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For an original illustrative readout model, take detected photon count $K$ in a fixed exposure to have a Poisson distribution, with stipulated means ${\lambda }_B=12$ for a bright state and ${\lambda }_D=0.2$ for a dark state. Classify $K\ge 3$ as bright. Direct summation givesWith equally likely preparations, the average misclassification probability is $0.0008354$. These numbers describe only the chosen count model. Actual fluorescence includes short-time photon correlations, detector effects and possible state changes; it need not follow a fixed-rate Poisson process. Longer exposure can improve separation of count distributions while increasing unwanted optical pumping, metastable decay or recoil.

If the prepared bright population is $p$, let $\alpha$ and $\beta$ denote the two conditional errors above. The observed bright-label probability is $q=(1-\alpha )p+\beta (1-p)$, hence $p=(q-\beta )/(1-\alpha -\beta )$ when the denominator is nonzero. Known-state controls must estimate $\alpha ,\beta$ in the actual apparatus; the formula cannot correct unknown drift. For $M$ independent preparations, the raw bright fraction has variance $q(1-q)/M$. Counting uncertainty, preparation error, pulse error and state-detection bias therefore enter different parts of the inference. Repeating a biased measurement reduces its sampling uncertainty without removing its bias.

A treatment contrast is a linear contrast of cell means comparing treatment responses, with coefficients adding to zero. Pairwise differences and marginal factorial differences are examples. Its variance depends on the true allocation units and their covariance matrix.

Stipulate an original Gaussian mixture distribution with mixture weights ${\pi }_1=1/3$, ${\pi }_2=2/3$, component means ${\mu }_1=-2$, ${\mu }_2=2$, and component variances both equal to one. These are chosen statistical parameters, with no observed sample or fitting. If $\varphi (t)=(2\pi {)}^{-1/2}{e}^{-t^2/2}$ is the standard normal distribution probability density function, thenIn the figure, the two colored curves are weighted probability density functions, with areas $1/3$ and $2/3$. Adding their heights at every $x$ gives the black curve. Neither colored curve stops at a dividing line.

The law of total expectation and law of total variance give the total mean and variance:All of the extra variance $32/9$ comes from variation between component means. Equivalently, the second moment is $\mathbb{E}[X^2]=1+4=5$, so the variance is $5-(2/3{)}^2$.

A mixture distribution is different from a deterministic linear combination of random variables. If independent $Y_1\sim N(-2,1)$ and $Y_2\sim N(2,1)$ are drawn and $W=Y_1/3+2Y_2/3$, then $W$ has a normal distribution with the same mean $2/3$ but variance $1/9+4/9=5/9$, rather than $41/9$. Every draw of $W$ uses both random variables. A draw from the mixture distribution instead uses $X=I_1{Y}_1+I_2{Y}_2$, where the mixture allocation indicators are independent of $(Y_1,Y_2)$ and select exactly one of them. The expression for $f$ adds densities, not simultaneous random outcomes.

For two centered predictors with squared norms $n$ and correlation coefficient $\rho$, the ordinary least squares estimate of either partial slope has variance ${\sigma }^2/[n(1-{\rho }^2)]$. Compared with the same predictor alone and the same error variance, the inflation factor is $1/(1-{\rho }^2)$. The eigenvalues of the slope Gram matrix are $n(1+\rho )$ and $n(1-\rho )$.

If an autoregressive model has white noise of positive variance, a unique weakly stationary process solving its equation on all integer times exists precisely when its autoregressive polynomial has no root on the unit circle. In that case $1/A(z)$ has an absolutely summable Laurent series on an annulus containing the unit circle; its coefficients define a bilateral linear filter of a stationary time series. Applying that inverse also proves uniqueness among bounded second-moment solutions. A root on the unit circle forces the candidate spectral density of a stationary process to have a nonintegrable singularity. The stronger causality root criterion for an autoregressive model requires every root outside the closed unit disk.

Two independent boson species occupy uniform single-particle modes on $L$ sites, with fixed occupation numbers $n_A,n_B$. This differs from a fixed-number two-mode condensate whose particles all occupy one superposition of the two modes. Here $⟨a_i^{†}{a}_j⟩=n_A/L$, $⟨b_i^{†}{b}_j⟩=n_B/L$, and the local total number operator has variance $(n_A+n_B)L^{-1}(1-L^{-1})$. The latter follows from two independent binomial distributions with success probability $1/L$. This state satisfies a prescribed Schwinger boson constraint only in the mean unless the local variance vanishes.

In a centered MA(1) process, $X_{T+2}$ is uncorrelated with every observation through time $T$. Its best linear predictor from that information is therefore zero, with error variance ${\sigma }^2(1+{\theta }^2)$. One-step innovation formulas using $X_{T+1}$ cannot be substituted when that observation is unavailable.

The count $K_n$ has binomial distribution with mean $n/4$ and variance $3n/16$. The information content per symbol is an affine function of $K_n/n$, so its variance isChebyshev's inequality therefore gives the finite-sample lower boundThe bound is often loose, but it proves that every fixed positive tolerance eventually captures nearly all the probability. Direct summation of the binomial distribution's probability masses gives $P(T_{0.1}^{(640)})\approx 0.999779$. A finite sample does not need to have a nearly full typical set: the 40-symbol example above captures only about $0.639$.

For independent $X_i=\theta +Y_i$ with $Y_i$ exponentially distributed at known rate $\lambda$, the joint likelihood factors through the sample minimum $X_{(1)}$. Its excess over the endpoint is exponential with rate $n\lambda$, because $\mathbb{P}(\min Y_i>t)=e^{-n\lambda t}$. Thus the displayed estimator is an unbiased estimator, with variance $(n\lambda {)}^{-2}$. Even if $\theta \ge 0$, the estimator can be negative; truncation at zero introduces bias.

An unbiased likelihood estimator has expectation equal to the likelihood function at each parameter. A pseudo-marginal algorithm requires this estimator to be nonnegative. Importance sampling provides $\hat{L}=M^{-1}{\sum }_j\ell (F_j)p_{\theta }(F_j)/g_{\theta }(F_j)$ for independent random variables $F_j\sim g_{\theta }$, provided the proposal covers the target support. Relative variance affects the mixing time even though unbiasedness guarantees the correct marginal target.

For $0\le q\le 1$ and $r>0$, this mixed probability distribution has probability mass $1-q$ at zero and ordinary probability density function $q/r$ on $(0,r)$. That ordinary density integrates to $q$, not to one; conditioned on $S>0$, the law is the uniform distribution on $(0,r)$. Its cumulative distribution function is zero for $s<0$, $1-q+qs/r$ for $0\le s<r$, and one for $s\ge r$.

For each positive integer $k$, integration gives $\mathbb{E}[S^k]=q{r}^k/(k+1)$. Thus the expected value is $qr/2$ and the variance is $q{r}^2/3-q^2{r}^2/4$. The nonzero atom contributes to normalization even though it contributes zero to positive moments. At $q=0$, conditioning on positive delay is undefined because the conditioning event has probability zero.

An unbiased estimator $T$ is uniformly minimum-variance unbiased if its variance is no larger than that of every other unbiased estimator of the same parameter, for every parameter value. The Lehmann–Scheffé theorem gives such an estimator whenever an unbiased function of a complete sufficient statistic exists. Attaining the Cramér-Rao bound is sufficient for this property under its hypotheses, but is not necessary: for $T\sim \Gamma (n,\lambda )$ with $n>2$, $(n-1)/T$ is unbiased and has variance ${\lambda }^2/(n-2)$, while the information bound is ${\lambda }^2/n$. Completeness follows from uniqueness of the Laplace transform of $h(t)t^{n-1}$, so the estimator is still uniformly optimal among unbiased estimators.

A UMVCUE has minimum conditional variance among all estimators unbiased in the selected experiment for every parameter. For independent Gaussian stage estimates $A,V$ with information $I_1,J$, continuation $A\ge c$, total information $I_2=I_1+J$ and pooled estimate $T$, Rao-Blackwellization of the fresh estimate gives $T-(I_{1}s/J)\varphi ((T-c)/s)/\Phi ((T-c)/s)$, where $s^2=1/I_1-1/I_2$. The conditional family has complete sufficient statistic $T$, so the Lehmann–Scheffé theorem establishes the optimum. A conditional-likelihood bias correction does not automatically have this unbiasedness property.

For independent and identically distributed random variables $U_1,\dots ,U_n$ with the continuous uniform distribution on $(0,1)$, the $r$th order statistic has Beta distributionIts probability density function, mean, and variance areThe probability density function is zero outside $[0,1]$; values assigned to its endpoints do not change any probability. Peter Kempthorne's MIT lecture notes derive the uniform order statistic law and these moments. ^[ref].

For $n=5$, the minimum, sample median, and sample maximum correspond to $r=1,3,5$. Their probability density functions are $5(1-t{)}^4$, $30t^2(1-t{)}^2$, and $5t^4$, respectively. The sample median therefore has Beta distribution $\mathrm{Beta}(3,3)$, mean $1/2$, and variance $1/28\approx 0.0357143$. Its cumulative distribution function isFor instance, $\mathbb{P}(U_{(3)}\le 1/4)=53/512\approx 0.103516$. The mean of the minimum is $1/6$, and the mean of the sample maximum is $5/6$. The fixed observed sample median $0.4$ in the order statistic example is one possible outcome; $1/2$ is the average over repeated statistical samples from this uniform distribution.

Each curve has area one. The extreme order statistics favor the ends, while the sample median favors the center. The endpoint density height $5$ is legitimate: the probability density function gives area per unit length, rather than point probability. In particular, a continuous uniform distribution has zero probability of producing an endpoint.

For a short proof, the binomial distribution counting identity under order statistic gives $\mathbb{P}(U_{(r)}\le t)={\sum }_{k=r}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k(1-t{)}^{n-k}$. Taking its derivative cancels consecutive terms, leaving the displayed probability density function. Recognize its normalizing factorial coefficient as that of the Beta distribution, then substitute $\alpha =r$, $\beta =n+1-r$ into the Beta distribution mean and variance formulas. Although the original observations are independent, their order statistics generally are dependent; see the positive covariance of two uniform order statistics.

Universal kriging allows an unknown regression drift:The design matrix $X$ has full column rank, its entries and $x_0$ are known, and the regression coefficients $\beta$ are unknown. For example, rows $(1,s_i)$ and target row $(1,s_0)$ allow a linear spatial trend. Requiring $w^{T}Z$ to have zero bias for every $\beta$ gives $X^{T}w=x_0$. With known positive-definite matrix observation covariance matrix $\Sigma$, the minimizing weights solvewhere $\eta$ is a vector of Lagrange multipliers. Equivalently, the generalized least squares estimate and prediction areThe first term is the estimated trend alone; the second predicts the correlated regression residuals. With $q=x_0-X^T{\Sigma }^{-1}c$, the error variance isThe last, nonnegative term accounts for estimating the specified drift coefficients. Merely inserting a fitted drift into simple kriging and keeping its known-mean error variance would omit this term. For a Gaussian random field, the same expressions arise by integrating over drift coefficients in a diffuse prior distribution limit; this is an additional inferential interpretation, not an assumption required by the linear optimization.^[ref]

For a normal distribution $N_n(\mu ,{\sigma }^{2}I)$ with unrestricted $\mu \in {\mathbb{R}}^n$ and unknown ${\sigma }^2$, an integrable data-only unbiased estimator of the mean-vector prediction risk of a fixed rank-$k$ orthogonal projection matrix exists exactly when $2k=n$. To prove necessity, randomize the mean by independent Gaussian variance $v$: conditioning adds $(n-k)v$ to the squared bias term, whereas the marginal variance identity would add $kv$. When $2k=n$, the residual sum of squares itself is unbiased.

The raw same-observable time correlation $C(t,\tau )=⟨X(t)X(t+\tau )⟩$. For a stationary process, it equals $⟨X{⟩}^2+\mathrm{Cov}(X(t),X(t+\tau ))$. Its connected part is the autocovariance; unlike the normalized statistical autocorrelation, no division by the variance is imposed. The distinction matters for fluctuations around a nonzero equilibrium point.

Child Development 89, 1821–1838; 198 children, five-year study starting at 7.5, shared language variance and nonlinear relations.

For Gaussian initial conditions and the Einstein-de Sitter second-order standard perturbation theory density kernel, the unsmoothed density contrast obeysThere are three choices of the second-order field and two connected Wick contractions; the angular average of $F_2$ is $5/7+(2/7)/3=17/21$. Thus the reduced skewness is $6(17/21)=34/7$. Smoothing changes this result, and the unsmoothed calculation requires a finite variance.

A Poisson distribution of parameter $a$ conditioned to lie in $\{0,\dots ,C\}$. It is the occupancy law of a unit-resource loss network. Its mean is the carried load of an Erlang loss resource, $a[1-E(C,a)]$. Differentiating the mean with respect to $\log a$ gives its variance, which is positive for $a>0$ and $C\ge 1$; this proves strict monotonicity of the carried load.