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Let $Y_i$ be iid with density $f$, set $H_i=h(Y_i)$, and put $I=E_{f}h(Y)$ and ${\tau }^2={\mathrm{Var}}_f(h(Y))$. The finite-second-moment assumption gives, for ${\tau }^2>0$,The sample variance $s_N^2=(N-1{)}^{-1}{\sum }_i(H_i-{\hat{I}}_N{)}^2$ consistently estimates ${\tau }^2$. The central limit theorem and Slutsky theorem therefore give the asymptotic confidence intervalIts confidence probability is approximately $1-\alpha$, conventionally written $100(1-\alpha )\%$; the printed $(1-\alpha )\%$ omits the factor $100$. For ${\tau }^2=0$, $h(Y)=I$ almost surely and the estimator has zero error. This interval concerns iid Monte Carlo estimators; correlated simulation output requires a long-run variance of a stationary process estimate instead of the iid sample variance.

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 $n-1$.

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.

The given identities implyThus, writingwe have the exact identityThe sample variance satisfies $s_n^2\stackrel{P}{\to }1$, soThe Slutsky theorem and part (c) now give

Let the columns of a centered design matrix $X\in {\mathbb{R}}^{n\times p}$ be variables and write $X^{T}X=V\Lambda V^T$ by the spectral theorem for real symmetric matrices. The columns $u_i$ of $U=XV$ are the sample principal-component score vectors. They satisfy $u_i^T{u}_j={\Lambda }_{ij}$, so distinct score vectors have zero sample covariance and $u_i$ has sample variance ${\Lambda }_{ii}/n$ under the divisor-$n$ convention.