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Write $\mu =\mathbb{E}[X]$ and $A=\{X>\theta \mu \}$. The finite second moment implies a finite first moment, and nonnegativity with $\mathbb{E}[X^2]>0$ implies $\mu >0$. On $A^c$, one has $0\le X\le \theta \mu$. ThereforeSubtract this from $\mathbb{E}[X]=\mu$ to obtain the truncated first moment bound:The indicator function uses the strict event specified in the question; any mass at the threshold belongs to $A^c$.

The Itô formula verifies the geometric Brownian motion solutionThis is valid for negative $x$ as well as positive $x$; $x=0$ gives the zero solution. Pathwise uniqueness follows from the Lipschitz continuity of the linear coefficients. The moment-generating function of a normal distribution gives $\mathbb{E}{e}^{k\sigma B_t}=e^{k^2{\sigma }^{2}t/2}$, and thereforeThese moments are finite on every finite horizon. If the convention $0\in \mathbb{N}$ is used, the zeroth moment is $1$.

Let $A=⟨M⟩$ be the quadratic variation and stop at $T_b=\inf \{u\ge 0:A_u>b\}$. Its continuity and monotonicity imply $A_{u\wedge T_b}\le b$ and $A_{T_b}=b$ whenever $T_b<\infty$. The standard martingale isometry after stopping by bounded quadratic variation gives $\mathbb{E}(M_{t\wedge T_b}^2)=\mathbb{E}{A}_{t\wedge T_b}\le b$; in particular the stopped local martingale is square-integrable. This criterion follows by localization and the Doob L2 maximal inequality, which bound the second moment of the stopped supremum by $4b$ and then permit removal of localization.

On $\{A_t\le b\}$ one has $T_b\ge t$, so the original and stopped paths agree through time $t$. Therefore the Doob L2 maximal inequality and the Markov inequality giveThus

The space-time Hermite polynomials obtained by scaling the Probabilists' Hermite polynomials are genuine polynomials in both variables, including at $t=0$:This expansion follows by differentiating the defining Gaussian exponential, and shows that $H_n(x,0)=x^n$ extends smoothly through $t=0$. Consequently the Itô formula is applicable from time zero. The given backward heat equation cancels the drift, and the derivative identity givesThe integrand is continuous and adapted, hence previsible, and stopping the Brownian motion and time makes it bounded. Thus $H_n(B_t,t)$ is a local martingale for every $n\ge 0$. The $n=0$ case is the constant $H_0=1$; in fact Gaussian moments make the displayed Itô integral square-integrable on every finite horizon, so these are true martingales as well.

The first three Probabilists' Hermite polynomials are $h_1(x)=x$, $h_2(x)=x^2-1$ and $h_3(x)=x^3-3x$. Hence

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

Let $S={\sum }_i{Y}_i$ and $V_H={\sum }_i(b_i-a_i{)}^2$. To prove the Hoeffding lemma, take a centered random variable $Y\in [a,b]$ and its cumulant-generating function $\psi (t)=\log \mathbb{E}{e}^{tY}$. Under exponential tilting, ${\psi }^{\prime \prime }(t)$ is the variance of $Y$. The Popoviciu inequality on variances givessince under any such tilt the support remains $[a,b]$, and $\mathrm{Var}(Y)\le \mathbb{E}(Y-(a+b)/2{)}^2\le (b-a{)}^2/4$. Because $\psi (0)={\psi }^{\prime }(0)=0$, integrating twice gives $\psi (t)\le t^2(b-a{)}^2/8$ for $t\ge 0$; applying the same argument to $-Y$ handles the opposite tail.

The independence of the $Y_i$ gives $\mathbb{E}{e}^{tS}\le \exp (t^2{V}_H/8)$. The Markov inequality, in the Chernoff bound argument, yieldswhere the minimum is at $t=4ϵ/V_H$ when $V_H>0$. Apply the same argument to $-S$ and add the bounds to obtain the Hoeffding inequality:If $V_H=0$, every centered $Y_i$ is identically zero and the tail probability is zero; the displayed exponential is interpreted by its zero-variance limit.

For the Bennett inequality, write $v=n{\sigma }^2$, and first assume $M>0$ and $v>0$. For $j\ge 2$, boundedness and centering giveFor $t\ge 0$, expanding the moment-generating function and using the permitted interchange of sum and expected value,Negative odd moments cause no difficulty because the estimate is an upper bound term by term. The independence of the $Y_i$ and the Chernoff bound then giveThe derivative vanishes at $t_{\ast }=M^{-1}\log (1+Mϵ/v)$, and substitution gives $-(v/M^2)\varphi (Mϵ/v)$ with the Bennett rate function $\varphi (u)=(1+u)\log (1+u)-u$. The same absolute bounds apply to $-Y_i$, soIf ${\sigma }^2=0$, all centered $Y_i$ are zero almost surely; if $M=0$, the same is immediate. Handle those cases directly rather than divide by zero. Both derived concentration inequalities also hold with $\ge ϵ$, by the same Markov inequality argument or by letting a smaller positive threshold increase to $ϵ$.

For the binomial distribution, write $X={\sum }_{i=1}^n{Z}_i$ with independent and identically distributed random variables $Z_i$ having the Bernoulli distribution with parameter $p=p_n$. Then $Y_i=Z_i-p$ has range $[-p,1-p]$, width one, and variance $p(1-p)$. For large $n$, $p<1/2$, so $M=1-p$ and $v=np(1-p)$. Set $ϵ=C\sqrt{v}$.

The Hoeffding inequality givesAfter truncation at one, this is asymptotically the trivial probability bound. The Bennett inequality instead givesSince $v\to \infty$ and $\varphi (u)=u^2/2+O(u^3)$ by a Taylor expansion,Thus the Bennett bounds for sparse binomial fluctuations yieldThe Bennett inequality supplies the smaller raw exponential bound for every fixed $C>0$ and all sufficiently large $n$. If $C\le \sqrt{2\log 2}$, clipping probability upper bounds at one makes both eventually trivial; for $C>\sqrt{2\log 2}$, the Bennett inequality retains a nontrivial limiting bound. This distinction avoids claiming a probability improvement after clipping when both raw bounds exceed one.

Taking expectations in the single-step discrete Tanaka formula, the martingale transform term vanishes because $f$ is bounded and ${\mathcal{F}}_T$-measurable. HenceAt a state with $p_K=\mathbb{P}(S_T=K)>0$, the tower property of conditional expectation gives $\mathbb{E}[S_{T+1}\mid S_T=K]=K$, so the conditional second moment of the increment is ${\sigma }^2(T,K)$. Part (a) gives $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)$. Combining these identities yields the discrete Dupire equationThe conditional variance needs a positive-probability conditioning state. The bound $|K-S_0|\le T$ does not guarantee $p_K>0$. If $p_K=0$, both the call-price curvature and the time increment above are zero; the equation can be extended by assigning an arbitrary finite value to ${\sigma }^2(T,K)$ at such a state, but its conditional variance is not determined there.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

With constant coefficients, $\lambda =(\mu -r)/\sigma$ is constant, so the stochastic exponential $Z_t=e^{rt}{Y}_t$ is a true martingale. Under its equivalent martingale measure $\mathbb{Q}$, the stock follows the Black-Scholes model $d{S}_t=S_t(rdt+\sigma d{W}_t^{\mathbb{Q}})$. The Gaussian exponential moment gives the value of the power optionThe option delta is ${\partial }_{s}V=V/(2s)$, so the replicating strategy holdsHalf the wealth value is in the stock and half in the bank account, with continuous rebalancing. To check the self-financing portfolio property under the original measure, the Itô formula gives $d{V}_t=\frac{1}{2}(r+\mu )V_{t}dt+\frac{1}{2}\sigma V_{t}d{W}_t={\theta }_{t}d{S}_t+{\eta }_{t}d{B}_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the Black-Scholes model has the necessary finite Gaussian moments, and the explicit strategy attains $V_0=\mathbb{E}[Y_T\sqrt{S_T}]$. The supermartingale bound from part (b) therefore proves its minimality. The cost is independent of the physical drift $\mu$.

Assume $m^2>0$ and take $\lambda \to 0^{+}$. The free Gaussian integral gives $Z_0=\sqrt{2\pi }/|m|$, and the normalized free Gaussian measure has covariance $C=m^{-2}$. Expanding only the interaction exponential gives the asymptotic expansionIn this zero-dimensional scalar field theory, the path integral has become an ordinary integral, but its combinatorics are precisely those of sextic scalar field theory.

The free generating functional is $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus the Feynman rules are a quantum field theory propagator $C$ for each internal line and a factor $-\lambda$ for each six-valent interaction vertex. The factorials $V!$ and $(6!{)}^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda {)}^V{C}^I/S$, where $S$ is the Feynman-diagram symmetry factor, the order of the automorphism group of the resulting diagram, including its half-edge symmetries. A tadpole diagram has an additional interchange symmetry of the two ends of its loop.

There is one empty Vacuum Feynman diagram, contributing $1$. At one interaction vertex, the six half-edges form three loops. Its Feynman-diagram symmetry factor is $2^{3}3!=48$, so its contribution is $-\lambda C^3/48$.

At two interaction vertices let $r$ count the lines joining them. Each interaction vertex has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all Vacuum Feynman diagrams at this order. Their Feynman-diagram symmetry factors areHere the leading $2$ exchanges the two interaction vertices, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a connected generating functional would discard it.

Consequently,Indeed ${\sum }_r{S}_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!{)}^2]$ from Wick theorem.

This is an asymptotic expansion, rather than a convergent perturbation series: the coefficient of ${\lambda }^V$ contains $(6V-1)!!/[V!(6!{)}^V]$ and grows too rapidly for a nonzero radius of convergence. For any fixed truncation order, the Taylor expansion remainder bound for $e^{-x}$ on $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

Use the imaginary-axis contour of the inverse transform of the moment-generating function, $J=iy$, so the Gaussian part gives the normal distributionBecause $J^3{e}^{-J\delta }=-{\partial }_{\delta }^3{e}^{-J\delta }$, the first non-Gaussian term isDifferentiating the Gaussian three times givesHence the Edgeworth expansion isEquivalently its correction is $(S_3\sigma /6){\mathrm{He}}_3(\delta /\sigma )$, with ${\mathrm{He}}_3(z)=z^3-3z$, a Probabilists' Hermite polynomial. It integrates to zero and reproduces the third moment $S_3{\sigma }^4$. The expansion is intended for weak non-Gaussianity in the bulk, rather than the far tails.

The linear density contrast is Gaussian, so its cubic moment vanishes. The leading contribution isTwo connected Wick contractions giveThe third pairing vanishes because $F_{2,s}(\mathbf{q},-\mathbf{q})=0$. Isotropy makes $⟨\mu ⟩=0$ and $⟨{\mu }^2⟩=1/3$, so the angular average of the kernel isSince ${\sigma }^2=D^2\int d^{3}q{P}_{\mathrm{lin}}(q)/(2\pi {)}^3$ to leading order, the unsmoothed matter-density reduced skewness isThis assumes finite unsmoothed variance; inserting a smoothing window generally changes $S_3$.

The Kantorovich–Rubinstein theorem concerns a Polish space $(X,\rho )$ and probability measures $\mu ,\nu$ with finite first moments, meaning $\int \rho (x,x_0)d\mu (x)+\int \rho (x,x_0)d\nu (x)<\infty$ for one, hence every, $x_0\in X$. It statesHere $\mathrm{Lip}(f)\le 1$ means $|f(x)-f(y)|\le \rho (x,y)$ for every $x,y$. The Wasserstein distance with ground cost $\rho$ therefore equals a supremum over functions with Lipschitz constant at most one. One may normalize $f(x_0)=0$, since adding a constant does not change the difference of integrals. This normalization bounds $|f(x)|$ by $\rho (x,x_0)$, so finite first moments ensure integrability. The supremum is unchanged if an absolute value is placed around the difference, because $-f$ is admissible whenever $f$ is.

Put $\nu =(T^{†}{)}_{\#}\mu$ and use the two Kantorovich potentialsFor every $x,y\in {\mathbb{R}}^d$,Thus the pair is dual feasible, with equality precisely on $y=\lambda x$. Finite second moments make both potentials integrable; compactness of $X$ is more than sufficient. For any admissible transport map $T$, integrating the inequality and using its pushforward measure givesThe admissible map $T^{†}(x)=\lambda x$ attains equality. ConsequentlyThe same certificate proves optimality of its graph transport plan among all transport plans. Notice that $u$ need not be a convex function when $\lambda >1$: it is a cost dual potential. The associated convex gradient potential is instead $\Phi (x)=\lambda |x{|}^2/2$.

For probability measures $\mu ,\nu$ on ${\mathbb{R}}^d$ with finite second moments, the Knott–Smith optimality criterion states that a transport plan $\pi \in \Pi (\mu ,\nu )$ minimizes the quadratic cost $|x-y{|}^2$ if and only if there is a proper convex function $\Phi :{\mathbb{R}}^d\to \mathbb{R}\cup \{+\infty \}$ that is sequentially lower semicontinuous and satisfiesThe subdifferential is characterized bywith $\Phi (x)<\infty$. Thus the transport plan is concentrated on the graph of the subdifferential. No absolute continuity of measures assumption on $\mu$ is needed. Multiplying the cost by $1/2$ leaves the criterion unchanged. The original quadratic optimal mapping result is Knott and Smith, On the optimal mapping of distributions.

A standard sufficient form of Brenier theorem assumes $\mu ,\nu \in \mathcal{P}({\mathbb{R}}^d)$ have finite second moments and $\mu \ll {\mathcal{L}}^d$, that is, absolute continuity of measures with respect to Lebesgue measure. For the cost $|x-y{|}^2$, there exists a unique optimal transport plan, and it is induced by a transport map:Here $\Phi$ is a proper convex function, which may be chosen sequentially lower semicontinuous, and its gradient exists $\mu$-almost everywhere. The map $\nabla \Phi$ is unique $\mu$-almost everywhere and is the unique minimizer of the Monge optimal transport problem; its cost equals the Kantorovich optimal transport problem minimum. Equivalently, it is the unique gradient of a convex function transporting $\mu$ to $\nu$.

The uniqueness claim concerns the map and the transport plan, not a globally unique potential. The potential may be shifted by a constant, and additional nonuniqueness away from the source can occur. No density assumption is required on $\nu$. A primary reference is Brenier's Polar factorization and monotone rearrangement of vector-valued functions.

Let $\Phi ={\phi }_{\epsilon }$ and let ${\Phi }^{\ast }$ denote its convex conjugate. Its Fenchel–Young gapis nonnegative by the Fenchel–Young inequality, and the hypothesis says $\int Hd{\pi }_{\epsilon }\le \epsilon$.

First justify the integrability needed for the certificate. Finite second moments and the Cauchy-Schwarz inequality imply, for every transport plan $\pi$,Since $\Phi \in L^1(\mu )$ and $H\in L^1({\pi }_{\epsilon })$, the identity ${\Phi }^{\ast }(y)=H(x,y)-\Phi (x)+x\cdot y$ proves ${\Phi }^{\ast }\in L^1(\nu )$ by the marginal distribution property. In particular, no subtraction of infinite integrals is being used.

SetThese are integrable Kantorovich potentials. The Fenchel–Young inequality gives $u(x)+v(y)\le \frac{1}{2}|x-y{|}^2$. More precisely, for every transport plan $\pi$,Taking the infimum in the first inequality therefore provesThe factor $1/2$ in the quadratic cost is essential for this exact gap identity.

The conditional mean estimator is the posterior meanprovided the first moment is finite. A maximum a posteriori estimate is any maximizer of the posterior density relative to Lebesgue measure:Existence of the posterior density alone does not guarantee a finite posterior mean or existence of a maximizing point.

For the stated multivariate normal distribution and independent standard Gaussian noise, the negative log of the posterior density, up to a constant, isIts precision matrix $A^{T}A+{\Sigma }^{-1}$ is a positive-definite matrix, so Gaussian conjugacy for a normal linear model givesThe two forms agree becauseThe covariance matrix is $(A^{T}A+{\Sigma }^{-1}{)}^{-1}$, and the maximum a posteriori estimate equals the posterior mean for this multivariate normal distribution.

Choose the common dominating probability measure $\nu =(\mu +{\mu }^{\prime })/2$ and let $p=d\mu /d\nu$, $q=d{\mu }^{\prime }/d\nu$ be the Radon-Nikodym derivatives. The assumed second moments imply first-moment integrability by the Cauchy-Schwarz inequality. Since $Y$ is a separable Banach space, the measurable $f$ is a strongly measurable function, and both expected values exist as Bochner integrals.

The difference of these Bochner integrals satisfiesThe first inequality is the norm bound for a Bochner integral, and the second is the scalar Cauchy-Schwarz inequality. Since $(\sqrt{p}+\sqrt{q}{)}^2\le 2(p+q)$, the first factor is at mostWith the Hellinger distance convention of the original PDF, the second factor is $\sqrt{2}{d}_{\mathrm{Hell}}(\mu ,{\mu }^{\prime })$. Multiplication proves the Hellinger bound for differences of expectations:The proof uses only the two laws and their common dominating measure; no absolute continuity of measures assumption between $\mu$ and ${\mu }^{\prime }$ is needed.

For any observable $f$, the Markov jump-process generator identity givesSet $m(t)=⟨x_2⟩$ and $s(t)=⟨x_2^2⟩$. With ${\alpha }_7=0$, reaction 2 changes $x_2$ by $+1$ and reaction 3 changes it by $-1$. Thereforeand, using $(x_2+1{)}^2-x_2^2=2x_2+1$ and $(x_2-1{)}^2-x_2^2=-2x_2+1$,The equation for the second moment contains the third, whose equation contains the fourth, and so on. This is an infinite moment hierarchy.