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The standardized observations $X_i/\sigma$ are independent standard normal random variables. By the definition of the chi-squared distribution,Its distribution is continuous, so $F_n(c_2)-F_n(c_1)=1-\alpha$ gives the required probability. Put $S={\sum }_i{X}_i^2$. Solving $c_1<S/{\sigma }^2<c_2$ for the positive variance reverses the reciprocal order. The exact confidence interval isThe mean is known to be zero, so the degrees of freedom are $n$, not $n-1$. Coverage is exactly $1-\alpha$ under the stated normal model.

The unrestricted likelihood is maximized at the empirical proportions, while the null likelihood is maximized at ${\pi }_i(\hat{\theta })$. Twice the log of their ratio isLarge values mean the fitted null likelihood is much smaller than the unrestricted maximum, so rejection is in the upper tail.

For a regular identifiable scalar parameter lying in the interior, positive true cell probabilities, and sufficiently large expected counts, Wilks theorem gives an approximate chi-squared distribution with $(k-1)-1=k-2$ degrees of freedom under the null. The unrestricted probability vector has dimension $k-1$ and the fitted null has dimension one. At significance level $\alpha$, reject if $\Lambda (\hat{\theta })>F_{k-2}^{-1}(1-\alpha )$, or report the approximate p-value $1-F_{k-2}(\Lambda )$.

Cells with very small expected counts may require a prespecified pooling scheme or simulation calibration; the asymptotic reference is not an exact finite-sample law. For a regular null with $d$ fitted parameters, the corresponding degrees of freedom are $k-1-d$.

For independent Gaussian measurements with correctly specified standard deviations, model residuals give the chi-squared test statisticFor a linear fit of $p$ identifiable parameters under the null model, it has a chi-squared distribution with $\nu =n-p$ degrees of freedom. The upper-tail p-value $P({\chi }_{\nu }^2\ge X_{\mathrm{obs}}^2)$ tests unusually large residuals; a nonlinear fit generally uses this distribution only approximately. Correlated Gaussian errors instead require $X^2={\mathbf{r}}^{\mathsf{T}}{\Sigma }^{-1}\mathbf{r}$.

For counts, Pearson's chi-squared test uses $X^2={\sum }_j(O_j-E_j{)}^2/E_j$, with $E_j$ calculated under the null hypothesis. For $k$ multinomial categories and $p$ fitted parameters the usual asymptotic degrees of freedom are $k-1-p$; the subtraction of one reflects the fixed total count. Sparse expected counts can invalidate this approximation, requiring a suitable exact or simulation-based test. Contingency-table independence gives $(r-1)(c-1)$ degrees of freedom.

A small p-value indicates tension with the specified model and error assumptions; it is not the probability that the null is true. Excessive $X^2$ may reflect a wrong model, underestimated errors, correlations or outliers. An implausibly small value may suggest overestimated errors or hidden constraints. For an adequate fitted model $X^2/\nu$ is typically near one, with fluctuations, rather than required to equal one.

A Gaussian distribution with mean $\mu$ and standard deviation $\sigma$ has probability densityIts symmetry, finite variance and simple behavior under linear combinations make it useful for repeated measurements. Roughly $68.3\%$, $95.4\%$ and $99.7\%$ of observations lie within one, two and three true standard deviations of the mean. These are coverage probabilities for a specified distribution, not guarantees that every experimental error is normal.

Why it appears. The central limit theorem explains why many sums of small independent finite-variance contributions approach a Gaussian law after centering and scaling, provided no single contribution dominates. Detector noise, small mechanical disturbances and multiple electronic fluctuations can combine approximately this way. The theorem does not remove a common calibration bias, and it does not justify assuming normal tails for rare large disturbances or arbitrarily correlated errors. A large-mean Poisson distribution also approaches a Gaussian with variance equal to its mean, making normal approximations useful for high-count photon or particle data; low counts require the discrete model.

A classical particle at thermal equilibrium in a one-dimensional harmonic potential has a Gaussian position distribution directly from the Boltzmann distribution:Thus a Gaussian law can arise from the energy dependence of equilibrium probability, rather than only from an informal appeal to many independent errors.

Precision of repeated measurements. For $N$ independent observations with the same mean and variance, the sample mean has variance ${\sigma }^2/N$. For Gaussian observations this mean is exactly Gaussian:Consequently quadrupling the number of independent readings halves the random standard error. When the true variance is unknown, a small-sample confidence interval for a normal mean uses a Student t-distribution with $N-1$ statistical degrees of freedom; replacing that by a normal quantile can overstate precision. Shared systematic errors and correlations do not decrease by the same averaging rule.

Combining results and fitting models. Independent Gaussian estimates $x_i$ of one quantity, with known variances ${\sigma }_i^2$, combine through the inverse-variance weighted meanFor a model $y_i=g_i(\theta )+{ϵ}_i$ with independent zero-mean Gaussian noise, maximizing the likelihood is equivalent to minimizing the weighted least-squares statisticFor a correct linear model with known independent Gaussian errors and $p$ fitted parameters of full rank, the minimized statistic has a chi-squared distribution with $N-p$ degrees of freedom. Nonlinear fits only inherit that familiar approximation under additional regularity and adequate-data assumptions. Known correlated Gaussian errors require the full inverse covariance matrix in the quadratic form.

These properties support confidence intervals, consistency checks, calibration and error propagation. They also give practical diagnostics: inspect residuals for trends, changing spread, correlations or outliers before trusting a Gaussian fit. Gaussian models connect physical fluctuations to quantitative uncertainty, but their assumptions must be checked rather than inferred from the shape of a convenient bell curve.

A chi-squared test compares residuals with the random scatter predicted by a model. For independent measurements $y_i$ with known Gaussian standard deviations ${\sigma }_i$ and predictions ${\mu }_i(\theta )$, formIf the mean model and error assumptions are correct, a full-rank linear fit of $p$ parameters gives a chi-squared distribution with $\nu =N-p$ statistical degrees of freedom. With a fully specified prediction and no fitted parameters, $\nu =N$. Nonlinear fits usually make this an approximation. Independent external calibration is not the same as fitting a parameter to these test data.

Under the reference distribution, the mean statistic is $\nu$ and its variance is $2\nu$. The upper-tail p-value $\mathrm{Pr}({\chi }_{\nu }^2\ge Q_{\mathrm{obs}})$ quantifies how unusual a discrepancy at least this large would be under the assumed model. A small value suggests model failure or incorrect uncertainties; it is not the probability that the model is true. An unusually small statistic can also reveal inflated uncertainties or correlations. A rejection threshold should be specified in advance.

For independent Poisson distribution counts, the model variance is its expected count ${\mu }_i$. The Pearson chi-squared statistic is $\sum (y_i-{\mu }_i{)}^2/{\mu }_i$. The estimated-error alternative ${\sigma }_i\simeq \sqrt{y_i}$ is common for large counts but can differ substantially when counts are small. In that regime the Poisson deviance or a direct likelihood calculation is preferable, with simulation if an accurately calibrated goodness-of-fit probability is required.

A chi-squared test compares squared deviations with the size expected from the assumed random errors. For independent observations with a normal distribution $y_i$ with known standard deviations ${\sigma }_i$ and predictions ${\mu }_i(\theta )$, useFor a correct fixed model, the standardized independent residuals are unit-variance normal distributions, so $Q$ has a chi-squared distribution with $\nu =N$. Fitting $p$ independent linear parameters removes $p$ residual directions, giving $\nu =N-p$. This distribution is exact for a full-rank linear model with known errors with a normal distribution; it is an approximation for many nonlinear fits. With known correlated errors with a multivariate normal distribution, replace the sum by $Q={\mathbf{r}}^{\mathsf{T}}{\Sigma }^{-1}\mathbf{r}$, where $\Sigma$ is the covariance matrix and its rank and fitted constraints determine the available statistical degrees of freedom.

Under the null hypothesis, $\mathbb{E}Q=\nu$ and $\mathrm{Var}Q=2\nu$. The reduced statistic $Q/\nu$ should therefore be near one, with standard deviation $\sqrt{2/\nu }$, rather than equal to one exactly. An upper-tail p-value isCompare it with a significance threshold chosen in advance. A small value indicates tension with the model together with its error assumptions; it can arise from a wrong mean model, underestimated uncertainties or neglected correlations. An unusually small statistic can suggest overestimated uncertainties, correlations or overfitting. A p-value is not the probability that the model is true, and failure to reject is not proof of the model.

For counts in categories, the Pearson form is $Q={\sum }_{j=1}^k(O_j-E_j{)}^2/E_j$. Its chi-squared distribution is an asymptotic approximation requiring adequate expected counts. For a fixed-total multinomial sample there is one normalization constraint, so the usual statistical degrees of freedom are $k-1-p$ after fitting $p$ other parameters. Independent Poisson bins instead start with $k$ statistical degrees of freedom before parameters are fitted. Sparse counts require an appropriate count likelihood or a calibrated exact or simulation-based test; plugging tiny expected counts into the asymptotic formula is unreliable. The error model, binning, fitted parameters and chosen test must therefore be reported with the result.

A chi-squared test compares the sum of squared standardized regression residuals with the appropriate chi-squared distribution. With both line parameters fitted, the data give ${\chi }^2=4.6537$ on $6-2=4$ degrees of freedom, with upper-tail P-value about $0.325$: the fitted exponential decay model describes the data adequately.

First consider fixing the alternative lifetime at its central value ${\tau }_a=0.478\mathrm{ps}$. Its slope is $m_a=-1/{\tau }_a=-2.09205{\mathrm{ps}}^{-1}$. One must refit the normalization rather than retain the earlier intercept. Minimization over the intercept gives $c_a=\bar{y}-m_a\bar{x}=7.29773$, andon $6-1=5$ degrees of freedom, with upper-tail P-value about $0.00130$. Thus that fixed central lifetime gives a substantially worse fit than $\hat{\tau }=0.4216\mathrm{ps}$. Equivalently, the increase after profiling over the intercept isThis fixed-value comparison does not yet include the uncertainty of the alternative measurement.

For the compatibility of independent lifetime measurements, include both errors. Under the approximately Gaussian lifetime-error model, equality of the two true lifetimes is tested byThis has one degree of freedom and P-value approximately $0.032$, equivalent to a $2.15$-standard deviation discrepancy. The two measurements are therefore in mild tension: equality is rejected at a $5\%$ significance threshold, but not at $1\%$. There is no unique yes-or-no compatibility verdict without choosing such a threshold. Treating the alternative central value as exact would overstate the disagreement between the experiments.

The lifetime that provides the best fit to these six rates remains $\boxed{\tau =0.4216\mathrm{ps}}$, not $0.478\mathrm{ps}$. The compatibility calculation above assumes the two experiments have independent uncertainties; a shared systematic error would require its covariance.

Assume independent normally distributed angle errors with known standard deviation ${\sigma }_y=0.002$. The relevant chi-squared test uses the sum of squared standardized residuals,For the model through the origin, the predictions and residuals, expressed in milliradians, areConsequently ${\chi }_0^2=30.0280$. There are $4-1=3$ statistical degrees of freedom. The upper-tail P-value is approximately $1.36\times 10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.

For the alternative linear regression, minimize ${\sum }_i(y_i-c^{\prime }{x}_i-d{)}^2$. Setting both partial derivatives to zero gives the normal equations, whose solution isFor these data $\bar{x}=2508.83699{\mathrm{m}}^{-2}$, $\bar{y}=0.0475$, and $S_{xx}=4.830232\times 10^6{\mathrm{m}}^{-4}$. This yieldsThe new predictions are $(66.6548,55.4171,41.4315,26.4966)\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\mathrm{mrad}$. ThereforeThe affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta {\chi }^2=27.668$ has a chi-squared distribution with one statistical degree of freedom under the nested zero-intercept hypothesis, corresponding to $p\simeq 1.44\times 10^{-7}$. Thus the model with a fitted intercept provides the better description, rather than merely lowering the residual by adding a parameter.

With the measurement variance treated as known, the coefficient uncertainties areHence $d=(13.11\pm 2.49)\mathrm{mrad}$, about $5.26$ standard deviations above zero. A positive constant systematic error in the angular zero, or an additional nearly constant torque, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred gravitational constant and cannot remove this residual pattern.

Using the fitted slope after allowing for this offset would give $G^{\prime }=\kappa {\hat{c}}^{\prime }/(mML)=(7.27\pm 0.51)\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.

For independent Poisson random variables with known expected counts ${\mu }_i$, the model variance is ${\mu }_i$. The Pearson chi-squared statistic is thereforeThere are six independently specified bins and no parameters fitted from them, giving approximately six degrees of freedom. This is far above its expected value six; the approximate chi-squared distribution tail probability is $9.5\times 10^{-6}$. The discrepancies are much too large for a good Poisson-noise fit. Replacing model variances by the observed-count estimates gives the alternative statistic $30.92$, with the same qualitative conclusion. There is no fitted normalization to justify subtracting a degree of freedom or rescaling the model to the observed total.

Let $N_{ij}$ be the observed cell counts, $N_{i+}$ the row totals, $N_{+j}$ the column totals and $N=33009$. The fitted expected frequencies under independence are $E_{ij}=N_{i+}{N}_{+j}/N$. For example the three-child, one-boy cell has expected count $7541\cdot 16472/33009=3763.1$.

The Pearson chi-squared statistic isUnder $H_0$ it has asymptotically a chi-squared distribution with $(2-1)(3-1)=2$ degrees of freedom, assuming the multinomial sampling model and sufficiently large expected counts. All expected counts here are large. The distribution is an asymptotic approximation, rather than an exact finite-sample identity.

Since the true mean is known to be zero, the chi-squared distribution gives $Q/\theta \sim {\chi }_n^2$. Let $q_{\beta }$ be the $\beta$ quantile of this law. The probabilityrearranges, using positivity, to the exact confidence intervalThere are $n$ degrees of freedom, because no unknown mean was fitted.

Up to a parameter-independent multinomial factor, the likelihood function is $L={\prod }_{i,j}{\theta }_{ij}^{n_{ij}}$. A Lagrange multiplier for $\sum {\theta }_{ij}=1$ gives $n_{ij}/{\theta }_{ij}=\lambda$, whence $\lambda =n$ andSince all counts are positive, the log-likelihood function is strictly concave on the positive simplex and tends to minus infinity at its boundary, proving this is the unique maximum.

Under independence of random variables, ${\theta }_{ij}={\varphi }_i{\rho }_j$, so the log-likelihood function separates as ${\sum }_i{n}_{i\cdot }\log {\varphi }_i+{\sum }_j{n}_{\cdot j}\log {\rho }_j$. Applying the same constrained maximization twice gives ${\hat{\varphi }}_i=n_{i\cdot }/n$, ${\hat{\rho }}_j=n_{\cdot j}/n$, andTwice the difference between the unrestricted and null maximized log-likelihood functions isthe required likelihood-ratio test statistic. Under the null hypothesis, its asymptotic chi-squared distribution has $(r-1)(c-1)$ statistical degrees of freedom, the parameter-count difference $(rc-1)-(r+c-2)$. Here that is four. Since $25.192>q_4=18.467$, reject independence at the 0.1 percent significance level. For four statistical degrees of freedom, the asymptotic upper-tail probability is $e^{-25.192/2}(1+25.192/2)\approx 4.60\times 10^{-5}$. This establishes an association in these data; it is not a causal conclusion.

The moment generating function is $M_X(t)=\mathbb{E}(e^{tX})$ wherever that expectation is finite. A useful continuity theorem says that if the moment generating functions $M_n$ are finite on a common open interval containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual continuity at zero, and obtain the uniquely determined limiting law.

For the gamma density, integrating the exponential against the density and changing variable to $(\lambda -t)x$ givesFor the waiting time to one head, independence of successive tosses gives $\mathrm{Pr}(Y=j)=p(1-p{)}^{j-1}$ for $j\ge 1$. Summing the geometric series givesThe waiting times between consecutive heads are independent copies of $Y$, so their sum $N$ has moment generating function $M_Y(t{)}^k$. ConsequentlyFor fixed $t<1/2$, numerator and denominator have leading terms $p$ and $p(1-2t)$ as $p\downarrow 0$. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small $p$, so the continuity theorem applies. The limiting distribution isequivalently a chi-squared distribution with $2k$ degrees of freedom.

Let $g_i=(X_{i1},\dots ,X_{in})$ be row $i$ of $X$, viewed as a real vector. These are independent standard Gaussian random vectors, and $(X{X}^T{)}_{ij}=g_i\cdot g_j$. The trace expression for the empirical spectral measure givesWe need the norm momentsFor completeness, $S=\Vert g_i{\Vert }^2$ has the chi-squared distribution with $n$ degrees of freedom. By independence and the Gaussian integral, its moment-generating function is $\mathbb{E}{e}^{sS}=(1-2s{)}^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,For the second moment, if $i\ne j$, conditional expectation given $g_i$ gives $\mathbb{E}[(g_i\cdot g_j{)}^2\mid g_i]=\Vert g_i{\Vert }^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields

For the third moment, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.

ConsequentlyThe three limiting expected values are therefore

For fixed $n$, the likelihood ratio is strictly decreasing as $t^2$ increases. A likelihood-ratio test rejects for small $\lambda$, so its rejection rule is equivalently large $|t|$. Under the null hypothesis, the standard normal distribution and chi-squared distribution results give independent quantitiesTheir ratio therefore has Student's t-distribution with $n-1$ statistical degrees of freedom. The exact level-$\alpha$ rule isThis is precisely the two-sided Student's t-test; choosing the corresponding cutoff for $\lambda$ yields the same rejection region and the same p-value.

Let $F$ be the cumulative distribution function of a gamma distribution with shape $A=a+{\sum }_i{x}_i$ and rate $B=b+n$; explicitly,The posterior distribution is continuous and has positive density on $(0,\infty )$, so its quantiles give the requested equal-tailed credible interval:Equivalently, $2B\theta \mid x\sim {\chi }_{2A}^2$, so the endpoints can be written as the corresponding chi-squared distribution quantiles divided by $2B$. Either form excludes posterior probability $\alpha /2$ from each tail.

Let $L(\vartheta ;x)$ be the likelihood function, with full parameter space $\Theta$ and null hypothesis $\vartheta \in {\Theta }_0\subseteq \Theta$. The generalized likelihood-ratio test compares the best null fit to the best unrestricted fit:The likelihood ratio $R$ lies between zero and one, and small values count against the null hypothesis; equivalently, large values of the likelihood-ratio test statistic $\Lambda$ count against it.

Compute restricted and unrestricted maximum-likelihood estimates, evaluate their log-likelihoods, and form $\Lambda =2(\ell (\hat{\vartheta })-\ell ({\hat{\vartheta }}_0))$. Under the regular conditions of Wilks theorem, the null limit isThese statistical degrees of freedom count independent parameter restrictions. The regularity includes a locally identifiable smooth model, nonsingular Fisher information, and an interior null value; parameter-dependent support can invalidate the result.

For a chosen significance level $\alpha$, obtain the $(1-\alpha )$ quantile $q_{d,1-\alpha }$ of the chi-squared distribution and useEquivalently, compare the asymptotic p-value $\mathrm{Pr}({\chi }_d^2\ge {\Lambda }_{\mathrm{obs}})$ to $\alpha$. This is a large-sample calibration; part (c) requires direct treatment because its support depends on a parameter.

The count vector has a multinomial distribution, with likelihood functionMaximize its log-likelihood using a Lagrange multiplier for the sum constraint. The equations $n_i/{\pi }_i=\lambda$ and ${\sum }_i{\pi }_i=1$ give $\lambda =n$ and ${\hat{\pi }}_i=n_i/n$. Since all counts are positive and the log-likelihood is strictly concave on the interior of the probability simplex, this is the global maximum-likelihood estimate.

Under the uniform null hypothesis, each category has expected countThe common multinomial coefficient cancels from the likelihood ratio:Taking minus twice its logarithm gives exactlyThere are $k-1$ free category probabilities, because their sum is one, and the null fixes all of them. Thus its regular asymptotic chi-squared distribution has $k-1$ statistical degrees of freedom.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.

Let $S={\sum }_i{X}_i$. For the exponential distribution with mean $\theta$, the likelihood function is $L(\theta )={\theta }^{-n}{e}^{-S/\theta }$ on positive observations. Fix any $0<{\theta }_1<{\theta }_0$. The alternative-to-null likelihood ratio isSince $1/{\theta }_1-1/{\theta }_0>0$, this is strictly decreasing in $S$. The Neyman-Pearson lemma says to reject for small $S$. Under the null hypothesis, the given sum rule yields $S\sim \mathrm{Gamma}(n,1/{\theta }_0)$, where the second argument is a rate, and henceLet $q_{\alpha }$ be the lower $\alpha$ quantile of this chi-squared distribution. The required critical region isIts test size is exactly $\alpha$ because the null distribution is continuous. Its cutoff is the same for every ${\theta }_1<{\theta }_0$, so it is simultaneously most powerful against every such value: it is a uniformly most powerful test. Equivalently, reject if $S$ is below the lower $\alpha$ quantile of the specified gamma distribution.