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For one small transverse displacement, the elastic energy of the elastic filament iswhere $A$ is the filament bending modulus. Its first variation isThe clamped boundary conditions remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h^{\prime }(L)$, so the natural boundary conditions for a free endpoint are $h^{\prime \prime }(L)=h^{\prime \prime \prime }(L)=0$: zero bending moment and transverse shear.

There is an important distinction between an energy minimum and a normal mode. The unloaded higher-order Euler-Lagrange equation is $h^{\prime \prime \prime \prime }=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the clamped--free bending modes, extremize the bending Rayleigh quotient, or equivalently $U[W]-\frac{1}{2}A{k}^4\int W^{2}dx$ with a fixed norm. Its Euler-Lagrange equation isThis is the constrained variational characterization of bending modes. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, henceThe free-tip conditions reduce toIts determinant is $-2(1+\cos q\cosh q)$, so nontrivial modes requireNo zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The clamped-free bending spectrum starts with $q_1\in (\pi /2,2)$; bisection or Newton iteration gives $q_1\simeq 1.875104$, so $k_1\simeq 1.875104/L$. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ givesThe second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.

The bending operator with these boundary conditions is positive and self-adjoint. Twice integrating by parts givesSymmetry implies orthogonality when $n\ne m$. Expand $h(x)={\sum }_n{a}_n{W}_n(x)$ and use $I_n=L^{-1}{\int }_0^L{W}_n^{2}dx$. Thenby the equipartition theorem. The supplied endpoint identity therefore yields the thermal bending fluctuations of a clamped filament:To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h^{\prime \prime }(L)=0$ and $-A{h}^{\prime \prime \prime }(L)=f$, and $h^{\prime \prime \prime \prime }=0$ in the interior. Integration gives $h(x)=f{x}^2(3L-x)/(6A)$, so the tip-force compliance of a cantilever is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f{\sum }_n{a}_n{W}_n(L)$ gives $a_n=f{W}_n(L)/(AL{k}_n^4{I}_n)$ and henceComparison evaluates the fourth inverse-power sum of the cantilever spectrum without truncating the modes:The boxed variance is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial ⟨h(L)⟩/\partial f=⟨h(L{)}^2⟩/(k_{B}T)$ at zero load, reproducing the same result from the static compliance. Keeping only the first normal mode gives about $0.32356k_{B}T{L}^3/A$, roughly $97.1\%$ of the exact variance. With two independent equivalent transverse directions, their summed variance is twice the boxed result. The small-slope model requires $k_{B}TL/A\ll 1$, or $L$ small compared with the usual three-dimensional persistence length $A/(k_{B}T)$.

Write $m=⟨E⟩$ and $L_0=i{\partial }_z^2/(2k)$. Substituting $n=1+\mu W$ into part (a), without prematurely replacing $n^2$ by its linear part, givesTaking the expectation yields the exact mean equation within the paraxial approximation:The correlation term cannot be discarded just because $⟨W⟩=0$: the field depends on the same random medium as $W$. Nor can $⟨W^{2}E⟩$ be factored exactly from the given variance alone. The printed statistics do not specify the autocorrelation function of a random field, so they do not determine a unique local closed equation for $m$.

A useful covariance-dependent equation follows to second order in $\mu$. Defineand letbe the Fresnel propagator kernel for one transverse coordinate. The Duhamel principle applied to the linear fluctuation term giveswhere $E^{(0)}=E_{\mathrm{free}}$. Multiply by $W(x,z)$ and average. Also $⟨W(x,z{)}^{2}E(x,z)⟩=E^{(0)}(x,z)+O(\mu )$. Since $m=E^{(0)}+O({\mu }^2)$, replacing $E^{(0)}$ by $m$ in terms already multiplied by ${\mu }^2$ preserves second-order accuracy. We obtain the weak-fluctuation memory equation for the coherent fieldThis perturbation statement is for fixed propagation intervals with sufficient covariance regularity and moment bounds. The kernel is understood as an oscillatory integral. If the field is jointly Gaussian, odd perturbative contributions vanish and the remainder can be improved under the corresponding expansion hypotheses. Joint Gaussianity itself still leaves the covariance unspecified. The term $ik{\mu }^{2}m/2$ is the quadratic refractive-index shift in the coherent field; it comes from the mean of $W^2$.

For the usual local answer, impose the additional Markov approximation for a random medium. Its longitudinal correlation length must be much shorter than the distance on which the envelope varies, and diffraction across that correlation distance must not resolve substantial transverse covariance variation. Away from the entry layer, replace the slowly varying factors in the memory integral by their local values and extend the longitudinal integral to infinity. Define the integrated covarianceFor an integrable real stationary covariance, $C(s,0)$ is even, and $B(0)\ge 0$ is its zero-longitudinal-frequency spectral density. The conventional leading weak-index model keeps only the linear random potential $ik\mu W$; in that model the local first-moment equation isThe factor $1/2$ can be checked independently in the white-noise model. If ${\mathcal{B}}_x(z)$ is a Brownian field with $⟨d{\mathcal{B}}_x(z)d{\mathcal{B}}_x(z^{\prime })⟩=B(z-z^{\prime })dx$, writeConverting the Stratonovich integral to an Itô integral adds $-k^2{\mu }^{2}B(0)Edx/2$. The remaining stochastic integral has zero mean, giving the boxed equation andIf the finite-correlation model retains $n^2$ consistently through order ${\mu }^2$, the local second-order expansion also has $+ik{\mu }^{2}m/2$, adding the phase factor $e^{ik{\mu }^2(x-x_0)/2}$. This phase is computed before the white-noise idealization: squaring ideal Gaussian white noise is not the finite-variance operation $W^2$.

The need for a correlation assumption can be seen without any closure argument. The stationary Gaussian random field $W(x,z)=Z$, with a single standard normal variable $Z$, has all the printed one-point statistics. In the linear weak-index model it givesIts attenuation is quadratic in propagation distance, whereas the Markov model gives linear-distance exponential attenuation. The local attenuation equation requires the additional correlation/Markov assumption; it does not follow from statistical homogeneity and unit variance alone.

For $n\ge 1$, the binomial log-likelihood is $X\log \theta +(n-X)\log (1-\theta )$ plus a constant. Its maximizer, including endpoints, is ${\hat{\theta }}_{\mathrm{MLE}}=X/n$. For $0<\theta <1$, the Fisher information isThis follows either from the expected negative second derivative or the variance of $(X-n\theta )/(\theta (1-\theta ))$. At $0,1$ the usual regular score formula does not apply; the interior information diverges as an endpoint is approached.

For a symmetric Beta distribution prior $\mathrm{Beta}(s,s)$, conjugacy gives $\theta \mid X\sim \mathrm{Beta}(X+s,n-X+s)$, with posterior mean $(X+s)/(n+2s)$. All three priors below are instances of this calculation.

Write $R_j=I-P_{-j}$. Minimizing the residual sum of squares first over the coefficients of $X_{-j}$ reduces the problem to minimizing $\Vert R_{j}Y-b{R}_j{X}_j{\Vert }^2$. Full column rank ensures $R_j{X}_j\ne 0$. Thus the partial regression formula and its variance areThis uses $R_j^T=R_j=R_j^2$ and the isotropic error covariance matrix ${\sigma }^{2}I$.

Since the span of $X_k$ is contained in the column space of $X_{-j}$, orthogonal projection onto the latter removes at least as much squared norm. Consequentlyand inversion proves the requested lower bound. Full rank makes the denominator positive.

Adding $c{1}_n$ to column $j\ge 2$ leaves the column space unchanged. Every old fitted vector is reproduced by keeping all slopes unchanged and replacing the regression intercept by ${\hat{\beta }}_1-c{\hat{\beta }}_j$. Uniqueness of the ordinary least squares estimators then proves that slope $j$ is unchanged. In particular all columns can be centered by such changes. Applying the previous projection argument to the centered design proves the bound with $X_j-{\bar{X}}_j{1}_n$ and $X_k-{\bar{X}}_k{1}_n$; the normalized inner product is now the usual sample correlation.

Each reported coefficient test is the two-sided Student's t-test of $H_0:{\beta }_j=0$ against $H_1:{\beta }_j\ne 0$, conditional on the other predictors. Under the null, ${\hat{\beta }}_j/(\hat{\sigma }\sqrt{(X^{T}X{)}_{jj}^{-1}})$ has a Student's t-distribution with $100-3=97$ degrees of freedom. At five percent the intercept is significant, but neither individual test slope is. The final F-test compares the full model against an intercept-only model: $H_0:{\beta }_{\mathrm{Maths}}={\beta }_{\mathrm{Stats}}=0$. It has $(2,97)$ degrees of freedom and rejects at five percent, since its $p$-value is $0.003166$. Thus the predictors jointly explain variation, without strong evidence separating their individual conditional effects.

The two test predictors have sample correlation $0.9371630$, although each has only moderate correlation with the final exam response. The variance inflation factor is $1/(1-0.9371630^2)\simeq 8.22$. Their nearly shared direction produces substantial multicollinearity, inflating the individual slope standard errors. The joint F-test can detect this common predictive direction even though either coefficient, given the other, has a large $p$-value.

The Chebyshev inequality states that for a random variable $Z$ with finite variance and $a>0$,Write ${\mu }_n=\mathbb{E}[T]$. The ratio in the hypothesis presupposes ${\mu }_n>0$ for all sufficiently large $n$. On the event $T=0$, the deviation from the mean equals ${\mu }_n$. Thus the Chebyshev inequality with $a={\mu }_n$ givesThis is the second moment method. It uses only the stipulated relative variance bound; no assumption that triangle indicators are independent is needed.

Let $X_s$ count the copies of the complete graph $K_s$ in the binomial random graph $G_{n,p}$. By linearity of the expected value and independence of distinct edges,Throughout, $n$ is sufficiently large that $k=\lfloor \log \log n\rfloor \ge 2$. We have $k\to \infty$, $k=O(\log \log n)$ and $k\log k=o(\log n)$.

First, since $p=n^{-2/k}$,The first moment method proves $\mathbb{P}(X_{k+1}>0)=o(1)$.

For the existence of a $k$-clique, put $X=X_k$ and $\mu =\mathbb{E}X$. Since $p^{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}=n^{-(k-1)}$,This alone is insufficient: we must control the variance via the overlap formula for the variance of a clique count. Let $I_S$ indicate that a given $k$-set $S$ spans a clique. If $|S\cap T|=j$, the two copies share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges, soFor $j=0,1$ their edge sets are disjoint and the covariance vanishes, even when they share one vertex. Counting ordered pairs of $k$-sets gives the exact formulaThis includes the diagonal case $j=k$.

To bound the overlap factor, regard it as the probability that a uniformly chosen $k$-set $T$ intersects a fixed $S$ in exactly $j$ vertices. A union bound over the possible specified $j$-subsets of $S$ yieldsHere $(a{)}_j=a(a-1)\cdots (a-j+1)$ is a falling factorial, and the last bound holds for $n\ge 2k$. ThereforeFor $2\le j\le k$, the concave quadratic $j(k-j+1)$ is minimized at an endpoint. Its endpoint values are $2(k-1)$ and $k$, both at least $k$. Thus every power of $n$ in the last sum is at most $n^{-1}$, andbecause the logarithm of its numerator is $O(k\log k)=o(\log n)$. The second moment method, or Chebyshev's inequality directly, gives $\mathbb{P}(X=0)=o(1)$. Finally, every larger clique contains a $(k+1)$-clique, so the absence of $K_{k+1}$ excludes all larger sizes. The clique number is exactly $k$ with high probability:

For an additive arithmetic function $f$, one standard form of the Turán-Kubilius inequality usesIt states, with an absolute constant,For real $f$, this also bounds the variance under the discrete uniform distribution, because centering at the actual expectation minimizes the mean square.

For the prime omega function, $f(p^k)=1$. The Mertens theorem for reciprocal primes states ${\sum }_{p\le N}1/p=\log \log N+O(1)$. The contribution of higher prime powers is bounded by ${\sum }_{p}1/(p(p-1))\le 1$, as proved in 1(a). It follows that $A_{\omega }(N)=\log \log N+O(1)$ and $B_{\omega }(N{)}^2\ll \log \log N$. Writing $L=\log \log N$, the Turán-Kubilius inequality and Chebyshev inequality therefore yieldThe bounded error in $A_{\omega }-L$ is absorbed by, for example, using the threshold $0.005L$ around $A_{\omega }$ for sufficiently large $N$.

For the multiplication table problem, call a pair $(a,b)\in [N{]}^2$ good when $\omega (a),\omega (b)\ge 0.99L$ and $\omega (\gcd (a,b))\le 0.1L$. The preceding Chebyshev inequality excludes only $O(N^2/L)$ pairs by either of the first two conditions. For the greatest common divisor condition, linearity of expectation under uniform sampling of the two coordinates givesMarkov inequality therefore excludes only another $O(N^2/L)$ pairs.

The prime factors of a product form the union of those of its factors. Consequently, for every good pair,On the other hand, apply the proved concentration estimate at $N^2$. Since $\log \log (N^2)=L+\log 2$, for large $N$ every integer $m\le N^2$ with $\omega (m)\ge 1.88L$ is exceptional there. There are only $O(N^2/L)$ such integers. Products of good pairs lie in this exceptional set; products having only bad representations are at most as numerous as the bad pairs. This proves the multiplication-table upper bound, using distinct prime factors alone:

A useful method of moments in probability states that if real random variables $Z_N$ have all moments, $\mathbb{E}{Z}_N^j\to m_j$ for every integer $j\ge 1$, and $(m_j)$ is the moment sequence of a moment-determinate probability distribution, then $Z_N$ converges to that distribution in the sense of convergence in distribution. In particular, convergence to the standard normal distribution follows from limiting odd moments zero and limiting even moments $(2k-1)!!$. The moment-generating function of a standard normal variable, finite in a neighbourhood of zero, ensures moment determinacy.

For the classical Erdős-Kac theorem, put $L=\log \log N$ and choose, for example, $\phi (N)=L^{1/4}$, with a harmless modification for small $N$. Truncate the prime omega function to primes $p\le y=N^{1/\phi (N)}$. The omitted number of distinct prime factors of any $n\le N$ is at most $\phi (N)$, since a product of $r$ omitted primes exceeds $N^{r/\phi (N)}$. Thus this truncation changes the normalized variable by $o(1)$ uniformly. Also, the Mertens theorem for reciprocal primes gives

Part (b) transfers every fixed normalized central moment of the truncated prime omega function to the independent Bernoulli random variables model. Indeed, its unnormalized error is at most $O_j(N^{2j/\phi (N)-1})$, using ${\sum }_{p\le y}1\le y$, and this tends to zero for each fixed $j$. In the independent expansion, any singleton index has zero expectation; the leading contributions are pairings, giving exactly the standard normal distribution moments, while blocks of size at least three are negligible after normalization. The method of moments in probability, followed by the uniformly negligible truncation and centering errors, proves the Erdős-Kac theorem.

The same strategy proves the bounded-prime-weight form in part (a). Choose $\phi (N)\to \infty$ slowly enough that $\phi (N)=o(B(N))$. The omitted weighted contribution is $O(\phi (N))$; its mean is likewise $O(\phi (N))$, and its independent variance is $O(\log \phi (N)+1)=o(B(N{)}^2)$ by the Mertens theorem for reciprocal primes. Hence the truncated model has the same normalization, and part (b) again transfers each fixed central moment.

The overshoot hypothesis as printed cannot hold for every $R>0$. For $R\le 1$, $T_R=0$ and $M_{T_R}=1$; choosing $R<1/\tau$ contradicts $M_{T_R}\le \tau R$. The intended and sufficient assumption is the overshoot bound for $R\ge 1$, on the event $\{T_R<\infty \}$. We use those thresholds below; when $R\le 1$, the requested stopped-time bound is trivial because $T\wedge T_R=0$.

Fix $R\ge 1$ and set $S=T\wedge T_R$. By part (a), the stopped martingale satisfies $0\le M_{S\wedge n}\le \tau R$: before stopping it is below $R$, at $T$ it is zero, and at $T_R$ the assumed overshoot bound applies. Thus the square of this stopped process is integrable. For $\Delta M_n=M_{n+1}-M_n$,The event $\{S>n\}$ is ${\mathcal{F}}_n$-measurable and contained in $\{T>n\}$. Conditional expectation of the linear term is zero, while the variance assumption makes the remaining conditional expectation nonnegative. The stopped increments here are bounded by the stopping and overshoot bounds, so all conditional expectations are legitimate. ConsequentlySince $X_0=1$, we have ${\sigma }^2\mathbb{E}(S\wedge n)\le \mathbb{E}{M}_{S\wedge n}^2-1$. Bounded-time optional stopping theorem also gives $\mathbb{E}{M}_{S\wedge n}=1$, soApply the monotone convergence theorem to $S\wedge n$. This proves the stronger estimate $(\tau R-1)/{\sigma }^2$ and hence the requested absorption-time bound from conditional variance and overshoot:In particular this stopping time is finite almost surely. For $0<R\le 1$, $S=0$ and $X_n=1$ is a constant submartingale, while the stopping-time expectation is zero.

Write $b=\mathrm{esssup}{X}_1$. The hypothesis $\mathbb{P}(X_1>a)>0$ implies $a<b$ and rules out a constant random variable. In particular $I(a)<\infty$. The cumulant-generating function is strictly convex, since its second derivative is the positive variance under exponential tilting, and ${\psi }^{\prime }(0)=\mu$, ${\psi }^{\prime }(\theta )\to b$ as $\theta \to \infty$.

If $a>\mu$, there is a unique ${\theta }_a>0$ with ${\psi }^{\prime }({\theta }_a)=a$, and the Legendre transform gives $I(a)={\theta }_{a}a-\psi ({\theta }_a)$. ThusIf $a=\mu$, use differentiability at zero: $\psi (\theta )=\mu \theta +o(\theta )$, so for some sufficiently small $\theta >0$,In either case choose a parameter $\theta \ge 0$ with $J=\theta (a+\epsilon )-\psi (\theta )>I(a)$, and choose $0<\eta <J-I(a)$. The Cramér theorem gives, for all sufficiently large $n$,The Chernoff bound at the chosen parameter gives $\mathbb{P}(S_n/n\ge a+\epsilon )\le e^{-nJ}$. The conditioning event has positive probability for every $n$, since all summands can simultaneously exceed $a$. ThereforeTaking the complement within the conditioning event proves the conditional concentration at a large-deviation threshold:The farther tail has a strictly larger exponential cost. The argument also works when $a+\epsilon$ reaches or exceeds the upper support endpoint, without requiring an interior maximizer there.

A standard real Brownian motion relative to a filtration $({\mathcal{F}}_t)$ is an adapted stochastic process with $B_0=0$ almost surely, continuous paths almost surely, andEquivalently, it has independent increments and stationary increments, each with a normal distribution of mean zero and variance equal to the elapsed time, together with continuous paths and the stated initial value. The filtration may be its natural filtration. A Brownian motion started at $z$ is $z+W_t$ for a standard process $W$; multidimensional standard Brownian motion has independent standard real coordinate processes. The normalization is variance $t$ at time $t$.

The $2^n$ increments are independent normal random variables, each distributed as $2^{-n/2}{B}_1$. Their squared means are $2^{-n}$ and, using the permitted variance, their squared variances are $2\cdot 2^{-2n}$. Independence therefore givesConsequently the dyadic quadratic variation converges to one:This is dyadic quadratic variation of Brownian motion on the time interval of length one. The calculation requires independence only within each partition; increments from different resolutions need not be independent.

Both Poisson processes start at zero and have càdlàg paths. On disjoint time intervals their increments are independent, and independence of the two processes makes the difference increments independent as well. Their increment laws depend only on interval length, so $X$ has stationary increments. For a time interval of length $h$, the probability that either Poisson process has a jump is $1-e^{-h}\to 0$. This proves stochastic continuity. Consequently the difference is a Lévy process, equivalently a continuous-time symmetric simple random walk with total jump rate one and jumps $+1$ and $-1$ each chosen with probability $1/2$.

The characteristic function of a Poisson variable of mean $\lambda$ is $\exp \{\lambda (e^{i\theta }-1)\}$. Independence therefore givesThus, in the negative-exponent convention of part (a),In the positive-exponent convention the answer is $\tilde{\psi }(\theta )=\cos \theta -1$. The Compound Poisson process has mean zero and variance $t$, consistent with the expansion $\psi (\theta )={\theta }^2/2+O({\theta }^4)$.

Use a common intercept and laboratory-specific slopes in the normal linear model. With $t_i$ denoting days and $I_{Bi},I_{Ci}$ the indicator variables for laboratories B and C, the design matrix has row $x_i^T=(1,t_i,t_i{I}_{Bi},t_i{I}_{Ci})$, andThe interaction terms change slopes, without introducing laboratory-specific intercepts. Thus the estimated daily growth rates arein millimetres per day, with fitted common intercept $0.6464$ mm. The residual degrees of freedom are $30-4=26$. The usual unbiased error variance estimate is the square of the residual standard error:The small difference from $5921.7/26\approx 227.758$ arises from rounding the standard error to two decimal places; the latter uses the more precise residual sum of squares supplied later. A maximum-likelihood estimate would instead divide that residual sum of squares by $30$.

For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0{)}^T$. Its fitted mean isWrite $h_0=x_0^T(X^{T}X{)}^{-1}{x}_0$. The prediction interval in a normal linear model iswhere $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha /2}$ is the indicated Student's t-distribution quantile. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient standard errors alone: the coefficient covariances are also needed.

To prove coverage, write $Y_0=x_0^T\beta +{\epsilon }_0$, with ${\epsilon }_0\sim N(0,{\sigma }^2)$ independent of the original sample. The error $Y_0-x_0^T\hat{\beta }$ has a normal distribution with mean zero and variance ${\sigma }^2(1+h_0)$. By the joint distribution of least-squares and variance estimators, $26s^2/{\sigma }^2\sim {\chi }_{26}^2$ and is independent of $\hat{\beta }$; it is also independent of ${\epsilon }_0$. Hence the pivotal quantityConsequently $\mathrm{Pr}(-t_{26,1-\alpha /2}\le T\le t_{26,1-\alpha /2})=1-\alpha$, which is exactly the claimed prediction interval coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean confidence interval.