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A random row vector has a multivariate normal distribution when every real linear combination of its coordinates is normally distributed, allowing a zero-variance constant. Its law is determined by its mean and covariance. For the conditional and testing formulas assume $\Sigma$ is positive definite, so both coordinate variances and the conditional variance are positive; the given ratio already requires ${\sigma }_{XX}>0$.

Put $Z=Y-\beta X$, with $\beta ={\sigma }_{XY}/{\sigma }_{XX}$. The pair $(X,Z)$ is jointly normal and $\mathrm{Cov}(X,Z)={\sigma }_{XY}-\beta {\sigma }_{XX}=0$, so $Z$ is independent of $X$. Its mean is ${\mu }_Y-\beta {\mu }_X$, and its variance is ${\sigma }_{YY}-{\sigma }_{XY}^2/{\sigma }_{XX}$. HenceStandardize each coordinate with its own positive standard deviation. The standardized pairs have mean zero and covariance $\left(\begin{array}{cc}1& \rho \\ \rho & 1\end{array}\right)$. Subtracting means and multiplying coordinates by positive constants leaves the sample correlation unchanged. Therefore the distribution of $r$ depends only on $\rho$.

For $n>2$, the stated statistic reduces algebraically toThis is strictly increasing from $(-1,1)$ onto $\mathbb{R}$, with derivative $\sqrt{n-2}(1-r^2{)}^{-3/2}$. Under $\rho =0$, change variables in the given Student density: $1+t^2/(n-2)=1/(1-r^2)$. HenceThe normalizing integral is $B(1/2,(n-2)/2)$. To test $\rho =0$ against a two-sided alternative at level $\alpha$, compute $t(r)$ and reject if $|t(r)|>t_{n-2,1-\alpha /2}$, equivalently ifA one-sided alternative uses the corresponding one-sided tail. Singular covariance gives $|r|=1$ almost surely when both variances are nonzero, and is outside the nondegenerate density calculation.

Group means are independent random variables with normal distributionsThe resulting univariate sampling distributions areThe same variance calculation holds for every $i\ge 2$, since the difference involves two independent group means. Therefore the standard errors satisfyReplacing $\sigma$ by a common estimated error standard deviation preserves this ratio. Although the individual group means are independent, the contrasts ${\hat{\alpha }}_i$ share the baseline mean and have covariance ${\sigma }^2/J$ for distinct $i\ge 2$.

The analysis of variance provides evidence of a chocolate effect both with day included ($F_{3,18}=3.9665$, p-value $0.02473$) and with day omitted ($F_{3,20}=4.039$, p-value $0.02133$). Because every chocolate–day cell has the same replication, balanced factorial orthogonality separates the two main effects; the chocolate row is meaningful despite being entered first.

The chocolate-only fitted group means are approximately $14.17,11.67,9.33,11.33$ boards for A, B, C, D respectively. Thus A has the largest fitted lecturing speed and C the smallest. Relative to A, the fitted differences are $-2.50,-4.83,-2.83$ boards. The printed individual Student t-tests give strong evidence for the A–C contrast (p-value $0.00245$); B and D versus A have p-values $0.08835$ and $0.05582$, respectively. Those latter contrasts are not significant at 5%, and the output does not test all other pairwise comparisons. Simultaneous claims would require accounting for multiple hypothesis testing.

There is little evidence of a day effect after adjusting for chocolate. A chocolate-only normal linear model is therefore a reasonable simpler summary, with residual standard deviation about $2.42$ boards and explained variation $R^2\simeq 0.377$. This is an association under the additive statistical model; a causal claim would additionally require an appropriate assignment of chocolate and checks of the regression residuals and possible interaction terms.

For subject $i=1,\dots ,100$ and $t_j\in \{0,2,4,6,8,10\}$, the random-intercept linear mixed model isAll subject random effects and measurement errors are mutually independent. The fixed effects ${\beta }_0,{\beta }_1$ describe the population mean baseline and weekly gain. Conditional on $b_i$, a person's observations have independent normal distributions; after integrating out $b_i$, their covariance is ${\tau }^2$ at distinct weeks and their common variance is ${\tau }^2+{\sigma }^2$. Therefore their correlation is ${\tau }^2/({\tau }^2+{\sigma }^2)$.

This Gaussian linear mixed model permits correlated repeated readings while keeping different subjects independent. The fitted standard deviations are $\hat{\tau }=21.67593$ kg and $\hat{\sigma }=14.23496$ kg, giving within-person correlation about $0.699$. All persons still have the same latent weekly slope in this model.

Allow each subject to have a separate baseline and slope through a correlated random-intercept and random-slope model:The subject random effects are independent of all measurement errors, and subjects are independent. The within-person covariance between times $s,t$ is ${\tau }_0^2+(s+t)\rho {\tau }_0{\tau }_1+st{\tau }_1^2$, with an additional ${\sigma }^2$ when the same reading is used twice.

Prefer the model with a random slope. The maximum likelihood estimation refits improve twice the log-likelihood by $227.9604$ while adding two covariance statistical parameters. Both the Akaike information criterion ($5165.779$ to $4941.819$) and Bayesian information criterion ($5183.367$ to $4968.200$) strongly favour it. The printed nominal likelihood-ratio test is overwhelming. The usual reference chi-squared distribution with two statistical degrees of freedom is not an exact regular calibration, since zero slope variance is a boundary and the intercept–slope correlation is then unidentified; a design-specific parametric bootstrap could calibrate it. This qualification does not undermine the substantial descriptive improvement shown by both information criteria.

The preferred fit estimates population mean baseline strength ${\hat{\beta }}_0=61.09286$ kg and weekly gain ${\hat{\beta }}_1=2.83143$ kg/week. The between-person baseline standard deviation is ${\hat{\tau }}_0=15.785582$ kg, and the between-person slope standard deviation is ${\hat{\tau }}_1=2.738666$ kg/week. Their estimated correlation $\hat{\rho }=0.117$ is weakly positive, giving random-effect covariance about $5.058$ kg${}^2$/week; its uncertainty is not supplied. The measurement-error standard deviation is $\hat{\sigma }=9.923280$ kg. The printed $-0.038$ instead describes the correlation between the estimated fixed effects, not between the subject random effects. These statistical parameter estimates come from restricted maximum likelihood; the model comparison uses the separate maximum likelihood estimation fits.

The population mean fitted trajectory is $61.09286+2.83143t$ kg. HenceUsing the reported slope standard error, an approximate 95% confidence interval for the population mean gain is $28.3143\pm 1.96(2.984464)=(22.46,34.16)$ kg; a Student t confidence interval with 499 statistical degrees of freedom is almost identical.

Successful individuals can improve far more than the average. Their latent ten-week gains have fitted normal distributionA clear estimate for an upper-performing group uses a between-person slope quantile: the 95th percentile is $28.3143+1.645(27.38666)\simeq 73.4$ kg, and the 97.5th percentile is about $82.0$ kg. The upper 5% of fitted underlying gains begin around 73 kg. These are person-to-person performance quantiles, not confidence intervals for the population mean. Identifying the best observed trainee would require that person's data or fitted subject random effects; the aggregate output cannot identify a literal maximum. If performance means the observed difference of endpoint readings, add the measurement-error variance $2{\sigma }^2$ to the latent gain variance.

At zero a centered normal distribution with standard deviation $\sigma$ has density $1/(\sqrt{2\pi }\sigma )$. The two prior standard deviations are $1/\sqrt{n_0}$ and $c/\sqrt{n_0}$, givingThe wider prior distribution has a lower density at its center.

Under the point hypothesis $\beta =0$, the test statistic $\sqrt{n}\overline{Y}$ has a standard normal distribution. The observed statistic is three, giving a two-sided p-value $2[1-\Phi (3)]\simeq 0.00270$. This is conventionally strong evidence against that point hypothesis. The narrow model $H_0$ is a continuous prior distribution around zero, rather than a point hypothesis.

For $n_0/n=100$ and $c=100$, normal-normal conjugacy givesThe narrow-model posterior mean is strongly pulled toward zero and its standard deviation is about $0.0995/\sqrt{n}$. The wide-model posterior mean is about $2.9703/\sqrt{n}$, very close to the observation, with standard deviation about $0.9950/\sqrt{n}$. Each model produces a markedly different posterior, so choosing between them requires their predictive evidence.

Use a proper normal distribution prior centered at zero on the log odds ratio. For example,puts approximately 95 percent of its mass between $-2$ and $2$, giving odds ratios roughly between $e^{-2}$ and $e^2$, close to $1/8$ and $8$. Centering at zero treats reciprocal odds ratios symmetrically. If the desired central 95 percent interval is exactly $(1/8,8)$, use standard deviation $\log 8/{\Phi }^{-1}(0.975)\simeq 1.0610$. A soft prior is appropriate for implausibility, whereas a bounded uniform prior would declare effects outside the limits impossible.

Conditional on $\mu ,\psi >0$, let $\lambda \sim {\chi }_4^2$ and independently $Z\sim N(0,1)$. The normal distribution with the stated precision parameter can be generated asThereforeby the defining normal distribution and chi-squared distribution representation of Student's t-distribution. Its density isThus $\mu$ is the location and $\psi$ is the scale, not the standard deviation: $\mathrm{Var}({\beta }_j\mid \mu ,\psi )=2{\psi }^2$. Each study gets its own independent chi-squared draw in this Student t random-effect model.

Fit both the normal and Student t random-effect model with comparable proper prior distributions. Compare priors on the same spread measure: a normal distribution scale $\tau$ is a standard deviation, whereas the $t_4$ standard deviation is $\sqrt{2}\psi$.

Use a posterior predictive check: draw study effects and binomial counts from each fitted hierarchy and compare replicated dispersion and extreme study contrasts with the observations. For predicting a new study, generate a new effect from the hierarchy rather than reusing an existing fitted effect. A Leave-one-out cross-validation with entire studies held out can compare integrated predictive probabilities for both arms of each omitted trial, averaging over hyperparameters and its unobserved study effect. Leave-one-study-out influence analysis also reveals whether the difference is driven by a single trial.

Prefer the heavier-tailed hierarchy if it improves the relevant predictive checks and held-out study predictions robustly to reasonable prior choices. The deviance information criterion can supplement the comparison, but its effective parameter count can depend on the latent-variable representation, and six studies give limited information about tail shape. A small numerical criterion difference alone is insufficient evidence.

Let $k=m/2$. After offset subtraction, each summed image has mean $N$ analogue-to-digital units per detector pixel. For a constant detector conversion gain $g$ and independent Poisson distribution detected counts, $\mathrm{Var}(A)=\mathrm{Var}(B)=N/g$: summing $k$ independent frames adds both their means and variances. Here $N$ is the mean of the sum, not one frame.

Write $A=N+a$, $B=N+b$. At high counts, the delta method gives $A/B\simeq 1+a/N-b/N$. Independence therefore givesThus the formula is a high-signal gain estimation from flat-field ratios, not an exact identity for a ratio of random variables. A Poisson distribution denominator can even be zero at low counts. The exposures must be linear and unsaturated, offsets removed, and read noise negligible. Being close to saturation increases the signal but does not establish linearity; clipping or charge-induced correlations biases a variance estimate. A large ensemble of suitable detector pixels supplies the measured standard deviation. In normal photodetector terminology $g$ is collected electrons per ADU, consistent with detected photons here only under the stated unit-yield convention; see the instrument gain definition.

If spatial illumination is stable between the two sums, it acts just like the sensitivity factor: the mean ratio remains one but each detector pixel has a different shot-noise variance. A whole-image standard deviation is disproportionately influenced by faint detector pixels, so substituting a global arithmetic mean signal generally underestimates detector conversion gain. If the illumination pattern differs between the sums, their noise-free ratio is not constant; its structure adds apparent variance and causes a further systematic error.

Use locally uniform signal bins or subregions, remove offsets, mask defective or saturated detector pixels, and propagate the local shot and read variances. In the stable-pattern case a useful alternative to pooling raw ratios is the normalized differenceAt high signal and negligible read noise, treating the denominator at its mean gives $\mathrm{Var}(U_i)\simeq (2N_i/g)/(2N_i)=1/g$, independent of the illumination. Thus a variance fit or $g\simeq 1/\mathrm{Var}(U)$ can use a spatially varying flat without the arithmetic/harmonic mismatch. Measured noisy denominators cause higher-order corrections, so independent or smoothed signal estimates can be preferable. Fit and subtract a genuine changing illumination pattern before noise analysis, using sufficient smoothing not to fit away the random noise. Arbitrary flat-field normalization without propagating its noise does not by itself recover the original gain.

Equal variances make every weight $1/N$, so the inverse-variance weighted mean reduces to the arithmetic mean. Independence givesThus the variance scales as $N^{-1}$, whereas the uncertainty expressed as a standard deviation scales as $N^{-1/2}$.

Independence makes the variances of the attendance indicators additive. The binomial distribution therefore gives $\mathrm{Var}(N)=np(1-p)=24(3/4)(1/4)=9/2$. Its positive square root is the standard deviation:

Repeated integration by parts gives, for every nonnegative integer $j$,With $A={\lambda }^2$, the first two moments are $\mathbb{E}[X]={\lambda }^2(2!/{\lambda }^3)=2/\lambda$ and $\mathbb{E}[X^2]={\lambda }^2(3!/{\lambda }^4)=6/{\lambda }^2$. Subtracting the square of the mean givesIf $X$ carries units, $\lambda$ has inverse units of $X$, so the mean and standard deviation have the units of $X$ and the variance has their square.

The second moment isSubtracting the squared expected value gives the variance and standard deviation:The variance is nonnegative because $0<r\le R$. Both $\mu$ and $\sigma$ have units of time. At $r=R$ the result reduces to $R/\sqrt{12}$, the standard deviation of a uniform distribution over one full cycle.

The interval within one standard deviation is $[3/5,9/5]$. Of the possible integer outcomes $0,1,2$, only $1$ lies in this interval. Hence

Compute the first two moments from the probability distribution:The variance is $E[X^2]-E[X{]}^2=9/5-36/25=9/25$, so the expected value and standard deviation are

Write $N={\sum }_{j=1}^{10}{X}_j$, where the independent Bernoulli random variables indicate rottenness. Since $X_j^2=X_j$, their variances are $E[X_j^2]-E[X_j{]}^2=p-p^2$. The independence makes all cross covariances zero, soThis standard deviation vanishes at $p=0$ and $p=1$, when the count is certain.

The expected value isThe second moment is $\mathbb{E}[X^2]=12(1/5-1/6)=2/5$. Hence the variance and standard deviation are