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Use $\hslash =c=1$. For a canonical light scalar field with effective mass $m^2\ll H^2$, define the canonical Fourier mode $u_k=a\delta {\Phi }_k$. In the nearly de Sitter approximation it obeysDeep inside the Hubble radius, $k\gg aH$, the mode behaves like a Minkowski oscillator. Choosing the Bunch-Davies vacuum sets $u_k\to e^{-ik\eta }/\sqrt{2k}$. The normalized de Sitter solution isInflation stretches the physical wavelength until it exits the Hubble radius. For $|k\eta |\ll 1$, the growing scalar solution freezes with $|\delta {\Phi }_k{|}^2=|u_k/a{|}^2\to H^2/(2k^3)$. The dimensionless power spectrum is thereforeHere $H_k$ is evaluated when $k=aH$, since slow evolution near horizon exit fixes the frozen amplitude. Nearly constant $H$ and a light field make this spectrum almost scale-invariant; slow background evolution and small mass corrections produce a small tilt. The stretched vacuum modes supply the primordial stochastic fluctuations.

Expanding the covariant derivative isolates the ghost-gluon vertex interactionThe derivative acts on the antighost. With all momenta incoming, let its momentum be $k$, the gluon's Lorentz index be $\mu$, and their color indices be $(a,b,c)$ in the order $(\bar{c},A,c)$. The convention $e^{-ikx}$ replaces ${\partial }^{\mu }$ by $-i{k}^{\mu }$, and multiplication by $i$ from $e^{iS}$ givesMomentum conservation is implicit. The structure constants reflect the Adjoint representation, and the antighost momentum follows directly from its differentiated field. Reversing the momentum convention changes the displayed vertex sign consistently; exchanging the ghost and antighost source order also carries a fermionic sign.

The scaling hypothesis for critical phenomena says that the singular equilibrium free-energy density is a generalized homogeneous function of the thermal and field variables, after analytic backgrounds are removed. To derive the ordinary mean-field form, put $r=r_{t}t$, $u>0$ and neglect the higher powers near the transition. Rescale $m=\sqrt{r_t/u}|t{|}^{1/2}\psi$. Minimizing the quartic potential becomesThis is the mean-field scalar free-energy scaling form with $a=r_t^2/u$, $b=\sqrt{u}/r_t^{3/2}$. The two functions correspond to $t>0$ and $t<0$; the latter has two pure-phase branches and a cusp at zero field. The question's $>$ and $<$ subscripts distinguish these temperatures above and below $T_c$.

Differentiating with respect to $h$ gives $m=-{\partial }_h{A}_s\propto |t{|}^{1/2}$ on an ordered pure branch, and differentiating again gives $\chi =-{\partial }_h^2{A}_s\propto |t{|}^{-1}$. Thus the displayed scaling form independently yields $\beta =1/2$ and $\gamma =1$.

To allow anomalous powers, replace it byThen $\beta =2-\alpha -\Delta$ and $\gamma =2\Delta -(2-\alpha )$. Taking $t\to 0$ at fixed small $h$ gives $A_s\propto |h{|}^{(2-\alpha )/\Delta }$, so $1/\delta =(2-\alpha -\Delta )/\Delta =\beta /\Delta$. Eliminating $\Delta$ proves the Rushbrooke scaling relation and Widom scaling relation:These power-law relations assume the simple scaling form; marginal corrections or dangerously irrelevant couplings can require additional scaling variables.

The PDF gives an inverse-gamma distribution with shape $k$ and scale $\theta$, and a gamma distribution prior with shape $\alpha$ and rate $\lambda$. Write $\mu (\theta )=\theta /(k-1)$. The supplied conditional moments, together with $\mathbb{E}\theta =\alpha /\lambda$ and $\mathbb{E}{\theta }^2=\alpha (\alpha +1)/{\lambda }^2$, give the three structural quantities of the Bühlmann model:Here $v$ is the variance of hypothetical means and $s^2$ the expected process variance. The law of total covariance and conditional independence of different years implyAll these quantities are finite because $k>2$.

For any coefficients $c_i$, minimizing the mean squared error over the intercept first gives $c_0=m(1-{\sum }_i{c}_i)$. Let $Y=\mu (\theta )-m$ and $W_i=X_i-m$. Expanding the remaining mean squared error givesIts normal equations are $s^2{c}_i+v{\sum }_j{c}_j=v$ for every $i$. Since $s^2>0$, subtracting any two equations shows that all $c_i$ agree. The quadratic part is strictly positive for every nonzero coefficient change, so the solution is the unique global minimizer. Solving givesA credibility estimate combines individual experience with the collective expected value, using an experience weight $Z\in [0,1]$. In this equal-exposure setting, the Bühlmann credibility premium isThus the computed coefficients give exactly a credibility estimate, with a fixed credibility factor determined by known parameters and the number of observed years. This is Bühlmann credibility for inverse-gamma observations.

For the final Bayesian calculation, the likelihood function in $\theta$ is proportional to ${\theta }^{nk}\exp (-\theta {\sum }_i{x}_i^{-1})$. Multiplication by the gamma distribution prior gives the Gamma prior for an inverse-gamma scale updateUnder quadratic loss, the conditional risk of an estimate $d$ is $\mathrm{Var}(\mu (\theta )\mid x)+(d-\mathbb{E}[\mu (\theta )\mid x]{)}^2$. Its minimizer is therefore the posterior mean:This is not a fixed-weight arithmetic-mean credibility estimate. For $n=2$, the observations $(1,3)$ and $(2,2)$ have the same sample mean, but reciprocal sums $4/3$ and $1$, hence different posterior means. For any $n>2$, append the same positive observations to these two samples. For $n=1$, the estimate is $Cx/(1+\lambda x)$, with $C=(\alpha +k)/(k-1)$; its second derivative is $-2C\lambda /(1+\lambda x{)}^3\ne 0$, so it is not affine either. Symmetry in the observations would force equal slopes in any affine representation, so these examples also exclude a general fixed-coefficient linear representation.

One can express the posterior mean as a broader data-dependent blend. With $A={\sum }_i{x}_i^{-1}$ and harmonic mean $H=n/A$, it isThis weights a harmonic-mean-based experience estimate, and the weight itself depends on the observations. It is not the credibility estimate based on $\overline{X}$ and the fixed Bühlmann credibility factor derived above.

Let $C_t$ be the cumulative claim amount and $U_t=u+ct-C_t$ the surplus in the classical risk model. The supplied adjustment coefficient satisfies $\lambda (M_X(R)-1)=cR$. The Compound Poisson process has independent increments, and the random-sum transform identity givesIt follows that $Z_t=\exp (R(C_t-ct))$ is a nonnegative martingale with $Z_0=1$: conditional on the past, the multiplicative increment has expected value one. This is the exponential surplus martingale.

Let $\tau =\inf \{t\ge 0:U_t<0\}$ be the ruin stopping time. Surplus decreases only at claim arrivals, so on $\{\tau \le t\}$ the deficit at ruin gives $C_{\tau }-c\tau >u$ and $Z_{\tau }>e^{Ru}$. The optional stopping theorem at the bounded stopping time $\tau \wedge t$ yieldsAs $t$ increases without bound, the events $\{\tau \le t\}$ increase to eventual ruin. Hence the ultimate ruin probability satisfies the Lundberg inequalityThe proof uses only bounded-time stopping; it does not presume that the martingale can be stopped with equality of expected values at an unbounded ruin time.

Condition first on the count in the random sum of independent claims. For $N=k$, independence of the summands and independence from $N$ giveThe empty product is one when $k=0$. Taking expected values proves the random-sum transform identity:For positive claim sizes this composition is always finite for $u\le 0$, and is an identity of Laplace transforms of nonnegative random variables after writing $u=-s$. For positive $u$, it holds wherever the composed series is finite; no positive exponential moment is implied merely by positivity of the summands.

The payoff $e(x,z)$ is the expected fitness of an individual using strategy $x$ against a population using strategy $z$. The resident $x^{\ast }$ is initially common; the mutant $y$ is introduced at small frequency $ϵ$. Convexity of the strategy set makes their population mixture admissible. If residents have higher expected fitness in every sufficiently small invasion by a fixed mutant, selection works against that mutant: this is an evolutionarily stable strategy. The allowed upper bound on $ϵ$ may depend on $y$.

For expected payoffs of mixed strategies, $e(x,z)$ is affine in the population strategy $z$; in a finite matrix game it is $x^{T}Az$. This property is the basis of the two-condition criterion for evolutionary stability. For fixed $y\ne x^{\ast }$, putAffineness gives the exact invasion differenceIf this is positive for all sufficiently small $ϵ>0$, taking $ϵ\downarrow 0$ gives $A_y\ge 0$. If $A_y=0$, the difference is $ϵB_y$, so necessarily $B_y>0$. Conversely, if $A_y>0$, it remains positive for sufficiently small $ϵ$: for example take $ϵ<A_y/(A_y+|B_y|)$. If $A_y=0$ and $B_y>0$, it is positive for every $ϵ>0$. Applying this argument separately to every $y$ provesThus an evolutionarily stable strategy is a symmetric Nash equilibrium with an additional condition against tied mutants. For a general nonlinear interaction function on a convex set, affineness would be an additional necessary hypothesis for this particular equivalence; here it follows from the mixed-strategy expected-payoff interpretation. For example, on $X=[0,1]$ take $e(x,z)=xz(x/2-z)$ and resident $x^{\ast }=0$. For every $y>0$, $A_y=0$ and $B_y=y^3/2>0$, but at population $ϵy$ the mutant earns $ϵy^3(1/2-ϵ)>0$ while the resident earns zero. Thus the two conditions alone would not suffice for an arbitrary nonlinear $e$; the expected-payoff assumption is used explicitly.

For the Hawk-Dove game, let $p$ be an individual's Hawk probability and $q$ the population's Hawk probability. The given matrix yieldsWe use this formula in each case. In the usual interpretation, resource value $V$ and damage cost $D$ are positive. The parts below also specify the boundary or negative-value alternatives because positivity is not written explicitly in the PDF.

A coalitional game with transferable utility has a finite player set $N$ and a characteristic function of a coalitional game $v:2^N\to \mathbb{R}$, normalized by $v(\varnothing )=0$. The value $v(S)$ is the total payoff the coalition $S$ can secure and redistribute among its members.

An imputation is an efficient, individually rational payoff vector:The core of a cooperative game consists of imputations satisfying every coalition constraint:A coalition assigned less than its characteristic value can leave and make every member better off, by sharing the positive surplus. Conversely a coalition with no such surplus cannot improve every member's payoff simultaneously. This explains the stability represented by the core of a cooperative game.

Use the optimization Lagrangian with the sensitivity sign conventionFor equality constraints the Lagrange multipliers have unrestricted signs. The Lagrangian sufficiency theorem says that if $h(x^{\ast })=b$ and, for some ${\lambda }^{\ast }$, $x^{\ast }$ globally minimizes $L_b(\cdot ,{\lambda }^{\ast })$ over $X$, then $x^{\ast }$ globally minimizes $f$ subject to the constraint. Indeed every feasible $x$ satisfiesThis proves the theorem without a convexity or differentiability assumption. Merely finding a stationary point of the optimization Lagrangian would not establish its required global minimum.

Let $\varphi (u)=\inf \{f(x):x\in X,h(x)=u\}$, with value $+\infty$ when the feasible set is empty, and suppose $\varphi (b)$ is finite. The problem has the Strong Lagrangian property if a finite multiplier $\lambda$ satisfiesThis includes attainment of the dual bound by a multiplier. It does not by itself assert attainment of the primal infimum. For every multiplier, restricting the infimum to feasible $x$ gives ${\inf }_X{L}_b(x,\lambda )\le \varphi (b)$, the relevant weak duality. A non-vertical supporting hyperplane of a value function is a planewith $\varphi (u)\ge \varphi (b)+{\lambda }^T(u-b)$ for every $u$. Its vertical coefficient is nonzero and has been normalized to one; the epigraph lies above it. The two directions of the equivalence are proved in parts (a) and (b).

For the sampling allocation put $A={\sum }_i\sqrt{a_i{v}_i}$. Finite variance requires every $x_i>0$; interpret the objective at a zero coordinate as $+\infty$. An optimum uses all the budget, because scaling every positive $x_i$ up would otherwise reduce the objective while remaining feasible. The Cauchy-Schwarz inequality gives the global boundEquality requires $v_i/x_i$ to be proportional to $a_i{x}_i$. The budget then determines the unique allocation:This is the optimal cost-constrained stratified sampling allocation.

For an inequality ${\sum }_i{a}_i{x}_i\le b$, our sign convention uses $\lambda \le 0$, so $L_b=f-\lambda ({\sum }_i{a}_i{x}_i-b)$ is a lower bound on $f$ at feasible points. Its coordinate stationary equations are $-v_i/x_i^2-\lambda a_i=0$, and the displayed allocation givesThe function $v_i/x_i+(-{\lambda }^{\ast })a_i{x}_i$ has a strict global minimum at $x_i^{\ast }$: its second derivative is $2v_i/x_i^3>0$ and its value diverges at both endpoints of $(0,\infty )$. Thus the multiplier also supplies the Lagrangian sufficiency theorem certificate, with the tight inequality giving complementary slackness.

Finally, for $b+\delta b>0$,Hence the first-order change in minimal variance is $\boxed{{\lambda }^{\ast }\delta b}$. Increased resources decrease the variance. With the alternative convention $L=f+\mu ({\sum }_i{a}_i{x}_i-b)$, the multiplier is $\mu =-{\lambda }^{\ast }>0$ and the sensitivity is $-\mu \delta b$.

For the distinct observed event times $a_k$, let $i_k=\pi (a_k)$ and let $R_k$ be the risk set immediately before that event. The Cox partial likelihood isThe different risk sets already incorporate each event and any intervening censoring. Differentiating this log-likelihood givesTherefore a finite interior maximum-likelihood estimate from the Cox partial likelihood satisfies the first-order conditionMoreoverEach summand is the conditional event-covariate variance, so the partial likelihood is log-concave. If some informative risk set has unequal covariates, the curvature is strictly negative and an existing finite root is the unique maximizer.

The finite-interior qualification is necessary. With two at-risk subjects having covariates zero and one, suppose the subject with covariate one is the sole observed event and the other is subsequently censored. Then $L_p(\beta )=e^{\beta }/(1+e^{\beta })$ increases strictly and has its supremum only as $\beta \to +\infty$; the score $1/(1+e^{\beta })$ has no finite zero. If every event risk set has identical covariates, the coefficient is instead unidentifiable and the score is identically zero. The displayed fitted-score equation applies to the regular case in which the proportional-hazards estimate is finite.

For individual $i$, let $N_i(t)$ count its observed event by time $t$ and let $Y_i(t)$ indicate membership of the risk set immediately before $t$. Set $N={\sum }_i{N}_i$ and $Y={\sum }_i{Y}_i$. Under a common hazard function $h(t)$ and independent censoring, a currently at-risk subject has event probability $h(t)dt+o(dt)$ in the next short interval. Summing over the current risk set gives the conditional intensityEquivalently, the event count decomposes aswhere $M$ is the mean-zero counting-process martingale obtained by subtracting its conditional compensator. Estimating the unknown hazard increment by the observed event increment divided by the risk set size gives the Nelson–Aalen estimatorHere $a_j$ are the distinct observed event times and $r_j=Y(a_j)$ includes the subject who is about to fail. A censored observation affects subsequent risk sets, but causes no event jump. This is the requested no-ties estimator; tied event counts would produce $d_j/r_j$.

The derivation also identifies the estimation error:Thus it estimates $H(t)$ on the time range where individuals remain at risk. It is not exactly unbiased for the entire cumulative hazard after the last risk set disappears. Its usual Nelson–Aalen variance estimator is ${\sum }_{a_j\le t}{r}_j^{-2}$, since the compensator variance increment of the no-ties event process is $YdH$.

For a series with constant mean $\mu$, write $X_t=x_t-\mu$. An autoregressive moving-average process has the formwhere $w_t$ is zero-mean white noise with variance ${\sigma }_w^2>0$. The backshift operator is $B{X}_t=X_{t-1}$. Define the autoregressive polynomial and the moving-average polynomial byThe autoregressive operator and moving-average operator give the concise equationOrders are normally specified minimally, with nonzero last coefficients and no common polynomial factor. A displayed unreduced equation can have larger orders than the process actually needs.

Causality means that the stationary process can be recovered from present and past noise:Invertibility means that the noise can be recovered from present and past observations:The causality and invertibility root criteria for an ARMA model, for the reduced representation, areIndeed the transfer series are $\sum {\psi }_j{z}^j=\Theta (z)/\Phi (z)$ and $\sum {\pi }_j{z}^j=\Phi (z)/\Theta (z)$. When all denominator roots lie outside the closed unit disc, each rational function is analytic on a larger disc, so its Taylor coefficients decay geometrically and are absolutely summable. Conversely, a genuine denominator zero in the closed unit disc gives a pole incompatible with such a convergent filter. The reduced-representation proviso prevents a cancelled root from being mistaken for an obstruction.

For order identification by autocorrelation cutoffs, first estimate the sample autocorrelation function and sample partial autocorrelation function from a suitably stationary, centred series. The autocorrelation function is $\rho (h)=\mathrm{Cov}(X_t,X_{t-h})/\mathrm{Var}(X_t)$. The lag-$h$ partial autocorrelation function is the last coefficient in the best linear regression of $X_t$ on $X_{t-1},\dots ,X_{t-h}$, equivalently the correlation left after removing the intervening lags.

For a pure moving-average model of order $q$, observations more than $q$ lags apart share no noise, so the population ACF is zero beyond $q$, while the PACF usually tails off. For a pure autoregressive model of order $p$, the PACF is zero beyond $p$, while the ACF usually decays, possibly with damped oscillation. Thus a clear sample ACF cutoff suggests an MA order and a clear PACF cutoff suggests an AR order. For a mixed ARMA process both usually tail off, so these plots guide a small set of candidate orders rather than determine $p,q$ uniquely. Sampling variation blurs cutoffs; compare fitted candidates and check that their residual ACF resembles white noise. Common-factor cancellation can also disguise the orders in a displayed equation.

One sufficient maximum likelihood estimation statistical consistency theorem is the following. Let the observations be independent random variables with a common density $f_{\theta }$ relative to a fixed measure, let $\Theta$ be a compact metric parameter set containing the true value ${\theta }_0$, and suppose that the expected log-likelihoodis finite, continuous, and uniquely maximized at ${\theta }_0$. Assume also the uniform law of large numbersand assume that a measurable global maximum-likelihood estimator exists. Then every such maximum-likelihood estimator converges in probability to the true parameter. A useful sufficient condition for this uniform law of large numbers is almost-sure continuity of $\theta \mapsto \log f_{\theta }(Y_1)$ on the compact parameter set and an integrable envelope ${\sup }_{\theta }|\log f_{\theta }(Y_1)|\le M(Y_1)$, ${\mathbb{E}}_{{\theta }_0}M(Y_1)<\infty$. Under these conditions, identifiability supplies uniqueness through the Kullback-Leibler divergence identitywith equality only at $\theta ={\theta }_0$.

To see the statistical consistency conclusion, fix $\epsilon >0$. Compactness and the unique maximum give a strictly positive separation gapwhenever the set outside the neighbourhood is nonempty. If ${\Delta }_n$ denotes the supremum discrepancy in the uniform law of large numbers, maximization of the empirical log-likelihood yieldsHence the probability of $d({\hat{\theta }}_n,{\theta }_0)\ge \epsilon$ is at most $\mathbb{P}(2{\Delta }_n\ge {\delta }_{\epsilon })$, which tends to zero. This is the argmin consistency under uniform convergence in probability argument applied to the negative log-likelihood. These are sufficient conditions, not necessary ones; parameter-dependent support can require a direct argument, and independent observations need not have a common distribution in other models.

Put $G=X^{T}X$ and $H=X{G}^{-1}{X}^T$. The rank of a matrix assumption makes $G$ invertible, and $H$ is the orthogonal projection matrix onto the column space of the design matrix. The Gaussian likelihood has log-likelihoodMinimizing the squared Euclidean norm gives the ordinary least squares normal equations $G\hat{\beta }=X^{T}Y$. Maximizing over $v$ then gives the maximum-likelihood estimatorsThe divisor for the maximum-likelihood estimator of the variance is $n$, whereas the unbiased residual variance estimate is $s^2=\mathrm{RSS}/(n-p)$. The exceptional event $\mathrm{RSS}=0$ has probability zero under the Gaussian linear model with ${\sigma }^2>0$; on that event the likelihood is unbounded as $v\downarrow 0$.

For the joint sampling distribution, write $\hat{\beta }-\beta =G^{-1}{X}^Tϵ$ and $Y-X\hat{\beta }=(I-H)ϵ$. Their cross-covariance matrix vanishes because $X^T(I-H)=0$. These two vectors are jointly multivariate normal, so they are independent random variables. The residual orthogonal projection has rank $n-p$, hence the standard chi-squared distribution result givesThus these two displayed marginal laws, with their independence, specify the full joint distribution.

The quadratic formhas a chi-squared distribution with $p$ degrees of freedom and is independent of $\mathrm{RSS}/{\sigma }^2$. Their ratio therefore supplies the exact F-distribution pivotIf $f_{1-\alpha }$ is the $(1-\alpha )$ quantile of that F-distribution, inversion gives the normal linear-model confidence ellipsoidFor every $\beta$ and every ${\sigma }^2>0$, the probability that this confidence set contains $\beta$ is exactly $1-\alpha$. To test the null hypothesis, reject precisely when ${\beta }_0$ lies outside this confidence ellipsoid, equivalently when $T({\beta }_0)>f_{1-\alpha }$. Under the null hypothesis the rejection probability is exactly $\alpha$, independent of the unknown variance.

For the numerical calculation, the second column of the design matrix sums to zero and its sum of squares is $60$. Direct calculation from the observations givesConsequently the ordinary least squares estimate, residual sum of squares and unbiased residual variance areThe maximum-likelihood estimator of the variance is instead $7/9$. At the proposed null value the F-distribution statistic isTherefore reject the null hypothesis at the 5% level. The relevant critical value has $2$ numerator and $7$ residual degrees of freedom.

Put $D=\mathbb{Q}/\mathbb{Z}$. The character module $M^{\ast }={\mathrm{Hom}}_{\mathbb{Z}}(M,D)$ has the natural actionAdditivity is immediate, and $a(b\varphi )(m)=\varphi (bam)=\varphi (abm)$, using commutativity of $A$. The identity of $A$ acts as the identity. We establish the extension and duality tools used in the four implications, so no flatness or injectivity criterion is assumed without proof.

An injective module $E$ is one for which every homomorphism from a submodule $U\subseteq V$ extends to $V$. The Baer criterion says it suffices to test the inclusions of ideals $I\subseteq A$. Necessity follows directly from the definition. For sufficiency, suppose every $I\to E$ extends to $A$. Given $h:U\to E$, use Zorn's lemma to choose a maximal extension $(W,h_W)$ inside $V$; extensions along a chain have a well-defined union. If $v\in V\setminus W$, letThis is an ideal and an $A$-linear map. Its extension to $A$ has the form $a\mapsto ae$ for some $e\in E$. Thenis well-defined: if two expressions represent the same element, their coefficient difference lies in $I$, where the two prescribed values agree. It extends $h_W$ to the strictly larger module $W+Av$, a contradiction. Thus $W=V$, proving the Baer criterion for arbitrary modules and ideals.

The group $D$ is divisible. A homomorphism $n\mathbb{Z}\to D$, for $n>0$, extends to $\mathbb{Z}$ by choosing $d\in D$ with $nd$ equal to its value at $n$; the zero ideal causes no restriction. The proved Baer criterion over $\mathbb{Z}$ therefore makes $D$ injective. It is also a cogenerator: a nonzero element $x$ of an abelian group generates either a finite cyclic group of order $n>1$, which maps $x$ to $1/n+\mathbb{Z}$, or an infinite cyclic group, which maps $x$ to $1/2+\mathbb{Z}$. Injectivity extends this character to the whole group. Thus every nonzero element is detected by an additive character.

It follows that a group homomorphism $u:T\to U$ is injective exactly whenis surjective. For an injection, extension into the injective group $D$ proves surjectivity. Conversely, a character nonzero on an element of $\ker u$ cannot be the pullback of a character on $U$. This is the detection property of the injective cogenerator of abelian groups.

Finally, for any $A$-module $N$ there is a natural character form of the Tensor-hom adjunction:An $A$-linear map $h$ gives the pairing $(n,m)\mapsto h(n)(m)$. It is balanced because $h(an)(m)=(ah(n))(m)=h(n)(am)$. Conversely a balanced additive pairing determines $h$ by this formula. The universal property of the tensor product of modules makes these constructions mutually inverse, and compatible with restriction maps.

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.