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For a parametric statistical model with probability mass function $q_{\theta }$ and an independent and identically distributed model, minimizing empirical cross-entropy is maximum likelihood estimation. The logarithm is strictly increasing, so maximizing ${\prod }_t{q}_{\theta }(X_t)$ is equivalent to maximizing ${\sum }_t{\log }_2{q}_{\theta }(X_t)$. Negating and dividing by $n$ gives the displayed minimization. Restricting the model family can prevent the optimum from matching the empirical distribution, and a finite optimizing parameter need not exist. This algebra does not by itself establish accuracy on new data.^[ref]

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 pairwise comparison model specifies the probability that one alternative beats another. Outcomes can be used to infer latent abilities through maximum likelihood estimation or another statistical method.

Put $S_1={\sum }_{i=1}^n{X}_i$, $S_2={\sum }_{i=1}^n{X}_i^2$. The joint normal distribution density can be written asThe Fisher-Neyman factorization theorem says that, for a family dominated by a common measure, a statistic $T$ is a sufficient statistic when its densities have the form $g_{\mu }(T(\mathbf{x}))h(\mathbf{x})$ with $h$ independent of the parameter. The displayed factorization therefore shows that $(S_1,S_2)$ is a two-dimensional sufficient statistic.

For maximum likelihood estimation, discard parameter-independent terms from the log-likelihood:If $S_2>0$, the quadratic equation has exactly one positive root. The numerator is positive before that root and negative after it, while the likelihood tends to zero at both ends of $\mu >0$. Hence the normal likelihood with variance equal to squared mean gives the global maximizerThis holds almost surely. The exceptional all-zero sample has $S_2=0$ and likelihood proportional to ${\mu }^{-n}$, so its supremum is approached as $\mu \downarrow 0$ and there is no positive maximizer.

The printed formula expands to the main effects of school type, acceptance and college, together with school-type-by-college and acceptance-by-college interactions. It contains neither a school-type-by-acceptance interaction nor the corresponding three-way interaction. Its Poisson regression model is thereforeFor every college the mean table has odds ratioThus it tests conditional independence of school type and acceptance given college, while permitting both margins to vary freely across colleges. It does not merely assert a common, possibly nonzero association.

The maximum likelihood estimation score equations match every school-type-by-college and acceptance-by-college margin. Solving them gives the stratified two-by-two conditional independence model's fitted countsIndeed these counts factor by row and column and have exactly the observed margins. For nondegenerate tables there are three free mean parameters per college: one total and two independent margin proportions. There are $100$ cells and $75$ parameters, hence 25 residual degrees of freedom. The residual deviance iswith zero-count contributions interpreted as zero. The Poisson linear terms cancel because the fitted totals equal the observed totals. Under the null and suitable large-count regularity, compare $G^2$ with ${\chi }_{25}^2$. Sparse cells, structural zeros or degenerate margins require a different calibration, such as an appropriate conditional test or simulated reference distribution.

A large residual deviance is evidence of school-type/acceptance association in at least some colleges. A small one indicates compatibility with conditional independence, not proof of identical admission processes or absence of selection on qualifications. Residuals and individual college contrasts locate departures. Adding the school-type-by-acceptance term gives a common odds ratio, one extra parameter and 24 residual degrees of freedom. Adding the full three-way term permits each college's association and saturates the tables. The first comparison tests a common association; the second checks whether a common association is adequate. No frequencies are supplied here to evaluate these tests numerically.

For the three-coordinate state realization of an ARMA(1,1) process, the observation row and transition matrix areThe first row of $G{S}_{t-1}$ is $x_{t-1}$ by the previous-time recursion; the second row is zero until the new white noise value is added; and the third row copies ${\epsilon }_{t-1}$. Hence the state equation has noise covariance

We use the causal state model, in which the new driving noise is independent of the previous state. The Gaussian filtering with exact observations form of the Kalman filter alternates prediction through the state equation with conditioning on the next observation. Given the posterior expected value and covariance at time $t-1$, independence of the new white noise value gives predictive expected value $a_t=G{\hat{S}}_{t-1}$ and predictive covarianceProject through the observation row to getThe residual $v_t=x_t-{\hat{x}}_t$ is the innovation. Conditioning the joint Gaussian state-observation vector adjusts the expected value according to this residual and reduces the covariance. In this exact-observation model, the updates can be writtenThus $F{\hat{S}}_t=x_t$ and $F{P}_t{F}^{\top }=0$: the observed linear combination of the state is now known exactly, though other state combinations remain uncertain.

The PDF's stated prediction variance omits $FQ{F}^{\top }={\sigma }^2$. This omission cannot be used in a valid likelihood: for the AR(1) specialization below it would give zero prediction variance after the first observation. The required likelihood calculation uses the corrected innovation variance just derived.

For stationary initialization of an ARMA(1,1) state with $|\varphi |<1$, take ${\hat{S}}_0=0$ and the unconditional stationary covariance. The causal coefficients are $1$ at lag zero and $(\varphi +\theta ){\varphi }^{j-1}$ at lags $j\ge 1$, givingSince $S_0=(x_{-1},{\epsilon }_0,{\epsilon }_{-1}{)}^{\top }$, the cross-covariance with ${\epsilon }_0$ is zero and that with ${\epsilon }_{-1}$ is ${\sigma }^2$. ThereforeDirect substitution verifies $P_0=G{P}_0{G}^{\top }+Q$. If prior state information is available, instead use its conditional expected value and covariance. A diffuse initialization is another option when the initial state is genuinely unknown, but it changes the finite-sample likelihood and should not silently replace the stationary initial law.

By the Gaussian prediction step,The chain rule for probabilities for the joint probability density function therefore givesTaking minus twice its logarithm shows that maximum likelihood estimation is equivalent to minimizingwith each $V_t$ and ${\hat{x}}_t$ evaluated at the candidate parameters and the specified initialization. The corrected $+{\sigma }^2$ is essential in every $V_t$.

For the stationary causal AR(1) case, $\theta =0$ and $|\varphi |<1$. The first observation is $N(0,{\sigma }^2/(1-{\varphi }^2))$, whereas each later conditional observation is $\varphi x_{t-1}+{\epsilon }_t$. HenceThis also gives a direct counterexample to the uncorrected PDF variance: with $\theta =0$, $FG=\varphi F$ and $F{P}_{t-1}{F}^{\top }=0$ after observing $x_{t-1}$, so its printed expression would give $V_t=0$ for $t\ge 2$.

To compare the conditional and stationary AR1 likelihood estimators, putThen its negative twice log-likelihood isFor fixed $\varphi$, $\widehat{{\sigma }^2}=Q_T(\varphi )/T$. Writingthe exact interior score equation isThe stationary initial contribution is order one for a fixed interior parameter, while $C_T$ is order $T$. Discarding this lower-order contribution givesConditioning on $x_1$ makes this ratio the exact unconstrained conditional maximum likelihood estimate. Under the stationary initial law it is only the leading approximation: at an interior optimum,This approximation presumes a nondegenerate sample denominator and a true coefficient fixed away from the unit-root boundary; a finite-sample constrained optimum must respect $|\varphi |<1$.

For independent observations from a normal distribution with unknown mean and positive variance, the maximum likelihood estimation of the mean givesIndeed ${\sum }_i(X_i-\mu {)}^2=S_{XX}+n(\bar{X}-\mu {)}^2$, which is minimized at $\mu =\bar{X}$. The likelihood isFor fixed $\mu$, differentiation in ${\sigma }^2$ gives the maximizer $[S_{XX}+n(\bar{X}-\mu {)}^2]/n$. Thus the unrestricted fit has ${\hat{\sigma }}^2=S_{XX}/n$, whereas under the null hypothesis $\mu ={\mu }_0$ the fitted variance is $[S_{XX}+n(\bar{X}-{\mu }_0{)}^2]/n$. Both maximize rather than minimize the likelihood: it tends to zero at either variance endpoint when the residual sum of squares is positive.

The restricted-to-unrestricted likelihood ratio is thereforeThe generalized likelihood-ratio test rejects for small $\Lambda$, equivalently large $|T|$. Under the null, $Z=\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal, $U=S_{XX}/{\sigma }^2$ has a chi-squared distribution with $n-1$ degrees of freedom, and they are independent. To see the independence, apply an orthogonal matrix to the standardized normal sample, taking its first row to be $(1,\dots ,1)/\sqrt{n}$: the transformed coordinates remain independent standard normal variables, and $U$ is the sum of squares of the other $n-1$ coordinates. Hence $T=Z/\sqrt{U/(n-1)}$ has Student's t-distribution with $n-1$ degrees of freedom.

For this two-sided test at $5\%$, the cutoff is the $97.5\%$ percentile of $t_9$, namely $2.26$. The data giveSince $|T|<2.26$, do not reject the manufacturer's mean-hardness claim at the $5\%$ level. This is a failure to reject, not proof that the mean equals $75$.

Work with a nonsingular multivariate normal distribution, so $V$ is positive-definite. The claimed finite maximum-likelihood estimates also require the centered sample covariance matrix to be nonsingular; the exceptional case is discussed below.

Independence gives, up to a constant, the log-likelihoodLet $\bar{Y}=n^{-1}{\sum }_i{Y}_i$ and $S=n^{-1}{\sum }_i(Y_i-\bar{Y})(Y_i-\bar{Y}{)}^T$. The centered sum is zero, and expansion therefore givesFor every fixed positive-definite matrix $V$, the extra quadratic term in the log-likelihood is $-n(\bar{Y}-\mu {)}^T{V}^{-1}(\bar{Y}-\mu )/2$, with unique maximum at $\mu =\bar{Y}$.

Assume $S$ is positive-definite. Maximizing over $V$ now means minimizing $\log \det V+\mathrm{tr}(V^{-1}S)$. Put $A=S^{1/2}{V}^{-1}{S}^{1/2}$, another positive-definite matrix. ThenIf $a_1,\dots ,a_p>0$ are its eigenvalues, the variable part is ${\sum }_j(a_j-\log a_j)$. The scalar inequality $a-\log a\ge 1$, with equality only at $a=1$, shows that this is minimized uniquely at $A=I$, or $V=S$. ThusThe divisor is $n$ because this is maximum likelihood estimation, not the unbiased covariance estimate with divisor $n-1$.

The nonsingularity condition matters. The centered scatter has rank at most $n-1$. If $S$ is singular, choose $\mu =\bar{Y}$, retain positive covariance eigenvalues on its range and let a covariance eigenvalue on its null space tend to zero. The quadratic likelihood term remains finite, while $-\log \det V\to +\infty$. This is singular covariance and nonexistence of a Gaussian maximum likelihood estimate: there is no finite maximizer over nonsingular $V$. Under a nonsingular Gaussian population, $S$ is positive definite almost surely when $n>p$; for $n\le p$ it is necessarily singular. The asserted estimate and the inverse in the next part have their usual meaning under $n>p$.

For independent individuals under fixed administrative censoring, an observed failure contributes the probability density function $\theta e^{-\theta x_i}$, while a censored observation contributes the survivor function $e^{-\theta x_i}$. Hence the survival likelihood and log-likelihood areFor $d>0$, the score function ${\ell }^{\prime }(\theta )=d/\theta -E$ vanishes at $d/E$, and ${\ell }^{\prime \prime }(\theta )=-d/{\theta }^2<0$. This is the unique maximum over $\theta >0$:Thus maximum likelihood estimation agrees with the observed-exposure estimating equation. If there are no failures, the likelihood decreases with $\theta$: its supremum is approached as $\theta \downarrow 0$, or attained at zero if that boundary is admitted.

Use the divisor-$n$ sample covariance matrix, since this is the normalization that enters the multivariate normal likelihood function:For a covariance matrix that is a positive-definite matrix $V$, the product of the multivariate normal densities givesTo separate the sample mean from the sample covariance matrix, expand $x_j-\mu =(x_j-\bar{x})+(\bar{x}-\mu )$. The cross terms sum to zero because ${\sum }_j(x_j-\bar{x})=0$. Also $u^T{V}^{-1}u=\mathrm{tr}(V^{-1}u{u}^T)$ by the trace identity. ThereforeThe printed expression suppresses the parameter-independent constant $np\log (2\pi )$; dropping this constant has no effect on maximum likelihood estimation or the likelihood-ratio test.

Assume $S_{xx}={\sum }_i{x}_i^2>0$, which is necessary to identify the slope in linear regression through the origin. The joint normal likelihood has logarithm, apart from a constant,Complete the square in $\beta$. With $\hat{\beta }={\sum }_i{x}_i{Y}_i/S_{xx}$ and $\mathrm{RSS}={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2$,The maximum likelihood estimation therefore first minimizes at $\hat{\beta }$; differentiating in ${\sigma }^2$ then maximizes atThis is a genuine finite maximum when $\mathrm{RSS}>0$. If all $x_i=0$, the slope is unidentifiable. If $\mathrm{RSS}=0$, the likelihood is unbounded as ${\sigma }^2\downarrow 0$, so no positive-variance maximum exists. For nonzero design and $n\ge 2$, zero residual has probability zero under the stated positive-variance model.

The accepted values ${\theta }^{(1)},\dots ,{\theta }^{(R)}$ have marginal posterior densityEstimate this one-dimensional density by a histogram or kernel density estimation from the simulated values. Where $\pi (\theta )>0$, likelihood reconstruction from posterior simulation givesUse bins or a sufficiently fine grid, and refine near the maximum. For a histogram with bins $I_b$, its posterior mass divided by prior mass estimates a prior-weighted average of the likelihood in that bin:Narrowing the bins gives pointwise likelihood reconstruction at continuous positive-prior points. A flat proper prior over the search interval makes the highest posterior-density bin approximately the highest-likelihood bin. With a nonconstant prior, maximizing the posterior alone gives a maximum a posteriori estimate, generally different from maximum likelihood estimation. Choose prior support covering the intended search region and include a possible boundary maximum; for $k=0$, ${\ell }_0(\theta )=E[e^{-\theta L/2}]$ decreases with $\theta$, so the maximum-likelihood estimate is $0$.

The sampled genealogies provide a second likelihood reconstruction when the one-dimensional prior integral is computable. PutTheir posterior density is $q(t)r_k(t)/m_k$. Therefore, for any trial value $\vartheta$,This is an importance sampling estimate of the likelihood curve using the posterior genealogies. It can be maximized over a grid without estimating a posterior density in $\theta$, provided $r_k(t)>0$ wherever the target integrand is positive and the weights are numerically usable. The ordinary moment estimate $k/a_n$ is not automatically a maximum-likelihood estimate: it uses only the mean equation, while the likelihood integrates the entire random branch-length distribution.

Let $K={\sum }_i{X}_i$ and $\bar{X}=K/n$. The likelihood is ${\theta }^K(1-\theta {)}^{n-K}$, so for $0<K<n$ its unique maximizer is $\hat{\theta }=K/n$. By invariance of maximum likelihood estimation,For $K=0$ or $n$, the strict parameter interval $(0,1)$ has no attained likelihood maximum; the usual extended estimate takes the limiting endpoint and gives $\hat{\varphi }=0$. Thus the displayed estimator is the conventional extended MLE on those samples, an important boundary qualification. This is the first estimator in estimating Bernoulli variance.

Assume $n\ge 2$, so the residual sum of squares is positive almost surely. Write $\overline{Y}=n^{-1}{\sum }_i{Y}_i$ and $S={\sum }_i(Y_i-\overline{Y}{)}^2$. The log-likelihood isThe decomposition ${\sum }_i(Y_i-\mu {)}^2=S+n(\overline{Y}-\mu {)}^2$ shows that, at every fixed positive variance, its constrained maximum-likelihood estimate of the mean is $\overline{Y}$. Substituting this mean and differentiating with respect to the variance givesThe denominator $n$ is required by maximum likelihood estimation; $S/(n-1)$ is the unbiased variance estimate, which is a different estimator.

Under the null variance ${\sigma }_0^2$, maximize over the same mean $\overline{Y}$. Subtracting the null maximum log-likelihood from the unrestricted maximum gives the likelihood-ratio test statisticBy Cochran's theorem, the residual sum of squares in a normal sample satisfies $U=S/{\sigma }_0^2\sim {\chi }_{n-1}^2$ under the null. Set $V=(U-n)/n$, so $U/n=1+V>0$. ThenThe normal mean is a nuisance parameter, but this null law does not depend on it.

A Bartlett correction removes the first-order discrepancy between the null mean of a regular likelihood-ratio test statistic and the degrees of freedom of its limiting chi-squared distribution. In the convention used here, if the limiting degrees of freedom are $d$ andthen $b$ is the Bartlett correction coefficient, also called the Bartlett correction factor, and the Bartlett-corrected likelihood-ratio statistic is $W_B=W/(1+b/n)$. Its null mean is $d+O(n^{-2})$. The finite-$n$ divisor is $1+b/n$, rather than $b$ itself. There is one restriction on the variance here, so $d=1$.

Expand the logarithm around $V=0$ and use the permitted termwise asymptotic integration:Using the supplied raw moments of the chi-squared distribution and expanding $(U-n{)}^r$ givesFor instance, $\mathbb{E}U=n-1$ and $\mathbb{E}{U}^2=(n-1)(n+1)$ give the first formula after subtracting $2n\mathbb{E}U-n^2$. The same binomial expansion, using the raw third and fourth moments, gives the other two formulas.

We must justify why higher moments cannot change the $n^{-1}$ coefficient. Represent $U={\sum }_{i=1}^{n-1}{Z}_i^2$ for independent standard normal variables, and put $T_i=Z_i^2-1$. Each $T_i$ is centered and has all absolute moments finite. With $H={\sum }_{i=1}^{n-1}{T}_i$, one has $V=(H-1)/n$. The supplied centered-sum bounds say, for each fixed integer $j\ge 2$,Also $\mathbb{E}H=0$ and $\mathbb{E}{H}^0=1$. Therefore, for every fixed $r\ge 2$, expanding $(H-1{)}^r$ givesThese are bounds on the absolute values of signed moments; the odd-moment estimate must not be misread as a bound on $\mathbb{E}|H{|}^j$. In particular, each fixed Taylor term of degree $r\ge 5$ has $n\mathbb{E}{V}^r=O(n^{-2})$ or smaller. Combined with the question's permission to integrate the asymptotic expansion term by term, this controls the higher-order contribution at the required order. It does not require a globally convergent Taylor series for every realized $V$.

Substituting the first three moment formulas now givesThus the Bartlett correction for a normal variance with unknown mean isIts null mean is $1+O(n^{-2})$, and the usual reference law is ${\chi }_1^2$. Keeping the estimated mean's loss of one residual degree of freedom, and using the maximum-likelihood variance divisor $n$, are both essential to the coefficient $11/6$.

Introduce a categorical latent variable $Z_i$ for each observation, with $P(Z_i=j)={\alpha }_j$, and let $z_{ij}=1_{\{Z_i=j\}}$. Conditional on $Z_i=j$, $X_i$ has a normal distribution of mean ${\mu }_j$ and the shared variance ${\sigma }^2$. The complete-data log-likelihood function isInitialize finite means, positive weights summing to one, and a positive common variance. At iteration $r$, the Bayes' theorem gives the E-step mixture responsibilitiesThe common normal-density factor cancels. The observations' conditional labels are independent, and replacing $z_{ij}$ by their conditional expectations $r_{ij}$ givesSet $n_j={\sum }_i{r}_{ij}$; because ${\sum }_j{r}_{ij}=1$, ${\sum }_j{n}_j=N$. The constrained weight maximization uses a Lagrange multiplier $\eta$: differentiating ${\sum }_j{n}_j\log {\alpha }_j-\eta ({\sum }_j{\alpha }_j-1)$ gives ${\alpha }_j=n_j/\eta$, and summing fixes $\eta =N$. Differentiating the residual quadratic in each mean gives ${\sum }_i{r}_{ij}(x_i-{\mu }_j)=0$. Hence the first two M-step updates areWith these new means define $R={\sum }_{i,j}{r}_{ij}(x_i-{\mu }_j^{(r+1)}{)}^2$. The remaining objective in $u={\sigma }^2$ is $-N\log u/2-R/(2u)$, whose derivative is $-N/(2u)+R/(2u^2)$. For $R>0$ it increases up to $u=R/N$ and decreases afterwards. Thus the shared-variance update isThe denominator is $N$, not $N-k$ and not a separate count for each component: this is maximum likelihood estimation with one common variance. These are the EM for Gaussian mixtures with a common variance updates. Recompute responsibilities with the new parameters and repeat until the observed-data likelihood changes sufficiently little.

The monotonicity claim also follows directly. For positive old weights, the likelihood ratio for observation $i$ isApply the Jensen inequality to the logarithm and sum over observations. This yieldsfor the maximizing M step. This proves likelihood increase, not convergence to the global maximum. Use several initializations to reduce sensitivity to local maxima. Compute responsibilities with a log-sum-exp function normalization to avoid numerical underflow.

If $n_j=0$, that component's updated weight is zero and its mean is unidentified; retain its old mean or remove the empty component. Strictly positive interior initialization avoids this case in exact arithmetic while the variance stays positive. If $R=0$, there is no positive finite variance maximizing this M step: the objective tends to infinity as ${\sigma }^2\downarrow 0$. Such boundary degeneracy needs a variance constraint or regularization, as discussed in the next solution. For $k=1$ the formulas reduce to the ordinary sample mean and the maximum-likelihood variance $N^{-1}{\sum }_i(x_i-\overline{x}{)}^2$.

Set $x_i=\sin (2{\theta }_i)$. The maximum likelihood estimation formulas for simple linear regression giveUsing the supplied rounded summaries gives $\hat{\beta }\approx 21082.7$ metres and $\hat{\alpha }\approx 312.0$ metres. Recomputing directly from the printed table gives approximately $21082.4$ and $312.0$ metres, respectively, so the small summary rounding differences do not affect the conclusion.

I would not accept the model merely because the Student t-test fails to reject a zero regression intercept. That hypothesis test tests one parameter restriction within the assumed model, not the adequacy of its error assumptions or its functional form. In particular, complementary elevations $\theta$ and $90^{\circ }-\theta$ have exactly the same $x_i$ and hence the same fitted mean, but their observed low-angle minus high-angle ranges arefor the four corresponding pairs. All four are positive, suggesting an elevation-dependent effect omitted by the mean model, such as air resistance, rather than random scatter determined only by $\sin (2\theta )$. This is a regression diagnostics concern, not a claim that these finite data logically rule out the normal-error model. The correct conclusion is nonrejection of the intercept restriction does not validate the regression model; further checks of the residual pattern and error assumptions are needed.

Assume the explanatory variables are not all equal, so $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2>0$. In the Gaussian linear model, the log-likelihood isFor fixed ${\sigma }^2$, maximum likelihood estimation therefore minimizes the residual sum of squares. Differentiating in $\alpha ,\beta$ gives the least-squares normal equationsThe first gives $\hat{\alpha }=\bar{Y}-\hat{\beta }\bar{x}$; substituting it into the second gives $\hat{\beta }=S_{xy}/S_{xx}$, with $S_{xy}={\sum }_i(x_i-\bar{x})(Y_i-\bar{Y})$. Let $\mathrm{SSE}={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$. Differentiating the profiled log-likelihood in ${\sigma }^2$ givesFor the variance estimator this is the interior maximum when $\mathrm{SSE}>0$, which occurs almost surely under the model when $n>2$.