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Use the independent increments of Brownian motion, rather than merely checking that an Itô formula drift vanishes. For $0\le s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of ${\mathcal{F}}_s$ and has normal distribution $N(0,h)$. Its first four moments are $0,h,0,3h^2$. ThereforeFor the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha (t)$, so $\alpha (t)=-3t$ makes it $\alpha (s)$. For the quartic expression, choose $\beta (t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomeswhich equals $3s^2$ when $\gamma (t)=3t^2$. Thus a standard choice isThe resulting stochastic processes are Hermite polynomial martingales $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable martingales, since Gaussian moments are finite at every finite time and the displayed conditional identities establish the martingale property directly.

The choice is not unique. Constants $c,d,e\in \mathbb{R}$ give the valid family $\alpha (t)=c-3t$, $\beta (t)=d-6t$, and $\gamma (t)=3t^2-dt+e$: these add $c{B}_t$ to the cubic martingale and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.

Sample Brownian motion at integer times and put $Z_n=B_n/\sqrt{n}$. Every $Z_n$ has normal distribution $N(0,1)$, although the $Z_n$ are correlated. LetFor every fixed integer $m$,because $B_m/\sqrt{n}\to 0$ almost surely. The right-hand expression depends only on the future independent increments $B_j-B_{j-1}$ with $j>m$. Thus $\{L\ge a\}$ is, up to a null set, a tail event of those independent increments, and its probability is zero or one by the Kolmogorov zero-one law.

For any finite real $a$, every $Z_n$ exceeds $a$ with the same positive probability $p_a=\mathbb{P}(N(0,1)\ge a)$. For each $N$,Taking the decreasing intersection over $N$ shows that $Z_n\ge a$ infinitely often with probability at least $p_a$. This event implies $L\ge a$, so $\mathbb{P}(L\ge a)>0$. The zero-one law makes it one. Intersecting over positive integers $a$ gives $L=+\infty$ almost surely. The continuous-time limit superior is at least the one along integers, henceThis proves that Brownian fluctuations exceed the square-root scale without needing the lower bound in the law of the iterated logarithm.

For the construction above, independence of the normal random variables makes $S_N(h)$ normal with mean zero and variance $v_N={\sum }_{j\le N}⟨h,e_j{⟩}^2$. Its characteristic function is $\exp (-{\theta }^2{v}_N/2)$. The $L^2$ convergence gives $L^1$ convergence, andSince $v_N\to \Vert h{\Vert }^2$ by the Parseval identity for a Hilbertian basis, the limiting characteristic function identifiesThis includes $h=0$, where the normal distribution is degenerate at zero.

For any $r_1,\dots ,r_n\in \mathbb{R}$, the linearity from part (a) givesalmost surely, and the right side has a normal distribution. This is the defining linear-combination criterion for a multivariate normal distribution; singular covariance matrices are allowed.

Passing to the limit in the inner products of the partial sums givesThe passage to the limit is justified by Cauchy-Schwarz inequality and $L^2$ convergence. Equivalently, the vector's characteristic function isThus both joint normality and the complete covariance matrix follow from the Hilbert-space inner product.

The density $Z=\exp (X(k)-\Vert k{\Vert }^2/2)$ is strictly positive. The Gaussian moment-generating function gives $\mathbb{E}Z=1$, so it defines an equivalent probability measure.

The joint normal distribution of $(X(h),X(k))$, with covariance $⟨h,k⟩$, gives the mixed exponential formulaMultiplying by the normalizing and centring factors therefore yieldsThe characteristic function identifies the answer:This is exponential tilting of an isonormal Gaussian process: the mean shifts by the inner product while its covariance remains unchanged.

Fix $T>0$. The given normal distribution and part (d) implyOne can deduce determinism without any moment assumption on the bracket. Put $Z=e^{-⟨X{⟩}_T}$. Taking $\lambda =1,2$ gives $\mathbb{E}Z=e^{-T}$ and $\mathbb{E}{Z}^2=e^{-2T}$, so $\mathrm{Var}(Z)=0$. Therefore $⟨X{⟩}_T=T$ almost surely. Applying this at every rational time and using continuity of quadratic variation gives $⟨X{⟩}_t=t$ simultaneously for all $t\ge 0$ outside a single null set.

The Lévy characterization of Brownian motion states that a continuous local martingale starting at zero with this bracket is Brownian motion in its filtration. To see the independent-increment conclusion directly, the Itô formula shows that $e^{i\theta X_t+{\theta }^{2}t/2}$ is a martingale on any fixed bounded time interval: it is a local martingale with a deterministic bound on its modulus. ThusThe deterministic conditional characteristic function identifies an $N(0,t-s)$ increment independent of ${\mathcal{F}}_s$. Together with the given path continuity and $X_0=0$, this proves $X$ is Brownian motion.

Use the normal linear modelHere $Y$ is the response random vector, $X$ is the known design matrix, $\beta \in {\mathbb{R}}^p$ contains the unknown regression coefficients, and ${\sigma }^2>0$ is the common error variance. Conditional on $X$, the errors have normal distributions and are independent random variables. Require $\mathrm{rank}X=p\le n$ for identifiability of $\beta$ and invertibility of $X^{T}X$. Usually $n>p$ is needed to estimate the error variance from the regression residuals. An intercept, when included, is represented by a column of ones in $X$.

Because both vectors are linear transformations of the same multivariate normal response, $(\hat{Y},e)$ is jointly multivariate normal. Its cross-covariance matrix isZero cross-covariance implies independence for jointly multivariate normal vectors, including singular ones. Therefore the fitted values and the entire vector of regression residuals are independent. The fitted-residual orthogonality identity gives the zero covariance; the normal distribution assumption is what upgrades it to independence.

The simple linear regression assumes a straight conditional mean, $E(Y_i\mid x_i)={\beta }_0+{\beta }_1{x}_i$, so that errors are centred at zero throughout the predictor range. The displayed regression residuals are predominantly positive at both ends and negative in the middle. This is a residual curvature diagnostic: the fitted straight line misses a curved conditional mean. The main concern is the shape of the mean function. The plot alone does not establish failure of the normal distribution assumption or a particular error variance model.

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 $i\ge 2$.

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.

The Poisson trick introduces ${\lambda }_i=\log {\mu }_{i1}$. With an exactly flat log-rate prior, the baseline prior measure is $d{u}_i/u_i$. For $n_i>0$,Thus flat log-rate marginalization gives the multinomial likelihood:Alternatively, independent Poisson counts conditional on their total are exactly multinomial distributions with probabilities $e^{{\eta }_{ik}}/S_i$. Locally uniform coefficient priors make the coefficient posterior proportional to this likelihood within their flat region.

The printed BUGS prior is a proper normal distribution on ${\lambda }_i$, with variance $s^2=10^5$, since BUGS uses precision as its second argument. It is broad but not exactly flat. Substitution $v=S_i{u}_i$ shows that, apart from a coefficient-independent constant, the integrated likelihood is the multinomial kernel times the Gaussian log-rate correction to the Poisson trickIt depends on the coefficients through $S_i$. Dominated convergence gives $R_i\to 1$ as $s^2\to \infty$, so the finite-variance code yields an approximation to the multinomial posterior, not exact equality. Large log rates can make prior sensitivity relevant.

The cell-wise factors form a standard Poisson regression, convenient for BUGS. They avoid an explicitly constrained count vector and permit scalar log-concave updates for intercepts and coefficients; an exactly flat-log-rate implementation also gives gamma conditional baseline rates. This can be computationally efficient, although the actual log-normal baseline prior is not gamma-conjugate and introduces $I$ nuisance intercepts. Efficiency depends on the update scheme.

For any $s\ge 0$, on the event $Y\ge 0$ one has $e^{sY}\ge 1$. The Markov inequality therefore gives $P(Y\ge 0)\le \mathbb{E}{e}^{sY}$, including the trivial value one at $s=0$. Taking the infimum proves the Chernoff bound.

For $s>0$ with finite moment-generating functions, independence givesConsequentlyAn effective bandwidth is an exponential-moment measure of demand at a chosen tail parameter $s$: independence makes these quantities additive. The spare capacity $\gamma /s$ pays for the desired exponential tail bound. For a cumulative-demand process over time $t$, the corresponding bandwidth would be $\log \mathbb{E}{e}^{sX(t)}/(st)$; here the time horizon is one. It is generally larger than mean demand because it charges for fluctuations, and mathematical optimization over $s$ selects the useful tradeoff.

For independent normal distributions, putThe Gaussian effective bandwidth is ${\alpha }_j(s)={\lambda }_j+s{\sigma }_j^2/2$. For $\gamma >0$ and $v>0$, the sufficient condition becomes $m+sv/2+\gamma /s\le C$. Its left side is minimized at $s_{\ast }=(2\gamma /v{)}^{1/2}$, givingThis is a sufficient Chernoff safety margin, not the exact normal tail quantile.

Indeed $X\sim N(m,v)$, so, writing $\Phi$ for the standard normal distribution function,The exact Gaussian chance constraint is thereforeThis is necessary and sufficient when $v>0$, even when $\varphi$ is negative. The Chernoff coefficient $\sqrt{2\gamma }$ is more conservative.

There is an important boundary qualification. If $v=0$, then $X=m$ deterministically, and for $\gamma >0$ the exact requirement is $C>m$, not $C\ge m$. For example, $X\equiv 0$, $C=0$, $\gamma =1$ satisfies the printed square-root condition but has $P(X\ge C)=1>e^{-1}$. Thus both Gaussian non-strict displayed forms require positive total variance, which follows if at least one flow is present and its variance is positive. With no positive variance, the deterministic boundary in an upper-tail chance constraint must be treated separately. If $\gamma \le 0$, the target upper bound is at least one and imposes no restriction; the displayed square-root discussion naturally assumes $\gamma >0$.

The independent, spatially homogeneous random-placement idealization gives a Poisson process of intersections along the sampling line. With line intensity $\mu$, an interval of length $s$ contains no ridge with probability $e^{-\mu s}$. The spacing $X$ thus has exponential distributionRandom orientation changes the intersection intensity, which has already been represented by the measured $\mu$. Merely saying that positions are random does not prove a Poisson process: independence and homogeneity are additional assumptions. Finite segments, clustering or excluded widths can invalidate them.

For a shifted lognormal distribution, write $Z=\log (X-\theta )\sim N(m,{\sigma }^2)$ with $\sigma >0$. Applying the change of variables formula, $dZ/dX=1/(X-\theta )$, givesThe threshold $\theta$ is a minimum observable separation. Finite keel widths impose geometric exclusion, and the sonar's footprint and ridge-identification criterion can suppress a shallow peak near a deeper peak. This sonar ridge shadowing or resolution effect means that $\theta$ need not be a universal physical distance between every pair of ridges; it can partly reflect how the profiles were processed.

A lognormal distribution is compatible with products of many positive factors: taking logarithms turns multiplicative changes into sums, which can approach a normal distribution. Repeated deformation, breakup, convergence and merging across scales are plausible contributors. The observed form therefore suggests correlated or multistage ridging rather than the simplest independent-intersection model. It does not identify a unique ridging mechanism. For example, directly generating $X=\theta +\exp Z$ with a normal distribution for $Z$ produces the same spacing law without specifying any particular mechanics. Distributional agreement must be supplemented by dynamical and spatial evidence.

For $\lambda \ge 0$, a Poisson distribution has mass $\mathbb{P}(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,\dots$; at $\lambda =0$ it is concentrated at zero. Its moment-generating function isFor independent $X,Y$, the moment-generating function of their sum is the product, $\exp [(\lambda +\mu )(e^s-1)]$. Uniqueness of the moment-generating function near zero therefore gives $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$. Repeating the argument gives $S_n={\sum }_{i=1}^n{X}_i\sim \mathrm{Poisson}(n)$ when every parameter is one.

The displayed sequence is $a_n=\mathbb{P}(S_n\le n)$. The classical central limit theorem states that independent identically distributed variables with finite mean $m$ and nonzero finite variance ${\sigma }^2$ have standardized sum $(S_n-nm)/(\sigma \sqrt{n})$ converging in distribution to the standard normal distribution. Here $m={\sigma }^2=1$, soConvergence of the distribution functions at the continuity point zero justifies the last step. This is the Poisson cumulative probability at its growing mean.

A finite right derivative at zero would make the difference quotients $B_{1/n}/(1/n)$ eventually bounded. Therefore that event is contained inFor fixed $K,N$, its probability is at most $\mathbb{P}(|B_{1/n}|\le K/n)$ for each $n\ge N$. By the normal distribution of the Brownian increment,Thus every event in this countable union has probability zero. With probability one, Brownian motion has no finite right derivative at zero, which is the appropriate derivative for its time domain. No independence of the quotients is assumed or needed. This proves the required endpoint case of nowhere differentiability of Brownian motion.

The same affine transformation of the normal distribution givesIts variance stays one while its mean moves positively. This is the exact alternative distribution used to calculate statistical power. The square, if used instead, has a noncentral chi-squared distribution with one degree of freedom and noncentrality $n({\delta }^{\ast }{)}^2/{\sigma }^2$.

For the normal linear model, the log-likelihood isThe full column rank of the design matrix makes $X^{T}X$ invertible. Differentiating in $\beta$ gives the normal equations $X^{T}X\hat{\beta }=X^{T}Y$. With the minimized residual sum of squares substituted, differentiation in ${\sigma }^2$ givesThese are the maximum-likelihood estimators almost surely; $RSS>0$ almost surely when ${\sigma }^2>0$ and $n>p$. The linear transformation of the multivariate normal distribution yieldsDefine the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. The fitted values are $\hat{Y}=HY=X\hat{\beta }$, the regression residuals are $e=Y-\hat{Y}=(I-H)Y$, and $RSS=e^{T}e$. The orthogonal projection $I-H$ has rank $n-p$ and annihilates $X\beta$. By Cochran's theorem,Consequently $E({\hat{\sigma }}^2)=(n-p){\sigma }^2/n$, with bias $-p{\sigma }^2/n$. The unbiased estimator isIts square root, the reported residual standard error, estimates $\sigma$; the assertion of unbiasedness applies to the variance, not generally to its square root.

For the paper-strength analysis, let $h_i$ be the measured percentage and $z_i=h_i-\overline{h}$. Both normal linear models use independent errors of common variance ${\sigma }^2$. Their mean functions are ${\beta }_0+{\beta }_1{z}_i$ for lm1, and ${\beta }_0+{\beta }_1{z}_i+{\beta }_2{z}_i^2$ for lm2, with separately fitted coefficients. The reported residual standard errors are $11.82$ and $4.42$, respectively.

For lm1, the estimated conditional expectation at a new percentage $x$ isThe original data mean $\overline{h}$ must be used for centering the new percentage; its numerical value is not supplied in the excerpt. Because ${\sum }_i{z}_i=0$, the two columns of this design matrix are orthogonal, and $\mathrm{Cov}({\hat{\beta }}_0,{\hat{\beta }}_1)=0$. Hence the coefficient standard errors in the output giveThis estimates uncertainty in the mean. Predicting an individual future batch would additionally require the new-error variance, estimated by $11.82^2$.

To compare the nested normal linear models, test $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$ within the quadratic model. The Student t-test statistic isIts two-sided p-value is $1.89\times 10^{-8}$. Equivalently $T^2$ is a partial F-test statistic with null distribution $F_{1,16}$. Reject the linear restriction and prefer lm2 at the 5% level. Its residual spread is much smaller; the increase in the coefficient of determination from $0.3054$ to $0.9085$ supports the same conclusion, although the test is the relevant complexity-adjusted comparison.

For regression diagnostics, inspect regression residuals against fitted means and hardwood percentage for omitted curvature, a scale-location plot for nonconstant variance, and a quantile-quantile plot against the normal distribution for departures from the error assumption. Check unusual observations using regression leverage and Cook's distance, and examine residuals in collection or batch order if dependence is plausible. Independence and a common variance require substantive justification as well as these plots; a small p-value for a polynomial term does not itself check the error model.