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Write $v={\sigma }_{\mathrm{int}}^2$, $r_s={\sigma }_{m,s}^2$ and $d_s=\mu (z_s;H_0,w,{\Omega }_M)$. The latent absolute magnitude and measurement error are independent normal variables, so their convolution is normal. After integrating out the individual intrinsic magnitudes, each apparent magnitude has distributionConsequently the likelihood function for the independent sample isThis is the marginal observational likelihood of a hierarchical Bayesian model, not the likelihood conditional on each unknown intrinsic magnitude. The individual errors remain heteroscedastic through their known $r_s$.

Whenever players 1 and 2 occur, the linear predictor contains their parameters only throughFor any constants $a,b$, replacingleaves every fitted probability and hence the likelihood function unchanged. The parameterization is therefore nonidentifiable: at most ${\gamma }_{12}$ can be estimated uniquely. Hence there is no unique maximum-likelihood estimate of the three separate parameters.

Let $S={\sum }_{i=1}^n{X}_i$ be the number of heads. The likelihood function for the Bernoulli distribution sample is proportional to ${\theta }^S(1-\theta {)}^{n-S}$. A uniform distribution prior on $[0,1]$ is the $\mathrm{Beta}(1,1)$ distribution, so Beta-binomial conjugacy gives the posterior distribution

LetSubtracting the measured distance modulus from the measured apparent magnitude givesFor the Hubble flow, defineThe Hubble law, with the Hubble constant parametrized by $\theta =5{\log }_{10}h$, gives $r_i\sim N(M_0-\theta ,H_i)$. After integrating over each intrinsic absolute magnitude and each unobserved true distance modulus, independence therefore gives the likelihood function

Let ${\pi }_0$ be the probability density function of the unknown and $\rho$ that of the noise, both relative to Lebesgue measure. By independence of random variables, the conditional distribution of the measurement given $u$ has density $\rho (m-Au)$. Thus its likelihood function and the joint density areThe Bayesian model evidence is the marginal data densityBayes' theorem gives the posterior density solving this Bayesian inverse problem:This formula applies when $0<Z(m)<\infty$. Since $\int Z(m)dm=1$ by Tonelli theorem, these conditions hold almost everywhere under the marginal data law. Values of a conditional distribution at exceptional data are not determined by the joint law. The posterior mean or a maximum a posteriori estimate can then give a point estimate, while the full posterior density describes the remaining uncertainty.

Write $y_{i+}={\sum }_j{y}_{ij}$ and ${\mu }_{i+}={\sum }_j{\mu }_{ij}$. Under the model in part (a),and $p_j$ is independent of the price category $i$. The Poisson-multinomial conditioning identity givesMore explicitly, the likelihood function for row $i$ factors asThe first factor is the likelihood of the row total and the second is the conditional multinomial likelihood. The free price effect ${\alpha }_i$ makes the first factor maximal at ${\hat{\mu }}_{i+}=y_{i+}$, while maximizing the remaining factors givesConsequently both fits have the same maximum-likelihood fitted valueHere every row total is $30$, the column totals are $31,28,31$, and $y_{++}=90$, so every fitted row isThis factorization is the Poisson trick.

For observations $x=(x_1,\dots ,x_n)$, the likelihood function isA maximum-likelihood estimator is any measurable choiceFor one observation in a regular scalar model, the score function isand the Fisher information iswhere the second identity requires the usual differentiation-under-the-integral regularity. For $n$ independent observations, Tensorization of Fisher information gives $I_n(\theta )=nI(\theta )$.

Let $M=X_{(n)}={\max }_i{X}_i$. The likelihood function isOn its feasible range $\theta \ge M/2$, this is decreasing in $\theta$. Therefore the maximum likelihood estimator is

Let $L(t)=f(x,t)$ be the likelihood function. Since the prior distribution is $N(0,I_p)$, the posterior density has the formwhere $Z$ is the normalizing constant.

Choose $0<\delta \le 1/2$ and use the Gaussian autoregressive proposal reversible with respect to a standard normal distributionThuswhose covariance matrix is the required $2\delta I_p$. Reversibility with respect to the standard normal distribution sayssoThe Metropolis–Hastings acceptance probability therefore reduces toThis is the Preconditioned Crank–Nicolson algorithm with proposal scale $\beta =\sqrt{2\delta }$.

A sufficient statistic $T$ is one for which the conditional distribution of the full sample $(X_1,\dots ,X_n)$ given $T$ does not depend on $p$. TakeFor a binary sample $x$ with ${\sum }_i{x}_i=t$, its likelihood function iswhich depends on the data only through $t$. By the Fisher-Neyman factorization theorem, $T$ is sufficient. Equivalently, conditionally on $T=t$, the sample is uniform over the $\left(\genfrac{}{}{0ex}{}{n}{t}\right)$ binary vectors containing $t$ ones, independently of $p$.

Here $X$ has the binomial distribution $\mathrm{Bin}(n,\theta )$, so, up to a factor independent of $\theta$, the likelihood function isFor $0<X<n$, differentiating the log-likelihood givesand the boundary cases give the same formula. Thus the binomial proportion maximum-likelihood estimator is

The noise has the product Laplace distribution densitywhich is normalized because ${\int }_{\mathbb{R}}{e}^{-2|t|}dt=1$. By translation invariance of Lebesgue measure, the conditional law of $A(u)+N$ has densityHence an associated likelihood function isThe map $(u,y)\mapsto y-A(u)$ is measurable because $A$ is measurable and vector subtraction is continuous. Composition with the continuous norm and exponential functions proves that $L$ is jointly measurable.

LetThus $d_t$ patients recover during month $t$, while $r_t-d_t$ survive that month without recovery. The likelihood function is

The independent $\mathrm{Beta}(T,1)$ prior densities are proportional to $q_t^{T-1}$. Beta-binomial conjugacy therefore leaves the coordinates posteriorly independent, with

LetIgnoring the immaterial choice of density values at the sample points, the likelihood function isBetween consecutive order statistics, $N(\theta )$ is constant while $(1+1/\theta {)}^{N(\theta )}$ decreases with $\theta$. A maximum can therefore be moved to the left endpoint of one of these intervals. With an equivalent version using $1_{\{x\le \theta \}}$, the maximum is attained at a sample order statistic. Thus the maximum-likelihood estimator coincides with one of $X_1,\dots ,X_n$.

Write the Cepheid measurement as $\hat{\mu }=\mu +{ϵ}_{\mu }$, where ${ϵ}_{\mu }\sim N(0,{\sigma }_{\mu }^2)$, and define the calibrated absolute magnitude dataThus $q=(q_1,\dots ,q_K{)}^T$ has a multivariate normal distribution with mean $M_{0}1$ and covariance matrix

The Hubble law and the definition of distance modulus giveConsequently, withthe Hubble-flow observations are independent random variables satisfying $x_i\sim N(M_0-\theta ,{\sigma }_{\mathrm{tot}}^2)$. Apart from a factor independent of the parameters, the joint likelihood function isThe expression extends by continuity to a singular limiting covariance such as ${\sigma }_I=0$.

The likelihood function and prior distribution give the posterior densityCompleting the square in the quadratic form givesThus Gaussian conjugacy for a normal linear model yields

Let $Y_i$ be the first team's goal count in match $i$, let $x_i^T$ be row $i$ of the design matrix, and let $\beta$ be the regression coefficient vector. The fitted Poisson regression assumes that the $Y_i$ are independent random variables withThe likelihood function maximized by glm is therefore

All three sets of statistics are independent. With $s_1={\hat{S}}_1$, $s_{2g}={\hat{S}}_2^g$, and $s_{3i}={\hat{S}}_3^i$, their likelihood function, up to a parameter-independent factor, is

For an observed random sample with all $x_i>0$, which happens almost surely, independence gives the likelihood functionIgnoring a term independent of $\theta$, the log-likelihood and its derivative areThis derivative is positive for $0<\theta <S/n$ and negative for $\theta >S/n$. Thus there is a unique global maximum, and the maximum-likelihood estimator is

Let $M={\max }_i{X}_i$. For an observed sample in the nonnegative orthant, the likelihood function under the uniform distribution isIt is zero for $\theta <M$, and decreases with $\theta$ on $[M,\infty )$. Thus the maximum likelihood estimation problem is solved at the smallest admissible value:Under the model $M>0$ with probability one. The all-zero sample is a null event on which the likelihood function has no finite maximizer in the parameter space $\theta >0$.