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Maximum likelihood estimation is a method of statistical estimation: within a specified statistical model, choose the statistical parameter values that maximize the likelihood function of the observed data. For a parameter set $\Theta$ and observations $x$, its exact requirement isThe equality requires an attained, finite maximum; a supremum approached only at an excluded boundary is not enough. The resulting rule is a maximum-likelihood estimator. The likelihood function uses a probability mass function for discrete data and a probability density function for continuous data. See MIT 18.05, Reading 10b, sections 3 and 3.2.

The direction of the comparison is the interesting part: hold the observations fixed and vary the statistical parameter. Ten independent Bernoulli trials with eight successes and two failures give the maximum-likelihood estimate $\hat{p}=0.8$, as derived in log-likelihood. This chooses the member of the statistical model that assigns the greatest mass to that outcome. It does not assert an $80\%$ posterior probability that this parameter value is true. Bayesian statistics combines the same likelihood function with a prior distribution to obtain a posterior distribution; maximum a posteriori estimation maximizes that posterior mass or density instead.

The method turns many fitting tasks into one maximization problem. In the normal distribution model with known variance, it recovers the sample mean and ordinary least squares, while binomial regression and Poisson regression use different sampling assumptions to build their own likelihood functions. The log-likelihood turns independent contributions into a sum, exposing score equations and observed information. A peak's location gives an estimate; its shape records how much the observed data distinguish nearby parameter values.

Existence, uniqueness, bias, and statistical consistency are separate questions. Normal mean and variance maximum-likelihood estimators include a variance estimate with bias, likelihood supremum at an excluded boundary gives nonexistence, and the Neyman-Scott incidental parameter problem gives inconsistency. The usual asymptotic normality of a maximum likelihood estimator requires assumptions: a consistent estimator at an interior true parameter in a fixed finite-dimensional model with identifiability, sufficient smoothness and control of log-likelihood derivatives, and finite nonsingular Fisher information. A common parameter-independent support of a probability distribution is part of a standard sufficient framework; see Rigollet, MIT 18.650, Maximum Likelihood Estimation, slides 22-23. This is a description of the regular setting, not a complete theorem's hypotheses. The exact endpoint limit for a shifted exponential distribution shows why changing support can produce a different rate and a nonnormal limit. These properties do not follow just from maximizing a function.

In his 1922 paper in the Royal Society's Philosophical Transactions A, R. A. Fisher explicitly distinguished likelihood from probability and described choosing the parameter values with greatest likelihood. Section 6 also discusses comparing likelihoods after giving the maximum value one, the convention used in the figures here. See Fisher, On the Mathematical Foundations of Theoretical Statistics, section 6, pp. 326-328, DOI 10.1098/rsta.1922.0009.

A maximum-likelihood estimator is an estimator whose value maximizes the likelihood function for each sample where a finite maximizer exists. Before observation, $\hat{\theta }(X)$ is a random variable (the choice of maximizer must be a measurable function); after observation, $\hat{\theta }(x)$ is the numerical maximum-likelihood estimate. A formula must specify the allowed parameter set, including whether its endpoints belong to it.

For ten independent Bernoulli trials, write $K={\sum }_{i=1}^{10}{X}_i$. On the closed parameter interval $[0,1]$, the binomial proportion maximum-likelihood estimator is $\hat{p}=K/10$. Eight successes give the estimate $0.8$. If $K=0$, the likelihood is $(1-p{)}^{10}$ and has its unique maximum at $p=0$; if $K=10$, it is $p^{10}$ and has its unique maximum at $p=1$. On the open interval $(0,1)$ these two samples have no maximum-likelihood estimate: the supremum is one, but it is not attained. This is the same issue as likelihood supremum at an excluded boundary.

A complete continuous example uses independent and identically distributed random variables $X_i\sim N(\mu ,1)$ with unknown mean $\mu \in \mathbb{R}$ and known standard deviation one. Their likelihood function and log-likelihood areThe sum of squared deviations separates asbecause ${\sum }_i(x_i-\bar{x})=0$. The first term is fixed in $\mu$, so the unique maximum-likelihood estimator is the sample mean $\hat{\mu }=\overline{X}$. This also proves the equivalence with ordinary least squares for this model. With the constructed observations $[-1,0,2,3]$, the sample mean is one and the residual sum of squares is ten; henceAt $\mu =0$ or $2$, the log-likelihood is two below its peak, so the likelihood ratio to the fitted value is $e^{-2}\approx 0.135335$. The absolute likelihood at the peak is $(2\pi {)}^{-2}{e}^{-5}\approx 0.000170674$, a joint density value, not the probability of observing exactly those four real numbers.

The following figure compares that sample with a constructed sixteen-observation sample consisting of four repetitions of the displayed four values, modeled as sixteen independent draws. Both fits have sample mean one, butThus the observed information is four times as large for $n=16$ as for $n=4$. The distance from the peak at relative height $e^{-2}$ decreases from one to $1/2$: the width scales as $1/\sqrt{n}$. This is a comparison of two stipulated samples; merely copying recorded observations would not create new independent evidence.

Even uniqueness needs checking. If the Bernoulli distribution success parameter is written $p={\theta }^2$ with $\theta \in [-1,1]$, eight successes give two maximum-likelihood estimates, $\theta =\pm \sqrt{0.8}$, because both give the same probability distribution. The problem is a failure of identifiability. The special cases below show other ways an MLE can have bias or fail the usual asymptotic normality.

In a regular model let ${\ell }_N(\theta )$ be the log-likelihood, $U_N(\theta )=\nabla {\ell }_N(\theta )$ its score, $I_N(\theta )$ its total Fisher information, and $\hat{\theta }$ the unrestricted maximum-likelihood estimator. For a point null $\theta ={\theta }_0$ in a p-dimensional parameter space, the standard statistics areThese define the Wald test, score test and likelihood-ratio test. With an interior true parameter, increasing information, suitable differentiability and asymptotic normality, all have asymptotic chi-squared distribution with p degrees of freedom under the null. Observed information can replace expected information when it is asymptotically equivalent.

For q smooth independent restrictions $g(\theta )=0$, replace the Wald statistic by $g(\hat{\theta }{)}^T[G{I}_N(\hat{\theta }{)}^{-1}{G}^T{]}^{-1}g(\hat{\theta })$, where G is the derivative of g at the estimate; compare with chi-squared q. The likelihood-ratio statistic uses the difference between unrestricted and null-restricted maxima. The score statistic is computed at the restricted maximum and uses efficient information for the tested coordinates: with nuisance coordinates lambda this is $I_{\psi \psi }-I_{\psi \lambda }{I}_{\lambda \lambda }^{-1}{I}_{\lambda \psi }$. Inverting the acceptance inequalities for point nulls produces approximate coverage-$1-\alpha$ confidence regions; for example the Wald region consists of theta satisfying $(\hat{\theta }-\theta {)}^T{I}_N(\hat{\theta })(\hat{\theta }-\theta )\le {\chi }_{p,1-\alpha }^2$. Analogous score and likelihood-ratio inequalities give their confidence regions.

For the Poisson family, put $Y_{-1}=1$ to include the initial observation, and defineMultiplying the initial and conditional Poisson probabilities gives, on the supported observation sequences,The parameter-free term is ${\sum }_{i=0}^n[Y_i\log Y_{i-1}-\log (Y_i!)]$, interpreting $0\log 0$ as zero. A positive count after a zero preceding count is outside the support.

For $S_n>0$, strict concavity gives unconstrained maximum $\tilde{\theta }=S_n/T_n$, and the upper endpoint restriction givesThere is an endpoint qualification in the printed question. With probability $e^{-\theta }$, $Y_0=0$ and all later counts are zero. Then $S_n=0$, $T_n=1$ and the likelihood is $e^{-\theta }$, which has no maximum on the stated space $0<\theta \le 1$. The formula produces zero only as an extended estimate on the closure $[0,1]$. This is a likelihood supremum at an excluded boundary, and zero cannot be called an actual maximum in the original space.

The observed negative second derivative is $S_n/{\theta }^2$. Conditional expectation gives $\mathbb{E}{Y}_0=\theta$ and $\mathbb{E}{Y}_i=\theta \mathbb{E}{Y}_{i-1}={\theta }^{i+1}$. Therefore for $0<\theta <1$,This is the bounded information in a subcritical Poisson branching process. Observing more generations of one family does not provide unbounded information. Indeed $\mathbb{P}(Y_n>0)\le \mathbb{E}{Y}_n={\theta }^{n+1}\to 0$, and zero is absorbing, so extinction occurs at a finite generation almost surely. The total observed progeny $S_{\infty }$ is finite; eventually $S_n=S_{\infty }$, $T_n=1+S_{\infty }$, and the extended estimate equals $S_{\infty }/(1+S_{\infty })$, a discrete nondegenerate random variable rather than a consistent increasingly precise estimate.

Here is a direct proof of the failure of a chi-squared Wald limit which does not rely solely on the information bound. Under any interior null ${\theta }_0$, the event $Y_0=1$, $Y_1=0$ has probability ${\theta }_0{e}^{-2{\theta }_0}>0$ and forces every later count to be zero. On it the genuine constrained estimate is $1/2$ for every $n\ge 1$, so the usual expected-information Wald statistic isThus it has a persistent atom tending to $4(1/2-{\theta }_0{)}^2$. The fact that atoms obstruct a continuous distributional limit rules out a chi-squared limit, including when ${\theta }_0$ equals one half and that atom is at zero. Any definition adopted for the separate all-zero sample event leaves this positive-count counterexample intact. Hence the Wald statistic does not have the usual asymptotic chi-squared null law for this increasing observation horizon.