that the "likelihood school" was basically the "frequentists who don't design samples school", while the "design school" was the rest of the frequentists. I actually find it hard myself to say which "school" I am, as I have a bit of knowledge from every school. The "Probability as extended logic" school is my favourite (duh), but I don't have enough practical experience in applying it to real problems to be dogmatic about it. – probabilityislogic Jan 28 '11 at 13:53
  • 4
    +1 for "the likelihood function does not obey the laws of probability (for example, it's not bound to the [0, 1] interval). However, the likelihood function is proportional to the probabiilty of the observed data." – Walrus the Cat Jun 13 '14 at 22:27
  • 9
    "the likelihood function does not obey the laws of probability" could use some further clarification, especialy since is was written as θ: L(θ)=P(θ;X=x), i.e. equated with a probability! – redcalx Apr 3 '15 at 19:53
  • Thanks for your answer. Could you please address the comment that @locster made? – Vivek Subramanian Jul 1 '15 at 11:48
  • 2
    To me as a not mathematician, this reads like religious mathematics, with different beliefs resulting in different values for chances of events to occur. Can you formulate it, so that it is easier to understand what the different beliefs are and why they all make sense, instead of one being simply incorrect and the other school / belief being correct? (assumption that there is one correct way of calculating chances for events to occur) – Zelphir Jul 26 '16 at 13:11

Given all the fine technical answers above, let me take it back to language: Probability quantifies anticipation (of outcome), likelihood quantifies trust (in model).

Suppose somebody challenges us to a 'profitable gambling game'. Then, probabilities will serve us to compute things like the expected profile of your gains and loses (mean, mode, median, variance, information ratio, value at risk, gamblers ruin, and so on). In contrast, likelihood will serve us to quantify whether we trust those probabilities in the first place; or whether we 'smell a rat'.


Incidentally -- since somebody above mentioned the religions of statistics -- I believe likelihood ratio to be an integral part of the Bayesian world as well as of the frequentist one: In the Bayesian world, Bayes formula just combines prior with likelihood to produce posterior.

  • This answer sums it up for me. I had to think through what it meant when I read that likelihood is not probability, but the following case occurred to me. What is the likelihood that a coin is fair, given that we see four heads in a row? We can't really say anything about probability here, but the word "trust" seems apt. Do we feel we can trust the coin? – dnuttle Jul 23 at 12:21

Suppose you have a coin with probability $p$ to land heads and $(1-p)$ to land tails. Let $x=1$ indicate heads and $x=0$ indicate tails. Define $f$ as follows

$$f(x,p)=p^x (1-p)^{1-x}$$

$f(x,2/3)$ is probability of x given $p=2/3$, $f(1,p)$ is likelihood of $p$ given $x=1$. Basically likelihood vs. probability tells you which parameter of density is considered to be the variable

http://yaroslavvb.com/upload/likelihood-vs-probability.png

  • Nice complement to the theoretical definitions used above! – Frank Meulenaar Sep 17 '11 at 10:47
  • I see that $C^n_kp^n(1-p)^{k-n}$ gives the probability of having $n$ heads in $k$ trials. Your $p^x(1-p)^{1-x}$ looks like $k$-th root of that: $x=n/k$. What does it mean? – Little Alien Sep 1 '16 at 13:29

If I have a fair coin (parameter value) then the probability that it will come up heads is 0.5. If I flip a coin 100 times and it comes up heads 52 times then it has a high likelihood of being fair (the numeric value of likelihood potentially taking a number of forms).

  • 1
    This and Gypsy's answer should be on top! Intuition and clarity above dry mathematical rigor, not to say something more derogatory. – Nemanja Radojković Jun 5 at 22:33

$P(x|\theta)$ can be seen from two points of view:

  • As a function of $x$, treating $\theta$ as known/observed. If $\theta$ is not a random variable, then $P(x|\theta)$ is called the (parameterized) probability of $x$ given the model parameters $\theta$, which is sometimes also written as $P(x;\theta)$ or $P_{\theta}(x)$. If $\theta$ is a random variable, as in Bayesian statistics, then $P(x|\theta)$ is a conditional probability, defined as ${P(x\cap\theta)}/{P(\theta)}$.
  • As a function of $\theta$, treating $x$ as observed. For example, when you try to find a certain assignment $\hat\theta$ for $\theta$ that maximizes $P(x|\theta)$, then $P(x|\hat\theta)$ is called the maximum likelihood of $\theta$ given the data $x$, sometimes written as $\mathcal L(\hat\theta|x)$. So, the term likelihood is just shorthand to refer to the probability $P(x|\theta)$ for some data $x$ that results from assigning different values to $\theta$ (e.g. as one traverses the search space of $\theta$ for a good solution). So, it is often used as an objective function, but also as a performance measure to compare two models as in Bayesian model comparison.

Often, this expression is still a function of both its arguments, so it is rather a matter of emphasis.

  • For the second case, I thought people usually write P(theta|x). – yuqian Jan 4 '16 at 20:40
  • Originally intuitively I already thought they're both words for the same with a difference in perspective or natural language formulation, so I feel like "What? I was right all along?!" But if this is the case, why is distinguishing them so important? English not being my mother tongue, I grew up with only one word for seemingly both of the terms (or have I simply never gotten a problem where I needed to distinguish the terms?) and never knew there was any difference. It's only now, that I know two English terms, that I begin to doubt my understanding of these things. – Zelphir Jul 26 '16 at 13:32
  • 3
    Your answer seems to be very comrphensive and is easy to understand. I wonder, why it got so few upvotes. – Julian Feb 16 '17 at 7:18
  • 4
    Note that P(x|$\theta$) is a conditional probability only if $\theta$ is a random variable, if $\theta$ is a parameter then it's simply the probability of x parameterized by $\theta$. – Mircea Mironenco May 9 '17 at 19:49
  • i think this is the best answer amongst all – Aaron Oct 11 '17 at 7:08

As far as I'm concerned, the most important distinction is that likelihood is not a probability (of $\theta$).

In an estimation problem, the X is given and the likelihood $P(X|\theta)$ describes a distribution of X rather than $\theta$. That is, $\int P(X|\theta) d\theta$ is meaningless, since likelihood is not a pdf of $\theta$, though it does characterize $\theta$ to some extent.

  • 1
    As the answer from @Lenar Hoyt points out, if theta is a random variable (which it can be), then likelihood is a probability. So the real answer seems to be that the likelihood can be a probability, but is sometimes not. – Mike Wise Dec 5 '17 at 17:47
  • @MikeWise, I think theta could always be viewed as a "random" variable, while chances are that it is just not so "random"... – Response777 Dec 6 '17 at 15:18

protected by whuber Jun 21 '16 at 13:31

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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