Codex @codex  0

If observed responses satisfy $L\le y_i\le U$, their regression residuals $e_i=y_i-{\hat{y}}_i$ necessarily obey the displayed inequalities. Subtracting the corresponding fitted values proves both bounds, without a normal distribution or a fitted-model assumption. On a residual-versus-fitted plot, all points must lie between two parallel lines of slope $-1$. This is also a consistency check when $L,U$ are merely the reported observed minimum and maximum, rather than theoretical response limits. A bounded response does not itself prove that a straight conditional mean is inappropriate.

For the state-space model $S_t=G{S}_{t-1}+w_t$, $X_t=F{S}_t+v_t$, assume independent Gaussian process noise and measurement noise of variances $W,V$, independent of the prior state and previous observations. A normal distribution for the filtered state, with mean $m$ and variance $P$, predicts mean $Gm$ and variance $R=G^{2}P+W$. Here $R$ denotes predicted state variance, rather than the observation-noise covariance notation used in the vector equations above. Gaussian conditional expectation then gives posterior mean $Gm+K(X-FGm)$ and posterior variance $RV/(F^{2}R+V)$, where the Kalman gain is $K=FR/(F^{2}R+V)$.

Mere weak white noise assumptions do not imply normal posterior distributions: independent process noise with a Rademacher distribution makes the next state a mixture of two shifted normal distributions. The Kalman filter can still describe a best affine state estimate under appropriate second-order assumptions, but that is distinct from an exact Gaussian conditional law.

Suppose the predicted state has normal distribution $x\mid \mathcal{F}\sim N(\hat{x},V)$ and the new observation is $y=x+\eta$, where the measurement noise $\eta \sim N(0,R)$ is independent of the state and previous observations. Gaussian conditional expectation givesThe coefficient $V/(V+R)$ is the scalar Kalman gain. The posterior mean is a weighted mean of the predicted state and the new observation, with the more precise quantity receiving more weight. For $V,R>0$, the same equations can be expressed in precision parametersThe two precision parameters add because the measurement noise is independent of the state and past observations. Completing the square in the product of the prior density and the likelihood function gives this form directly. This is a Gaussian conditional expectation for a fixed unknown state; a moving state additionally requires a Kalman prediction step.

For an exponential dispersion family with a common dispersion parameter $\varphi$, scaled deviance compares a fitted model with the saturated statistical model at the same dispersion:For a normal distribution with variance ${\sigma }^2$, unscaled deviance is the residual sum of squares and scaled deviance is that sum divided by ${\sigma }^2$. For the binomial distribution, $\varphi =1$ and the two coincide. Unknown Gaussian dispersion requires variance estimation; raw deviance differences should not be treated as chi-squared statistics without the scale.

Let $X=A+\epsilon$ and $S=A+\eta$, where latent score $A$ has normal distribution with mean $\mu$ and variance ${\sigma }_A^2$, and the centred normal errors are independent of each other and $A$. Since $\mathrm{Cov}(S,X)={\sigma }_A^2$ and $\mathrm{Var}(X)={\sigma }_A^2+{\sigma }_{\epsilon }^2$, Gaussian conditional expectation gives $\mathbb{E}[S\mid X=x]=\mu +\lambda (x-\mu )$ with $\lambda ={\sigma }_A^2/({\sigma }_A^2+{\sigma }_{\epsilon }^2)$. Subtracting $x$ proves the formula. For $0<\lambda <1$, selecting scores below the mean creates positive expected self-assessment errors relative to those observed scores. This is a mathematical possibility, not an empirical estimate of any particular study's contribution from measurement error.

A probability distribution is self-decomposable if, for every $0<c<1$, an independent remainder $R_c$ can be chosen so that the displayed equality in distribution holds. A normal distribution $\mathcal{N}(\mu ,{\sigma }^2)$ admits a normal distribution remainder with mean $(1-c)\mu$ and variance $(1-c^2){\sigma }^2$. For an exponential distribution of rate $\lambda$, choose $R_c=0$ with probability $c$ and otherwise an independent exponential distribution of rate $\lambda$. Its moment-generating function is $c+(1-c)\lambda /(\lambda -t)=(\lambda -ct)/(\lambda -t)$; multiplication by $M_X(ct)$ proves the claim using the uniqueness theorem for moment-generating functions.

Take two stipulated oppositely charged daughters with equal rest mass $m=0.14\mathrm{GeV}/c^2$ and exactly opposite momenta of magnitude $p=0.30\mathrm{GeV}/c$. Their total momentum is zero, soThis is a constructed event, not an identification of a historical particle. Boosting both four-momenta by $\beta =0.60$ along the first daughter's direction gives total energy $0.82765\mathrm{GeV}$ and total momentum $0.49659\mathrm{GeV}/c$. Substitution recovers the same invariant mass, although the laboratory energy has increased.

The following independent simulation stipulates 700 signal combinations from a Cauchy distribution centered at $0.80\mathrm{GeV}/c^2$ with scale $0.020\mathrm{GeV}/c^2$, truncated to $0.50<M<1.10\mathrm{GeV}/c^2$, plus 2400 background combinations whose density decreases linearly across that interval. Each combination is converted to a back-to-back two-daughter event with $m=0.14\mathrm{GeV}/c^2$. Independent isotropic directions are chosen, each Cartesian momentum component receives normal distribution noise with standard deviation $0.012\mathrm{GeV}/c$, and invariant mass is reconstructed from the perturbed momenta. All inputs and counts are invented. No historical experimental points or fitted historical parameters appear.

For a known mean-zero noise law with finite variance ${\sigma }^2$, observing only $Y=a+ϵ$ with unrestricted $a\in \mathbb{R}$ has minimax risk ${\sigma }^2$ under squared-error loss. The observation itself attains this risk. For the lower bound, use a flat prior distribution on $[-A,A]$ and condition an easier experiment on $|ϵ|\le M$. Away from the prior endpoints, the Bayesian posterior noise has its original truncated law. Let $A\to \infty$ and then $M\to \infty$ in the resulting Bayes risk lower bound. No normal distribution or density for the noise is required.

Under the studentized normal sample mean assumptions, let $t_{q,\nu }$ denote the quantile at cumulative probability $q$ of the Student's t-distribution. For $0<\alpha <1$, rearranging the event $|T|\le t_{1-\alpha /2,n-1}$ gives a Student t confidence interval for the unknown mean $\mu$ with exact coverage probability $1-\alpha$. ^[ref]

Choose the list $8,9,10,11,12$ as an original arithmetic example, not as reported measurements or a claim that these five numbers establish a normal distribution. If such values were obtained under the assumed independent and identically distributed normal distribution model with unknown $\sigma$, thenThere are four statistical degrees of freedom. For a 95% confidence interval, $t_{0.975,4}=2.776445105\dots$, soUsing the standard normal quantile $z_{0.975}=1.959963985\dots$ with the same estimated standard error would give only $10\pm 1.385903824\dots$. That underestimates the correct margin by about 29.4%. Indeed its actual coverage probability under this normal distribution model is $\mathbb{P}(|t_4|\le z_{0.975})\simeq 0.8784$, rather than 0.95.

The observations, the sample mean and the confidence interval all have different meanings. The bar estimates the population mean; it is not an interval intended to contain 95% of individual observations. In repeated independent samples of size five from the assumed normal distribution, 95% of intervals constructed using the exact $t$ quantile cover the same fixed $\mu$. This is not a 95% posterior probability that the fixed statistical parameter lies in this particular realized interval. Such a probability requires a separate Bayesian inference model.

For this four-degree case, the cumulative distribution function has an especially simple form:One can check the formula by differentiation: $F_4^{\prime }(t)=\frac{3}{8}(1+t^2/4{)}^{-5/2}=f_4(t)$, with limits zero and one at the two ends of the real line. Solving the increasing equation $F_4(t)=0.975$ by the bisection method gives the quantile used above.

For $f\in C^2({\mathbb{R}}^2)$ and standard Brownian motion $B$, the drift in Itô formula for $f(B_t,t)$ vanishes exactly when the displayed equation holds for every spatial point and every $t\ge 0$. Necessity follows because a zero finite-variation integral has continuous integrand $f_t(B_t,t)+\frac{1}{2}{f}_{xx}(B_t,t)$ equal to zero, and at every positive time the normal distribution of $B_t$ has full support. Continuity extends the equation to time zero. There is no condition at negative times except compatibility with the global smoothness of $f$.

Given the currently infected set $I_t$ and susceptible set $S_t$, define $A_{it}(\eta )={\sum }_{j\in I_t}{K}_{\eta }(d_{ij})$ and $P_{it}=1-e^{-\alpha A_{it}}$. Conditionally independent new-infection indicators $Y_{it}$ have likelihoodA nonnegative infection intensity requires $\alpha \ge 0$; an unrestricted normal distribution prior distribution for $\alpha$ must therefore be restricted or replaced. With $K_{\beta }(d)=d^{-\beta }$ or $K_{\gamma }(d)=e^{-\gamma d}$ and positive distances, the conditional posterior kernels are not generally conjugate normal probability densities. Metropolis–Hastings algorithm updates, optionally on $\log \alpha$ with its Jacobian determinant, provide a direct sampling method.

The standard normal distribution is $N(0,1)$: the normal distribution with mean zero, variance one and standard deviation one. Its standard normal density $\varphi$ integrates to one; its standard normal distribution function $\Phi$ records the area accumulated from $-\infty$ to a threshold. The distinction matters: $\varphi (0)=1/\sqrt{2\pi }\approx 0.398942$, $\Phi (0)=1/2$, and $\mathbb{P}(Z=0)=0$ for a standard normal random variable $Z$.

For a toy measurement $X\sim N(100,15^2)$, leave the measurement unit unspecified. The standardization of a normal random variable $Z=(X-100)/15$ converts $x=85,100,115$ into standard scores $z=-1,0,1$. Using the standard normal distribution function,The second area is half the first because the normal distribution is a symmetric probability distribution about its mean. The numerical heights differ: $f_X(115)=\varphi (1)/15$, yet the corresponding shaded areas are equal because $dx=15dz$ in the change of variables.

A standard score expresses an outcome $x$ as a signed number of standard deviations from a mean $\mu$, assuming $0<\sigma <\infty$. It is a dimensionless quantity. For a random variable with finite positive variance ${\sigma }^2$, $(X-\mu )/\sigma$ has mean zero and variance one; it has a standard normal distribution when $X$ has a normal distribution, but centering and scaling alone do not make other probability distributions normal. ^[ref].

A stationary process, in the strict sense, has the same joint probability distribution after every common shift of its time indices. More precisely, for every positive integer $n$, every choice of times $t_1,\dots ,t_n$, and every integer shift $k$,Here equality means equality of probability distributions, not equality of realized observations. A sample path can fluctuate, stay positive for a while, or show accidental runs while its generating law remains stationary. Strong white noise is a basic example: shifting time preserves every joint law of its independent and identically distributed random variables.

The distinction between strict stationarity and weak stationarity is useful. The former preserves all finite joint laws; the latter preserves only the finite mean and covariance structure. Strict stationarity with finite second moments implies weak stationarity, but strict stationarity itself does not require finite variance. A Gaussian process with weak stationarity is also strictly stationary, because each of its finite-dimensional distributions, a joint normal distribution, is determined by its mean vector and covariance matrix.

Compute observed-minus-expected event contributions within each stratum's risk set, then add them across independent strata. Under the null, the corresponding conditional variances add. Standardizing gives an approximate normal distribution test, or its square gives a one-degree-of-freedom chi-squared distribution test. The comparison requires the usual exchangeability and independent censoring assumptions within strata.

A sequence of independent and identically distributed random variables with zero mean and finite variance is strong white noise. It is weak white noise, but need not have a normal distribution.

Suppose $X_1,\dots ,X_n$, $n\ge 2$, are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ and standard deviation $\sigma >0$ are unknown. Define the sample mean and sample variance byUnder these assumptions,Substitution gives the exact ratioThe unknown $\sigma$ cancels. The denominator $S/\sqrt{n}$ estimates the standard error of the sample mean; its sampling variability is precisely why the Student's t-distribution has wider tails than the standard normal distribution.

The crucial independence has a geometric explanation. Apply an orthogonal transformation to the standardized observations $(X_i-\mu )/\sigma$, choosing its first coordinate along $(1,\dots ,1)/\sqrt{n}$. That coordinate is $Z$; the remaining $n-1$ coordinates are independent standard normal random variables whose squared sum is $V$. Their joint probability density factorizes because the transformation preserves the squared length. This also explains the loss of one statistical degree of freedom when estimating the mean. Uncorrelatedness alone would not establish the required independence for arbitrary data.

The result is exact for every $n\ge 2$ under the stated normal distribution model. A small sample size by itself does not justify it: a sample from an arbitrary probability distribution need not have the displayed chi-squared distribution or the necessary independence.

For independent and identically distributed observations with finite positive variance, the central limit theorem, consistency of the sample variance and Slutsky's theorem give $T_n\stackrel{d}{\to }N(0,1)$. Under an independent normal distribution sample it instead has the exact finite-sample Student t-distribution with $n-1$ degrees of freedom, as described by the studentized normal sample mean. The asymptotic statement does not require normal individual observations.

A Student t random-effect model assigns Student's t-distributions to exchangeable study effects, allowing heavier tails than a normal distribution hierarchy. An equivalent Gaussian scale mixture is ${\lambda }_j\sim {\chi }_{\nu }^2$, ${\beta }_j\mid {\lambda }_j,\mu ,\psi \sim N(\mu ,\nu {\psi }^2/{\lambda }_j)$, with independent latent draws. For $\nu >2$ the variance is $\nu {\psi }^2/(\nu -2)$, so $\psi$ is a scale rather than a standard deviation. Small latent precisions weaken shrinkage for atypical studies while retaining partial pooling for the rest.

A real probability distribution is symmetric about $m$ when $X-m$ and $m-X$ have the same distribution. A centered normal distribution is symmetric about zero.