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For an estimator with finite second moment, its bias and mean squared error areWrite $\hat{\theta }-\theta =(\hat{\theta }-{\mathbb{E}}_{\theta }\hat{\theta })+b_{\theta }$. On squaring and taking expectations, the cross term vanishes. This proves the bias-variance decomposition of mean squared error:For the sample mean, independence gives $\mathbb{E}\bar{X}=\theta$, $\mathrm{Var}(\bar{X})={\theta }^2/n$. Thus $k\bar{X}$ has bias $(k-1)\theta$. Unbiasedness throughout the parameter family requiresAt the single parameter value $\theta =0$, all these estimators are zero almost surely; that special value does not change the requirement of unbiasedness for all $\theta$.

For a one-sided Wald test use the signed normal Wald statistic, rather than its square. The maximum-likelihood estimate is the sample mean ${\bar{Y}}_n$, with exact variance ${\sigma }^2/n$, soA sum of independent normal random variables is normal, giving ${\bar{Y}}_n\sim N(\delta ,{\sigma }^2/n)$ and hence $W_n\sim N(\sqrt{n}\delta /\sigma ,1)$. Under the null hypothesis its mean is zero:Thus its null distribution is exactly the standard normal distribution, without an asymptotic approximation. If the squared Wald statistic convention is used, $W_n^2$ has a chi-squared distribution with one degree of freedom; the signed form is needed to distinguish the two directions.

Let $X$ and $Y$ denote the first- and second-stage sample means, let $s=\sigma /\sqrt{n}$, and write $S=(X+Y)/2$. Continuation is the selection event $\mathcal{C}=\{X\ge sf\}$. The second-stage sample mean stays independent of $\mathcal{C}$, so $E(Y\mid \mathcal{C})=\delta$. The first-stage sample mean has a truncated normal distribution. With $a=f-\delta /s$ and the upper-tail Inverse Mills ratio $\lambda (a)=\varphi (a)/(1-\Phi (a))$,The positive conditional selection bias after futility continuation comes from selecting unusually large first-stage outcomes. The unconditional sample mean of a fixed $2n$ observations would be unbiased; that is a different sampling distribution from the one restricted to continued trials.

As $\delta \to \infty$, $a\to -\infty$, $\varphi (a)\to 0$ and $1-\Phi (a)\to 1$, so the estimator bias tends to zero. It decreases with $\delta$: differentiating gives ${\lambda }^{\prime }(a)=\lambda (a)(\lambda (a)-a)>0$, because $\lambda (a)=E(Z\mid Z>a)>a$ for a standard normal random variable. Thus

Use a Rao-Blackwell estimator after interim selection, which is exactly conditionally unbiased. The second-stage sample mean $Y$ alone is unbiased conditional on continuation, but discards the earlier observations. Apply the Rao-Blackwell theorem by averaging $Y$ conditional on the combined sample mean $S$ and the fact of continuation.

Set $c=sf$ and $v=s/\sqrt{2}=\sigma /\sqrt{2n}$. Before truncation, $X\mid S$ is $N(S,v^2)$, a distribution whose mean no longer involves the unknown $\delta$. After imposing $X\ge c$, its mean is $S+v\lambda ((c-S)/v)$. Since $Y=2S-X$, the resulting estimator isIt uses the outcomes from both stages through their combined sample mean. By iterated expectation, $E(\tilde{\delta }\mid \mathcal{C})=E(Y\mid \mathcal{C})=\delta$, so its conditional estimator bias is zero, compared with the strictly positive estimator bias above. Its conditional variance is no larger than that of the second-stage-only estimate. This does not assert a smaller mean squared error than every biased estimator.

In the Bühlmann model, a latent risk parameter $\Theta$ is drawn from a population distribution. Conditional on $\Theta$, the yearly observations are independent and identically distributed random variables, with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. Define the structural parametersHere $v$ is the expected process variance, while $a$ is the variance of hypothetical means. The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ from the observed claims, under mean squared error. Predicting the next claim gives the same affine estimate: the extra conditional observation noise contributes the constant $v$ to the prediction error.

The law of total variance and conditional independence giveAn affine estimate can be written as $\hat{m}=m+{\sum }_{j=1}^n{b}_j(X_j-m)$: for any chosen $b_j$, optimizing the constant makes its expected value equal to $m$. The normal equations for the linear least-squares projection areFor $v>0$ they force all $b_i$ to agree, with $b_i=a/(v+na)$. Thus the credibility factor and premium areThe credibility factor increases with the observation count and between-risk variance, and decreases with within-risk variance. If $a=0$, the risk mean is known and $Z=0$; if $v=0$ and $a>0$, one observation reveals it and $Z=1$. If both vanish, the premium is the fixed value $m$ and the factor is immaterial.

In the specified model, the conditional law is a gamma distribution with shape $\alpha$ and scale $\theta$. ThereforeThe prior is an inverse-gamma distribution with shape $k$ and scale $\lambda$. To obtain its moments directly, substitute $y=\lambda /\theta$ in the defining integral, obtainingThe Gamma function recurrence yieldsThe assumption $k>2$ makes both structural variances finite. HenceIt follows that the model-specific credibility estimate is

For the Bayes estimator under squared error loss, the quantity to estimate is $\alpha \Theta$, so the optimum is its posterior mean. The likelihood function, viewed as a function of $\theta$, is proportional toMultiplication by the prior shows gamma scale inverse-gamma conjugacy:Although the printed hint only mentions integer shapes, the same substitution and gamma integral normalize this posterior for every positive real shape, so no integrality of $\alpha$ is needed. Its posterior mean gives the Bayesian estimate and comparisonThis exact Bühlmann credibility for gamma claims holds for every observed sample, not merely on average. Here the posterior mean is affine in the sample mean, so the best affine Bühlmann credibility premium is also the unrestricted Bayes estimator under squared error loss.

Use the usual clinical trial sampling assumption that all patients' responses are independent, with fixed positive arm sizes and positive known variances. Marginal normal distributions alone would not specify the variance of a treatment contrast without this assumption. Put ${\overline{Y}}_i=n_i^{-1}{\sum }_j{Y}_{ij}$, $v_i={\sigma }_i^2/n_i$, and $s_i=\sqrt{v_i+v_0}$. The sample mean contrast estimates ${\delta }_i$ and has standard error $s_i$. Thus the one-sided Wald statistics areLarge positive values of these Wald test statistics provide evidence against the respective null hypotheses.

The three independent sample means satisfy ${\overline{Y}}_i\sim \mathcal{N}({\mu }_i,v_i)$. A linear transformation of their joint multivariate normal distribution is again normal. The two treatment contrasts have covarianceWriting $m_i={\delta }_i/s_i$ and $\rho =v_0/(s_1{s}_2)$, their joint distribution isThe positive correlation coefficient arises from the shared control arm. Under positive variances, $0<\rho <1$.

With angular frequency $\lambda$, the periodogram isAt Fourier frequencies $2\pi j/T$ it is the squared modulus of the normalized discrete Fourier transform. If a nonzero mean is unknown, subtract the sample mean first.

For a zero-mean stationary process, expanding the square givesUnder absolute summability of the covariances this converges to the spectral density of a stationary process. Thus the periodogram is generally biased at finite $T$, but asymptotically unbiased under this short-memory time series condition.

It is nevertheless not a pointwise estimator with statistical consistency unless it is smoothed. For Gaussian white noise, at a nonzero Fourier frequency other than the Nyquist frequency, the real and imaginary Fourier components are independent normal variables. Exactly,where $f={\sigma }^2/(2\pi )$. The variance does not decrease with $T$. Under usual short-memory time series assumptions the same exponential limit is asymptotic for general processes. Averaging nearby frequencies or using a lag-window estimator reduces variance; a frequency bandwidth tending to zero while $T$ times that bandwidth tends to infinity can give statistical consistency. The raw plot remains useful for detecting strong periodic peaks, but increasing the record length alone does not remove its pointwise noise.

Under fit.2, observations in each period are independent Poisson random variables with a common mean, so each sample variance should fluctuate around its sample mean. The plot instead has variances roughly 130–300 for means roughly 80–120, well above the line variance equals mean. This is overdispersion.

Day-to-day traffic variation can produce a Poisson mixture: a random daily intensity increases the marginal variance and can correlate observations from the same day. A useful replacement is a negative binomial regression, with variance $\mu +{\mu }^2/\kappa$, or a Poisson generalized linear mixed model with a day random intercept. A day-factor effect can also be included if inference is restricted to the observed days. A Quasi-Poisson regression estimates a dispersion multiplier for uncertainty, but does not itself explain the shared-day dependence and has no ordinary likelihood-based AIC. The means can retain the period-factor or selected jump structure.

An unbiased estimator $T$ of a parameter $\theta$ satisfies ${\mathbb{E}}_{\theta }[T]=\theta$ for every permissible value of $\theta$. Since the sample mean is linear,so $\hat{\mu }$ is unbiased.

Use the identityThe two expectations on the right are $n{\sigma }^2$ and $n\mathrm{Var}(\overline{X})={\sigma }^2$, respectively. HenceThusThe unbiased sample variance instead divides the same sum of squares by $n-1$.

The sample mean isWrite the Gaussian sample vector as its orthogonal projection onto the span of $(1,\dots ,1)$ plus its projection onto the orthogonal complement. These two Gaussian projections are independent. Their squared standardized lengths give Cochran's theorem:

Because every column of $X$ has sample mean zero, every column $u_i$ of $U=XV$ also has sample mean zero. The spectral theorem for real symmetric matrices givesHence for $i\ne j$ the sample covariance is $n^{-1}{u}_i^T{u}_j=0$, while the sample variance of $u_i$ isThus the $u_i$ are pairwise uncorrelated sample principal components, ordered by decreasing sample variance.

A statistic $T$ is sufficient for $\theta$ if the conditional distribution of the full sample given $T$ does not depend on $\theta$. It is minimal sufficient if it is a function of every sufficient statistic.

The likelihood factors asso the Fisher-Neyman factorization theorem shows that $M$, and hence its one-to-one transform $\hat{\theta }=M/2$, is sufficient. Moreover, for two samples $\mathbf{x}$ and $\mathbf{y}$, the ratio $L(\theta ;\mathbf{x})/L(\theta ;\mathbf{y})$ is independent of $\theta$ exactly when their maxima agree. The likelihood-ratio criterion for minimal sufficiency therefore shows that $M$ and $\hat{\theta }$ are minimal sufficient.

For $n\ge 2$, the sample mean $\tilde{\theta }$ is not sufficient: two samples can have the same mean but different maxima, and their likelihood ratio then depends on $\theta$ through the support indicators. It is consequently not minimal sufficient. For the degenerate special case $n=1$, the sample mean and maximum coincide and both conclusions reverse.

Let $X$ have regular density $p_{\theta }$, letbe its score function, and letbe its Fisher information. Regularity gives the mean-zero score identity ${\mathbb{E}}_{\theta }{S}_{\theta }=0$.

Suppose an estimator $T$ has mean ${\mathbb{E}}_{\theta }T=\psi (\theta )$. Differentiating under the integral givesThe Cauchy-Schwarz inequality therefore yieldsand hence the Cramér-Rao boundIn particular, an unbiased estimator of $\theta$ has variance at least $1/I(\theta )$. For an independent sample, the information is the sum of the individual informations.

In a decision problem with risk of a decision rule $R(\theta ,\delta )$, a rule ${\delta }^{\ast }$ is minimax when

Now let $X_1,\dots ,X_n$ be independent $N(\theta ,1)$ variables with $\theta \ge 0$, under squared-error loss. The sample mean is unbiased with variance $1/n$, sofor every $\theta \ge 0$. Thus the minimax value is at most $1/n$.

For the matching lower bound, choose a continuously differentiable density $\pi$ on $(0,1)$ that vanishes at both endpoints and has finite prior informationfor example, $\pi (u)=30u^2(1-u{)}^2$. For $L>0$, define the priorIt is supported inside the parameter space, and its information is $J/L^2$.

Here the likelihood score iswhose Fisher information is $n$. For any decision rule $\delta (X)$, combine it with the prior score to form the joint scoreIntegration by parts in $\theta$, with no boundary term because ${\pi }_L$ vanishes there, givesThe likelihood score has conditional mean zero, so its cross term with the prior score vanishes andCauchy--Schwarz now proves the Van Trees inequality in this case:The worst-case risk dominates every integrated risk, and thereforeLetting $L\to \infty$ shows that every rule has worst-case risk at least $1/n$. Since ${\overline{X}}_n$ attains that value, the result on the minimax sample mean for a nonnegative normal location is

Use the sample mean to estimate each magnetic susceptibility, and the unbiased sample variance with denominator $n-1=4$ to estimate the repeatability of individual readings. In the numerical scale supplied by the paper,If the repeated measurements have independent random errors, their mean estimates have standard errorsThe difference and its estimated standard uncertainty are thereforeThe separation is about $2.53$ standard errors. Under an approximately normal distribution model with no experimenter-dependent bias, Welch's t-test gives about $7.34$ statistical degrees of freedom and a two-sided p-value of $0.038$. Thus there is modest evidence against equal underlying susceptibilities at a conventional $5\%$ significance level; overlap of the ranges of individual readings is not an adequate comparison of the means. Five readings per sample provide limited information about the error distribution.

Crucially, material and student exhibit complete confounding: each student measured only one sample. An experimenter-dependent systematic error could mimic the difference. Paramagnetism also depends on conditions such as temperature, so equal material alone does not guarantee equal readings without controlling those conditions. The defensible physical conclusion is the data suggest a difference, but do not establish different materials. Both students should measure both samples in randomized or alternating order, with calibration and temperature controlled, to separate a sample effect from an operator effect. No unit or scale factor absent from the original table has been invented.

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.

All four angle measurements have the same standard uncertainty, so ordinary least squares minimizesDifferentiation givesThus the unique minimum isThe overbars denote sample means over the four observations. Numerically,soThe fixed $x_i$ have negligible uncertainty. Assuming independent errors in the four $y_i$, propagation of uncertainty in quadrature givesHenceIf the period measurement is independent of the angle measurements, including its calibration uncertainty givesThis last uncertainty includes the stated random measurement errors only. Whether the model describes the data must be checked before treating it as a reliable measurement of the gravitational constant.

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

For a simple pendulum in the small-angle approximation, the oscillation period is $T=2\pi \sqrt{L/g}$, soInterpret the quoted random timing error and length error as standard uncertainties, with independent errors and no additional systematic error. For $N$ independent measurements, the standard error of the sample mean of the period is ${\sigma }_{\overline{T}}/T=0.02/\sqrt{N}$. The length has relative standard uncertainty ${\sigma }_L/L=0.1/25.0=0.004$. By propagation of uncertainty,Requiring a relative standard uncertainty no greater than $0.01$ givesFor $N=20$, the combined relative standard uncertainty is $0.009798$, or $0.980\%$; $N=19$ gives $1.001\%$ and just fails. Repeating the timing does not reduce the fixed length measurement uncertainty. The stated “accuracy” is consequently a statistical uncertainty target, not a guarantee for every experimental outcome. This is an example of repeated-timing uncertainty in a pendulum gravity measurement.

Treat the repeated angles as independent measurements of one fixed angle. Their sample mean, sample standard deviation and standard error of the mean areFor $y=A\cos \theta$, uncorrelated standard uncertainties giveInterpreting the stated 1% amplitude accuracy as a standard uncertainty,Use radians in the derivative. This is uncertainty in the result using the mean angle, not the spread of individual angle measurements; the latter convention would use $s_{\theta }$ and give about $2.34\%$. An accuracy bound or systematic angular calibration error would require a different uncertainty model.