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The Gaussian free field is a random distribution, so its purported level curve cannot be defined by evaluating its height at every point. A coupling must instead be stated using conditional means and covariances. The SLE4 coupling with a Gaussian free field does exactly this.

Choose the Green-function normalizationA Zero-boundary Gaussian free field $H_0$ is characterized on real compactly supported smooth test functions byEquivalently its energy normalization is the Dirichlet inner product $(2\pi {)}^{-1}\int \nabla f\cdot \nabla gdA$. This fixes the otherwise convention-dependent height constant.

Set $\lambda =\pi /2$ andThe field $H$ has prescribed Dirichlet values $-\lambda$ on the negative half-line and $+\lambda$ on the positive half-line. Subtracting $m_0$ recovers the zero-Dirichlet field $H_0$. It is the shifted field $H$, rather than an unshifted zero-boundary field, whose zero-height interface has ordinary chordal ${\mathrm{SLE}}_4$ law.

Let $\eta$ be chordal ${\mathrm{SLE}}_4$ from $0$ to $\infty$, with Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-W_t)$ and $W_t=2{\beta }_t$. Write $D_t=\mathbb{H}\setminus \eta [0,t]$ and $f_t=g_t-W_t$. The coupling statement is:where, conditionally on the curve, $H_0^{D_t}$ is an independent zero-boundary field in the slit domain. The equality is for restrictions to test functions in that domain, and extends to suitable stopping times. Its covariance is $G_{D_t}(z,w)=G_{\mathbb{H}}(f_t(z),f_t(w))$ by conformal invariance. The two sides of the revealed slit have heights $-\lambda$ and $+\lambda$, in the order specified by their real images under $f_t$. This is the continuum meaning of a Gaussian free field level line.

The fundamental calculations explain why the parameter is four. For $Z_t=f_t(z)$, the Itô formula givesAt $\kappa =4$ the drift vanishes. ThusEach harmonic mean is a bounded martingale before approaching its point. Differentiating the explicit conformally transformed Green function gives the Loewner variation of the Dirichlet Green functionThereforeIn words, the variance learned from the evolving mean exactly equals the covariance lost when the slit is removed. For a Green kernel $cG$ the same calculation gives $\lambda =(\pi /2)\sqrt{c}$; using a different Green normalization changes the height constant, not the SLE parameter.

To construct the coupling, first sample the SLE curve. In the two components to its left and right, sample independent zero-boundary fields and add the corresponding constant heights. Extend these as distributions to obtain the candidate full field. The following martingale identity proves its marginal law rather than merely matching its first two moments.

For a real test function $\phi$, put $M_t=(m_t,\phi )$ andLocalization permits integration of the pointwise identities. They yield $d⟨M{⟩}_t=-d{V}_t$. Henceis a complex local martingale: the $-\frac{1}{2}d{V}_t$ drift cancels the $-\frac{1}{2}d⟨M{⟩}_t$ Itô correction. Since $V_t\ge 0$, $|{\mathcal{Z}}_t|\le 1$, making it a true martingale. At the complete-curve limit, $m_t$ becomes the constant height in each component and the remaining Green kernel becomes that of those components. Thus ${\mathcal{Z}}_{\infty }$ is the conditional characteristic function of the sampled candidate field. Taking expectations givesApplying this to every linear combination of test functions proves the entire Gaussian law, not only its covariance. The conditional version $\mathbb{E}({\mathcal{Z}}_{\infty }\mid {\mathcal{F}}_t)={\mathcal{Z}}_t$ proves the displayed conditional-field statement. Exhaustion by compact test supports justifies passage across the slit and the limiting distributional extensions.

This revealed curve is a local set of a Gaussian free field: conditionally on it, the remaining field is a zero-boundary field plus a specified harmonic function. The spatial Markov property is thereby preserved at random domains, a property not available for arbitrary field-dependent sets. One can strengthen the coupling to a curve measurable from the field. The proof explores compatible interfaces in small subdomains and uses the conditional boundary heights and monotonicity to show that two such interfaces for the same field cannot separate; a countable exhaustion gives uniqueness. This supplies a rigorous replacement for the informal phrase “draw the zero contour.” It does not assert that the distribution has pointwise values.

The coupling is useful in both directions: SLE martingales give exact conditional Gaussian free field data, while the Gaussian Markov structure explains the SLE domain Markov property and its distinguished parameter. With the above normalization, the height jump is $2\lambda =\pi$, and the corresponding interface is chordal SLE4.

Work under the money-market risk-neutral measure, with $r_0\ge 0$. The zero-coupon bond price is $P(0,T)=\mathbb{E}[\exp (-{\int }_0^T{r}_{s}ds)]$. For this CIR model, write $p(\tau ,r)=A(\tau )e^{-B(\tau )r}$. The Feynman-Kac formula givesMatching the constant and linear coefficients of $r$ yieldsFor CIR bond pricing, solving this Riccati equation and integrating the scalar equation for $A$, put $\gamma =\sqrt{b^2+2{\sigma }^2}$ and $D(\tau )=(\gamma +b)(e^{\gamma \tau }-1)+2\gamma$. ThenSubstitution verifies both equations and their initial conditions. The discounted candidate solves the martingale pricing equation; its boundedness on finite horizons, or Feynman-Kac formula, identifies it with the conditional expectation. The answer is $\boxed{P(0,T)=A(T)e^{-B(T)r_0}}$. More generally the same coefficients give $P(t,T)=A(T-t)e^{-B(T-t)r_t}$.

The CIR model mean reverts to $a/b$ and has volatility $\sigma \sqrt{r}$, so it remains nonnegative and its fluctuation size depends on the rate level. For positive initial rate, zero is inaccessible if $2a\ge {\sigma }^2$; below this threshold zero is reachable but the process remains nonnegative. Its stationary law is a gamma distribution, of shape $2a/{\sigma }^2$ and rate $2b/{\sigma }^2$. Explicit affine bond prices and nonnegative rates are useful features. Limitations include exclusion of negative rates, restricted volatility behavior and a single stochastic factor; the time-homogeneous parameter family cannot fit an arbitrary initial term structure exactly.

A futures contract fixes a maturity and underlying quantity, but its quoted delivery price is marked to market: gains and losses from quote changes are settled through the margin account, typically daily. A newly settled position has zero contract value; the quoted futures price is not an upfront purchase price for that position. At maturity the quote equals the spot price. In the continuous-settlement idealization, discounted futures gains must be local martingales under the money-market risk-neutral measure. Thus the quote has zero pricing-measure drift. Assuming the relevant integrability makes it a true martingale, $\boxed{F_{tT}={\mathbb{E}}_Q[S_T\mid {\mathcal{F}}_t]}$.

This is a futures pricing formula. In contrast, an unsettled forward contract has delivery quote ${\mathbb{E}}_Q[D_{tT}{S}_T\mid {\mathcal{F}}_t]/{\mathbb{E}}_Q[D_{tT}\mid {\mathcal{F}}_t]$, where $D_{tT}=e^{-{\int }_t^T{r}_{u}du}$. Stochastic interest rates can make the two quotes differ. Contango means the futures quote is above spot for the maturity considered, while backwardation means it is below spot; futures curves are correspondingly described as rising or falling when comparing maturities.

For the Multivariate Ornstein-Uhlenbeck process, multiplication by the matrix exponential $e^{At}$ gives $d(e^{At}{X}_t)=e^{At}d{B}_t$, henceFor deterministic $X_0=x_0$, the Gaussian distribution has mean $e^{-At}{x}_0$ and covarianceEquivalently $\dot{C}=I-AC-C{A}^{\mathsf{T}}$, $C_0=0$. No symmetry or invertibility of $A$ is required. This covariance can also be evaluated from a single block matrix exponential: if the upper-right block of $\exp (t\left(\begin{array}{cc}-A& I\\ 0& A^{\mathsf{T}}\end{array}\right))$ is $H_t$, then $C_t=H_t{e}^{-A^{\mathsf{T}}t}$. If $X_0$ is random, these assertions hold conditionally on $X_0$; the unconditional law need not be Gaussian. A stationary Gaussian law exists when the eigenvalues of $A$ have positive real parts, but stability is unnecessary for the finite-time formulas.

For $\tau =T-t$, the conditional mean of $X_T$ is $e^{-A\tau }{X}_t$ and its conditional covariance is $C_{\tau }$. The Gaussian exponential-moment formula therefore gives the explicit quoteIt tends to $e^{b\cdot X_t}=S_t$ as $T\downarrow t$, as required.

Yes: different adapted volatilities can give perfectly correlated terminal prices. The counterexample in part ii already establishes this with one nonconstant volatility. Both can be nonconstant as well. Put $h(x)=1+\frac{1}{2}\tanh x$ and takeThe Doléans-Dade exponential has volatility $h(B_t)$, bounded between $1/2$ and $3/2$. The second stock has volatility $h(B_t)S_t/(1+S_t)$, also bounded and nonconstant. Their squared volatility difference integrates toThe Novikov condition and boundedness give true square-integrable martingales, and the Itô isometry gives positive variance for each $t>0$. Yet $S_t^{\prime }=(1+S_t)/2$ gives correlation one. The mechanism is positive affine dependence, not equality of the multiplicative volatility coefficients; it also explains the failure of the literal part-ii premise.

The principle terminal correlation does not identify adapted stock volatility exposes a false printed assertion for the correlation coefficient defined in the introduction. Take $r=0$, $S_0=S_0^{\prime }=1$, andThen $d{S}_t=S_{t}d{B}_t$ andHence ${\sigma }_t=1$ and ${\sigma }_t^{\prime }=S_t/(1+S_t)$ are bounded. Both terminal prices have positive finite variance and are perfectly correlated, because one is a positive affine function of the other. Nevertheless $\boxed{{\int }_0^1({\sigma }_t-{\sigma }_t^{\prime }{)}^{2}dt={\int }_0^1\frac{dt}{(1+S_t{)}^2}>0}$ on every continuous finite stock path. Equal initial prices fix the means, not the affine slope.

Here is the exact general conclusion and the intended qualified proof. Bounded volatilities make these zero-rate stochastic exponentials true square-integrable martingales. Writing $s_0=S_0=S_0^{\prime }$, perfect terminal correlation gives $S_1-s_0=c(S_1^{\prime }-s_0)$ for some $c>0$. Conditional expectation therefore gives $S_u-s_0=c(S_u^{\prime }-s_0)$ for every $0\le u\le 1$. Comparing stochastic integrals, the Itô isometry yieldsThis relation need not imply equality of the volatilities.

The terminal proportionality identifies bounded stock volatility criterion repairs the statement. If one additionally assumes equal terminal variances, then $c=1$ and the whole stock paths coincide, so positivity gives $\sigma ={\sigma }^{\prime }$ almost everywhere and the requested integral is zero. The same repair follows if terminal proportionality $S_1=c{S}_1^{\prime }$ is assumed: equal martingale means force $c=1$, after which the conditional-expectation and Itô isometry argument applies. This proves the intended conclusion under a sufficient additional hypothesis without treating centered correlation as uncentered proportionality.

For the quadratic Ornstein-Uhlenbeck state-price density, apply the Itô's formula under the reference probability measure. Since $f^{\prime }(x)=x$ and $f^{\prime \prime }(x)=1$,The stochastic integral is a true martingale on finite horizons by the finite moments of the Ornstein-Uhlenbeck process. A uniform sufficient condition is $\boxed{\alpha \ge {\sigma }^2/(2a)}$, making the finite-variation drift nonpositive. This is also the condition for nonpositive drift at every state.

Use the positive state-price density $\zeta$ to price a terminal payoff $H$ by ${\Pi }_t(H)=\mathbb{E}[{\zeta }_{T}H\mid {\mathcal{F}}_t]/{\zeta }_t$. The riskless account must cancel the drift of $\zeta$, giving the short rateThis pricing system is consistent with a money-market risk-neutral measure: with $M_t=e^{{\int }_0^t{r}_{u}du}$, the process ${\zeta }_t{M}_t/{\zeta }_0$ is a stochastic exponential whose diffusion coefficient $\sigma X_t/f(X_t)$ is bounded. The Novikov condition makes it a true density martingale. The original Brownian motion is for the reference measure, not automatically for this new pricing measure.

For $X_0=x_0$, the OU second moment is $\mathbb{E}{X}_T^2=x_0^2{e}^{-2\lambda T}+{\sigma }^2(1-e^{-2\lambda T})/(2\lambda )$. Thus the zero-coupon bond price isConditionally at time $t$, replace $T$ by $T-t$ and $x_0$ by $X_t$.

The short rate is nonnegative under the condition above, increases with $X_t^2$, and lies between $\alpha -{\sigma }^2/(2a)$ and $\alpha +2\lambda$. The model is tractable but imposes fixed rate bounds, identical rates for opposite values of $X$, and a restricted one-factor term structure. Its long-maturity yield is $\alpha$. These structural restrictions, including exclusion of negative rates under the supermartingale condition, may be unsuitable for a general interest-rate fit.

For zero drift, the reflection principle gives the known Brownian first-passage time densityTo obtain a general drift, on a fixed finite horizon use the exponential martingale $M_t=e^{c{B}_t-c^{2}t/2}$ as change-of-measure density. The Novikov condition holds for constant $c$, and the Girsanov theorem says that the coordinate process $B$ has drift $c$ under the new measure. Thus the hitting law of $B$ under that measure is the law of $B+ct$ under the original measure. Conditional expectation of $M_T$ at a hitting time bounded by $T$ equals its stopped value, by the martingale optional sampling theorem. At $H_a=t$ this value is $e^{ca-c^{2}t/2}$. The drifted Brownian first-passage density is obtained by multiplying $h_0$, giving $\boxed{h_c(t)=a{e}^{-(a-ct{)}^2/(2t)}/\sqrt{2\pi t^3}}$.

To verify the integrated formula, put $u=(a-cT)/\sqrt{T}$ and $v=(-a-cT)/\sqrt{T}$. The identity $e^{2ac}\varphi (v)=\varphi (u)$ for the standard normal density givesBoth terms vanish as $T\downarrow 0$, so $\boxed{\mathbb{P}(H_a\le T)=\overline{\Phi }((a-cT)/\sqrt{T})+e^{2ac}\Phi ((-a-cT)/\sqrt{T})}$. The distribution has total mass one for $c\ge 0$ and mass $e^{2ac}$ for $c<0$; in the latter case the remaining probability is an atom at infinity.

Use the usual normalization $Z_0=1$, finite deterministic initial wealth $x=X_0\ge 0$, positive state-price density, and nonnegative consumption. The deflated asset prices $ZP$ are local martingales, so predictable stochastic integrability makes $M=\int H\cdot d(ZP)$ a zero-starting local martingale. Integrating part (b) givesA nonnegative local martingale is a supermartingale: localize to martingales, apply Conditional Fatou lemma, and obtain the conditional expectation inequality and integrability. Therefore $N$, and hence $M=N-x$, are supermartingales. In particular,This is supermartingale control of deflated consumption gains. Nonnegative consumption is essential to the lower bound; it is the ordinary meaning of a consumption rate here. Wealth nonnegativity alone would not supply that bound if arbitrary signed cash flows were allowed.

Write $\mu =1+\epsilon$, $p=\mu /n$, $q=1-p$, and let $T$ be the total progeny of a binomial branching process. The preceding binomial branching survival correction gives $\rho =(2+o(1))\epsilon$. For a branching process conditioned on extinction, the offspring probability generating function is $f((1-\rho )s)/(1-\rho )$, with meanfor sufficiently large $n$. Summing expected generation sizes therefore gives $\mathbb{E}[T;T<\infty ]=(1-\rho )/(1-f^{\prime }(1-\rho ))=O(1/\epsilon )$.

Put $K=\lceil {\epsilon }^{-3}\rceil$ and let $N_{\ge K}$ count vertices in graph components of order at least $K$. A breadth-first exploration of a binomial random graph is dominated by $T$. The Markov inequality applied to finite total progeny givesAlways $L_1\le K+N_{\ge K}$. Since $n{\epsilon }^4\ge n^{1/3}\to \infty$, we have $K=o(\epsilon n)$ and $n/(\epsilon K)=O(n{\epsilon }^2)=o(\epsilon n)$. Hence $\mathbb{E}{L}_1\le (2+o(1))\epsilon n$. This bound controls the expected value directly, including rare large graph components.

For the lower bound, use the breadth-first exploration of a binomial random graph with $U_t$ unseen vertices, $A_t$ active vertices and $t$ explored vertices, starting from $U_0=n$, $A_0=0$. Let $b_t=1_{\{A_{t-1}=0\}}$ indicate a new root. Conditional on the past, the newly discovered count $Z_t$ has a binomial distribution $\mathrm{Bin}(U_{t-1}-b_t,p)$. Put $D_t=Z_t-p(U_{t-1}-b_t)$. The unseen-count recursion givesHere $M_t$ is a martingale. Up to $T_0=\lceil 3\epsilon n\rceil$, martingale-difference orthogonality gives $\mathbb{E}{M}_{T_0}^2\le C\epsilon n$, since each conditional variance is at most $np<2$ and $q^{-2T_0}$ is bounded.

Take $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$. Then $h\to 0$, $\epsilon =o(h)$ and $h^{2}n{\epsilon }^3\to \infty$. The Doob L2 maximal inequality impliesThe Taylor expansion, uniformly for $t\le T_0$, gives $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$. Thus $F(t)\ge h{\epsilon }^{2}n/3$ throughout the integer interval $[\lceil h\epsilon n\rceil ,\lfloor (2-h)\epsilon n\rfloor ]$ for large $n$. The new-root sum is nonnegative, so on the preceding event $A_t>0$ throughout this interval. No graph component finishes there: all these explored vertices belong to one graph component, of order at least $(2-2h)\epsilon n-O(1)$, with high probability. Its expected value is therefore at least $(2-o(1))\epsilon n$. Combining both inequalities provesThe barely-supercritical largest-component expectation uses both a positive exploration window and a finite-progeny bound; a bare convergence-in-probability assertion would not by itself justify this expectation.

The Martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.

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.