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Financial data show approximate, scale-dependent regularities rather than an exact universal scaling law. Heavy return tails can resemble a power law over an intermediate range, but the fitted exponent, cutoff and range depend on instrument, period and sampling. A straight segment on a log-log plot alone is not enough to establish a power law.

For independent finite-variance increments, variance grows linearly with horizon and typical fluctuations scale as $\Delta t^{1/2}$. Central-limit aggregation can make longer-horizon returns more Gaussian even when short-horizon returns are heavy-tailed. Real data can depart from a single self-similar distribution because of volatility clustering, changing regimes, asymmetric responses, intraday seasonality and market microstructure. Apparent different scaling of higher moments may reflect intermittent volatility, but moments estimated from rare tails can also be unstable or may not exist in an idealized model.

Useful tests compare rescaled distributions, structure functions and volatility correlations across several scales, with uncertainty estimates and checks for nonstationarity. Scaling is a valuable approximate modelling guide over a documented range, not proof that one exponent describes every asset and horizon or that historical tail behaviour fixes future extreme losses.

Write the displacement after $m$ independent isotropic zero-mean steps as $\mathbf{R}={\sum }_j{\mathbf{s}}_j$. Cross terms vanish, giving $\boxed{⟨R^2⟩=m\Delta }$ when $\Delta =⟨|\mathbf{s}{|}^2⟩$. Isotropy makes each Cartesian variance one third of this. Since $m\simeq kt$ and $⟨R^2⟩=6Dt$, $\boxed{D=k\Delta /6}$. If “step size” were instead defined as a single-coordinate mean square, the corresponding expression would be $k\Delta /2$; the factor of three is purely the stated three-dimensional convention.

The probability density is even, so its first moment vanishes. Its second moment follows by integrating the polynomial:Using the normalization from part a givesThe variance in position is consequently $a^2/7$.

The moment-generating function of a standard normal distribution gives $\mathbb{E}[e^{t{X}_j}]=e^{t^2/2}$. Using independence of the coordinates,The accepted draws are independent and identically distributed random variables, so their sample mean is an unbiased estimator and has varianceThis counts $n$ accepted draws, requiring $n{\sigma }^d$ proposals in expectation.

An importance weight has expected value ${\mathbb{E}}_q[w(Z)]=\int p(z)dz=1$. From the ratio in (a),The supplied normal distribution transform applies with quadratic coefficient $-(1-s^{-1})$, giving each factor $(2s-1{)}^{-1/2}$. ThusFor $s>1$, the base $A:=s/\sqrt{2s-1}$ exceeds one because $s^2-(2s-1)=(s-1{)}^2>0$: importance weight variance increases exponentially with dimension. For $s=1$, the importance weight is identically one and its variance is zero. This calculation concerns the importance weight, rather than the weighted integrand.

The integrand is positive, so the minimum-variance importance distribution is its target-weighted probability density function, normalized by $\theta =\sqrt{e}$:By completing the square,Thus $q^{\star }$ is the multivariate normal distribution $N(-e_1,I_d)$, where $e_1=(1,0,\dots ,0)$ is the first standard basis vector. Its importance weight and weighted integrand areA variance cannot be negative, and this proposal makes every summand constant, proving optimality directly. It is admissible under the arbitrary-proposal condition even though its mean vector is not zero. Although optimal proposals usually involve an unknown normalizing integral, here both that integral and the resulting normal distribution are explicit.

At equal sample count, rejection sampling eventually has smaller variance for fixed ${\sigma }^2>1$; at equal expected proposal cost, importance sampling eventually wins. The distinction matters because a sample of the rejection sampling estimator means an accepted draw.

For the importance sampling estimator, all coordinates contribute to the importance weight although only one enters the integrand. The same normal distribution transform as in (c), now with linear coefficient $-2$ for coordinate one, givesBoth estimators are unbiased estimators of $\sqrt{e}$, andFor fixed $s>1$, the first variance grows exponentially while the second stays constant. This is the benefit of returning exact target draws in rejection sampling. However, obtaining those draws costs $s^{d/2}$ proposals each on average.

To compare computational work, multiply each variance by its expected proposal count. For importance sampling that count is $n$; for rejection sampling it is $n{s}^{d/2}$. The ratio of the resulting cost-normalized variances isbecause $A/\sqrt{s}=\sqrt{s/(2s-1)}<1$. Thus importance sampling eventually gives better precision for the same expected number of proposals, assuming comparable work per proposal. This is an asymptotic cost comparison, not a guarantee that a fixed small computing budget produces any accepted draws. Both methods still deteriorate exponentially in this comparison. For $s=1$ they coincide and every proposal is accepted.

A practical improvement is to exploit the integrand's one-coordinate structure directly: average $e^{-G_1}$ for standard normal distribution draws $G_1$, without generating unused coordinates. The optimal change of proposal distribution in (iii) improves further.

Let $m=\mathbb{E}X$ and $v=\mathrm{Var}(X)$, evaluated from the given polylogarithm formulas. Generate a large sample of independent and identically distributed random variables with r_sample_f, and compute its sample mean ${\overline{X}}_n$ and sample variance $S_n^2$ using mean and var. The central limit theorem gives the first standardized check:For the second check, the correct asymptotic variance of the sample variance involves the fourth central moment ${\mu }_4=\mathbb{E}[(X-m{)}^4]$:The denominator is not $\sqrt{2v^2}$ unless the probability distribution has the requisite fourth central moment, as does the normal distribution. Here the probability density function decays exponentially, so all required moments are finite.

To compute ${\mu }_4$, expand the denominator of the probability density function as a geometric series and integrate each nonnegative term. The gamma function integral gives the raw momentsThus $v=m_2-m_1^2$ and ${\mu }_4=m_4-4m_1{m}_3+6m_1^2{m}_2-3m_1^4$. The following R code implements both checks for chosen interior parameters and large n:

p <- 0.4
beta <- 1.3
n <- 100000
raw_m <- vapply(1:4, function(j) {
  -factorial(j) * polylog(j + 1, 1 - p) / (beta^j * log(p))
}, numeric(1))
m <- raw_m[1]
v <- raw_m[2] - m^2
mu4 <- raw_m[4] - 4 * m * raw_m[3] +
  6 * m^2 * raw_m[2] - 3 * m^4
x <- r_sample_f(n, p, beta)
c(mean_check = sqrt(n) * (mean(x) - m) / sqrt(v),
  variance_check = sqrt(n) * (var(x) - v) / sqrt(mu4 - v^2))

For each check separately, an absolute value below $1.96$ is consistent with its approximate $95\%$ confidence interval. Repeat with many independent samples: the two standardized statistics should each resemble a standard normal distribution, with about $5\%$ outside that range. Also vary $p$, $\beta$ and n; a systematic discrepancy that grows with $\sqrt{n}$ suggests a bias. The two checks from the same sample are generally dependent, so passing both is not itself a joint $95\%$ statement, nor can a finite test prove that a generator is correct.

The second central limit theorem follows from the central limit theorem for the sample variance: $\sqrt{n}(S_n^2-v)$ differs by a quantity tending to zero in probability from $n^{-1/2}{\sum }_i[(X_i-m{)}^2-v]$.

The central limit theorem gives $\sqrt{n}({\hat{\theta }}_n^{\mathrm{MC}}-\theta )\Rightarrow N(0,1)$, so, with $\Phi$ the cumulative distribution function of the standard normal distribution,The error has two tails. Using the supplied $2.58$ quantile, whose two-sided coverage is approximately $0.9902$, givesThe supplied $2.33$ quantile would be appropriate for a one-sided $99\%$ calculation; it would give only approximately $98.02\%$ two-sided coverage. Using the exact $0.995$ quantile instead of $2.58$ would give $66,349$ draws; we consistently use the paper's rounded value $N=66,564$ below.

For routine Monte Carlo integration, the central limit theorem calculation is preferable as a practical budget: it uses about one fifteenth as many draws as the Chebyshev's inequality bound. However, its coverage is approximate. Finite variance alone supplies no universal rate of convergence to the normal distribution; an integrand with large skewness or a heavy-tailed distribution can make the approximation poor even at a large sample count. If an unconditional finite-sample $99\%$ guarantee is required under only the stated assumptions, use the Chebyshev's inequality budget.

Apply Chebyshev's inequality to the Monte Carlo estimator:Thus $10^4/n\le 0.01$ suffices, givingThis is a finite-sample probability guarantee for every target satisfying the given variance assumption. It is a sufficient sample count supplied by the inequality, rather than a claim that every such target actually needs a million draws.

Generate independent and identically distributed random variables $X_1,\dots ,X_n$ with probability density function $p$, evaluate $\varphi (X_i)$ for each draw, and return the Monte Carlo estimatorLinearity of expectation makes this an unbiased estimator: $\mathbb{E}{\hat{\theta }}_n^{\mathrm{MC}}=\theta$. Independence and the specified variance give $\mathrm{Var}({\hat{\theta }}_n^{\mathrm{MC}})=1/n$. Its Monte Carlo standard error is therefore $n^{-1/2}$, and the strong law of large numbers gives convergence to $\theta$ almost surely.

The definition of $a$ and (c)(i) give $a\le M$. Nonnegativity of the variance strengthens the lower bound toIn fact the variance of one weighted observation is exactlyIt is finite by the envelope. For $v_{\mathrm{IS}}>0$, the central limit theorem gives approximate two-sided coverage $2\Phi (0.01\sqrt{n/v_{\mathrm{IS}}})-1$, so the paper's rounded quantile yieldsIf $v_{\mathrm{IS}}=0$, the weighted observation equals $\theta$ almost surely, and one draw is sufficient. These are theoretical budgets: $a$ and $\theta$ may themselves be unknown and need preliminary estimation in an implementation.

Compare with the average rejection sampling cost $66,564M$. Ignoring integer rounding, importance sampling uses fewer proposal draws at the same approximate precision precisely whenFor literal integer budgets, the comparison is $\lceil 66,564v_{\mathrm{IS}}\rceil <66,564M$. The simpler strict inequality is the intended leading-order condition. The bound $a\le M$ alone does not ensure an improvement when $\theta \ne 0$; this is a proposal-budget variance comparison of importance and rejection sampling.

There is a source error to resolve first. For a drag coefficient defined by $F_{\mathrm{drag}}=-\xi v$, the velocity relaxation time is $\tau =m/\xi$. The mass and drag coefficient in the PDF's exponential are interchanged: $\xi t/m$ is dimensionless, whereas $mt/\xi$ is not. The consistent mean relaxation law is therefore $⟨v(t)⟩=v_0{e}^{-t/\tau }$.

A mean relaxation law alone does not determine a variance. Use the usual thermal Langevin equation, with a bath at temperature $T$ and the fluctuation-dissipation relation for a Langevin particle:Here the calculation is for one component in Cartesian coordinates. With fixed initial position and velocity, integration gives the mean displacementIntegrating the stochastic part of the velocity once more gives the centered displacementThe delta correlation reduces its variance to a single integral:This is the integrated Ornstein-Uhlenbeck displacement for a prescribed initial velocity. Its short-time variance is $2D{t}^3/(3{\tau }^2)+O(t^4)$; at long times it is $2Dt+O(1)$. Thus the diffusion coefficient isIf the requested mean-squared fluctuation instead refers to the conventional equilibrium mean squared displacement, the initial velocity is also averaged over its thermal distribution, with $⟨v_0⟩=0$ and $⟨v_0^2⟩=k_{B}T/m$. Its contribution adds $D\tau (1-e^{-t/\tau }{)}^2$, givingThis has the ballistic limit $(k_{B}T/m)t^2$ and the same diffusive limit $2Dt$. For fixed nonzero $v_0$, the uncentered mean square instead equals the first boxed variance plus $v_0^2{\tau }^2(1-e^{-t/\tau }{)}^2$. In three independent spatial dimensions, the centered variances add, and the long-time result is $⟨|\mathbf{r}-⟨\mathbf{r}⟩{|}^2⟩\sim 6Dt$. Stating the initial-velocity convention is necessary to distinguish these finite-time formulas.

TODO: biomolecular force assays are user-deferred. The following notes cover the generic physical apparatus, calibrated dielectric handles and synthetic attachments only.

In optical tweezers, a tightly focused laser produces a restoring gradient force on a suitable transparent bead. Scattering also transfers momentum along propagation, so a stable three-dimensional trap needs axial as well as lateral restoration. An objective with high numerical aperture, correctly filled entrance pupil, controlled beam polarization and low-aberration sample chamber are practical design elements. Collect transmitted/scattered light with a condenser and measure bead displacement with calibrated imaging or back-focal-plane photodetection. A steerable beam or translation stage sets relative positions; the force signal comes from bead displacement, not commanded stage displacement alone.

Near a stable operating point, use a harmonic optical trap with $U(x)=k{x}^2/2$. For a slowly changing external force, $F_{\mathrm{ext}}=k⟨x⟩$ in that coordinate; the optical restoring force is $-kx$. Calibrate each axis and the range over which its response is linear. Detector sensitivity must also be known: write $x=\beta (s-s_0)$, with measured detector voltage $s$, offset $s_0$ and position sensitivity $\beta$ in $m/V$. A spring constant in arbitrary detector units is not a force calibration.

Three complementary physical calibrations are useful. At thermal equilibrium the equipartition theorem givesThis needs an independently calibrated position scale and a variance corrected for drift, detector noise and exposure averaging. In an overdamped Newtonian medium, the optical trap Langevin calibration obeysUsing a one-sided position power spectral density per hertz,The corner frequency and spectral amplitude test both relaxation and fluctuation size. If drag is known, $k=2\pi \zeta f_c$. The isolated-sphere value $\zeta =6\pi \eta a$ must be corrected near boundaries; otherwise measured height dependence can be mistaken for a changed trap stiffness. An independent steady optical trap drag calibration moves fluid past the trapped sphere: $k⟨x⟩=\zeta u$. Use small positive and negative speeds to verify slope, offset and force range.

One can also determine position sensitivity without assuming the bulk drag. In a low-frequency sinusoidal-stage optical trap calibration, let the known local fluid velocity amplitude be $u_0$ at drive angular frequency $\Omega$. The coherent bead displacement amplitude isIf the corresponding detector amplitude is $A_s$, thenAfter removing the coherent drive line from the thermal record, combine this sensitivity with equipartition theorem to recover $k$, and then infer $\zeta$ from $f_c$. A published calibration combines thermal spectra and stage-driven motion without taking bead drag as an input.^[ref]

Controls determine whether these formulas measure the apparatus rather than its noise. Record blocked-light offsets and independent fixed-bead noise; calibrate length with known motion; keep sampling and electronics bandwidth adequate around $f_c$; and include exposure filtering and aliasing in a spectral fit. motion blur reduces measured variance and therefore makes naive equipartition theorem overestimate stiffness; additive position noise has the opposite bias. Changing exposure time provides a direct diagnostic.^[ref] Check temperature and laser heating, wall separation, bead size and refractive index, alignment and power stability. Repeat after changing power or optical geometry and compare thermal, spectral and drag estimates. For a synthetic force handle, also measure the attachment's compliance and geometry so that bead motion is not automatically identified with its extension.

Only the separable instrument and mechanical tasks are treated here; biological operation and stepping energetics remain deferred.

For generic optical tweezers, a tightly focused laser and objective with high numerical aperture create an intensity gradient that attracts a suitable dielectric bead toward the focus. Momentum transfer also creates a scattering force; stable three-dimensional trapping requires sufficient restoring gradient force. A condenser and photodetector or camera track displacement. Near a stable focus, $F_x=-kx$ defines the stiffness and requires both a position calibration and a force calibration.^[ref]

At thermal equilibrium in a harmonic well, the equipartition trap calibration is $k⟨(x-⟨x⟩{)}^2⟩=k_{B}T$. Alternatively, fitting the overdamped harmonic-trap power spectrum gives $f_c=k/(2\pi \gamma )$, hence $k=2\pi \gamma f_c$. Detector noise, finite sampling and exposure time, wall corrections to drag, and drift must be controlled rather than attributed to stiffness.

For an isolated sphere in a Newtonian fluid, define $\mathrm{Re}={\rho }_{f}v(2a)/\eta$ using its diameter. The supplied viscosity converts to $10^{-3}\mathrm{Pa}\mathrm{s}$. The generic Stokes drag calculation givesHere $\mathrm{Re}=2\times 10^{-6}[{\rho }_f/(10^3\mathrm{kg}{\mathrm{m}}^{-3})]$; this retains the unspecified fluid density as a parameter rather than treating a density as supplied data. Stokes drag assumes an isolated no-slip sphere; nearby surfaces change the coefficient. TODO: the requested assessment of biological motor operation is deferred.

TODO: the biological stepping-energy and step-detection estimate is deferred in its entirety; no biological energy source or motor mechanism has been researched.

Ignoring biological coupling, the mechanical thermal model is the Langevin equation for a harmonic trap,For independent repeats with the specified initial conditions, the ensemble mean $X=⟨x⟩$ obeys the homogeneous damped-oscillator equation. An average over $N$ independent zero-mean force samples reduces its variance by $N$, with a finite detector bandwidth needed to define samples of white noise. High bandwidth alone does not remove random forces; averaging independent records or sufficiently separated samples is the additional operation. A single trajectory retains thermal fluctuations. For stationary overdamped position fluctuations, a long time average over duration $\mathcal{T}\gg {\tau }_k=\gamma /k$ has approximately $\mathrm{Var}(\bar{x})\simeq 2(k_{B}T/k){\tau }_k/\mathcal{T}$.

The initial-value solution of a damped harmonic trap has characteristic roots $r_{\pm }=[-\gamma \pm \sqrt{{\gamma }^2-4mk}]/(2m)$. For distinct roots,For critical damping, ${\gamma }^2=4mk$,For the underdamped case the same real solution can be written $e^{-\gamma t/(2m)}[x_0\cos {\omega }_{d}t+(v_0+\gamma x_0/(2m))\sin {\omega }_{d}t/{\omega }_d]$, with ${\omega }_d^2=k/m-{\gamma }^2/(4m^2)$. These describe the deterministic response or conditional mean, not the entire noisy trajectory.

For overdamped trap relaxation after an inertial boundary layer, let $ϵ=km/{\gamma }^2\ll 1$. The slow and fast roots areAfter $t\gg {\tau }_m$, the fast exponential is negligible and the surviving amplitude is $x_0+(m/\gamma )v_0+O(ϵx_0,ϵm{v}_0/\gamma )$. If also $|(m/\gamma )v_0|\ll |x_0|$, then to leading overdamped orderSmall Reynolds number justifies the linear viscous force; it does not alone prove the separate small inertial-trap parameter $ϵ$. Neglecting $v_0$ requires both rapid momentum relaxation and a negligible slow-mode amplitude shift. If $x_0=0$, a nonzero initial velocity can leave a small slow displacement even after its own fast transient decays. The result concerns generic bead mechanics only.

Write $I_1(\theta )$ for the Fisher information in one observation:Under the stipulated regularity assumptions, the maximum-likelihood estimator is consistent and has asymptotic normality:Equivalently, its large-sample normal approximation has mean $\theta$ and variance $1/[n{I}_1(\theta )]$. The information for the whole random sample is $n{I}_1(\theta )$; keeping the one-observation convention explicit prevents an extra factor of $n$.

Define $R={\sum }_i(Y_i-\overline{Y}{)}^2$ and $s_Y^2=R/(n-1)$, with $n\ge 2$. The unrestricted maximum-likelihood estimates are $\hat{\mu }=\overline{Y}$ and ${\hat{\sigma }}^2=R/n$. Under the null hypothesis, maximizing the normal distribution's likelihood over the remaining variance givesbecause the cross term vanishes when ${\sum }_i(Y_i-\overline{Y})=0$. At either fitted variance, the exponential factor in the likelihood function is $e^{-n/2}$, so it cancels in the ratio:The standardized test statistic isThe sample variance $s_Y^2$ uses divisor $n-1$, whereas the maximum-likelihood estimate uses divisor $n$; this explains the $n-1$ in the final expression. Under a nondegenerate normal distribution, $R>0$ with probability one.

Prefer Student 2's paired analysis, because the observational units are matched pairs: two shoes exposed to the same child's activity. A paired t-test uses $Y_i=A_i-B_i$, so shared child-to-child variation is removed before estimating the standard error. The independent units are the ten pairs, giving $9$ statistical degrees of freedom under the model of independent normally distributed pair differences.

The role of covariance is explicit:With common marginal variance ${\sigma }^2$ and within-pair correlation coefficient $\rho$, the true variance of the average difference is $2{\sigma }^2(1-\rho )/10$. The independent-groups calculation ignores the covariance term and uses $2{\sigma }^2/10$. Positive within-pair association can therefore make the paired standard error much smaller, explaining how the reported p-values can differ without arithmetic mistakes. The preference follows from the paired design, not from selecting the smaller p-value. Marginal normality alone need not imply normally distributed differences; exact paired inference uses the normal-difference assumption, for example justified by a joint multivariate normal distribution.

The equal-variance two-sample Student's t-test pools the within-group sums of squares. If $s_A^2$ and $s_B^2$ are the two unbiased sample variances, the required positive square root isFor ten observations in each group, $S^2=(s_A^2+s_B^2)/2$ and the usual unpaired reference Student's t-distribution has $18$ statistical degrees of freedom, provided the assumed independence between groups is justified.

The sample mean $\overline{X}=T/n$ has mean $\theta$ and variance $\sigma (\theta {)}^2=\theta (1-\theta )/n$. The central limit theorem gives a normal approximation for interior $\theta$ when both expected binomial counts are sufficiently large. Under the null hypothesis, a level-$\alpha$ upper-tail cutoff is approximatelyTo obtain the power function, evaluate this same cutoff under the true statistical parameter $\theta$, standardizing with $\sigma (\theta )$:Here $\Phi$ is the standard normal distribution's cumulative distribution function. The formula treats the binomial threshold as continuous; exact integer cutoffs and boundary randomization are finite-sample refinements.