MAR is plausible if clinic withdrawal is driven by recorded QoL, observed side effects, treatment, and other measured history. It is doubtful if patients leave because of an unrecorded deterioration, an imminent recovery, treatment toxicity not included in the analysis, or dissatisfaction that predicts their unseen 12-month QoL. The large dropout fraction makes such missing not at random mechanisms a serious concern, so MAR should be supported by rich predictors and sensitivity analysis rather than assumed without examination.

In this trial, MAR means that among patients assigned the same treatment who have the same recorded QoL trajectory up to a visit, the probability of dropping out next is unrelated to what their later QoL values would have been. Dropout may depend strongly on previous poor QoL, treatment assignment, and other observed history; MAR only rules out residual dependence on the unobserved outcomes.

Let $A_i$ be treatment, $Y_i=(Y_{i0},\dots ,Y_{i,12})$, and $R_i\in \{0,\dots ,12\}$ the last observed visit. The missing at random assumption isfor every feasible $r$. Equivalently, each conditional dropout hazard may depend on treatment and observed QoL history but not on current or future unobserved QoL after conditioning on that history.

The mediation estimate relies on several strong assumptions. Possible failures include mediator-outcome confounding, residual exposure-outcome or exposure-mediator confounding, and an exposure-induced mediator-outcome confounder. Smoking duration may itself be part of the causal pathway, making adjustment inappropriate. Self-reported cigarettes per day has measurement error and does not fully measure tobacco exposure. The case-control sampling can create selection bias; population stratification can confound the genotype relations; cancer may alter reported smoking; and the product-of-coefficients calculation can be inappropriate for a binary outcome because odds ratio effects are nonlinear and noncollapsible. Any of these can attenuate or distort the indirect effect.

The significant additive-scale interaction and the association of the variant with cancer among smokers but not nonsmokers indicate effect modification: the joint effect of genotype and smoking exceeds additivity on the risk scale. Under adequate control of confounding and selection, that pattern supports a causal role for smoking in activating or amplifying the genetic pathway. Interaction alone is not proof that smoking is causal, because smoking was not randomized and the stratum-specific estimates can be affected by confounding, selection, and low power among nonsmokers.

For the vertical slit $A=(1,1+iϵ]$, choose the square-root branch asymptotic to $z-1$ at infinity. The normalized map isfor which ${\psi }_A(0)=0$ andTherefore

Let $A=\mathbb{H}\cap B(1,ϵ)$. Its normalized mapping-out map isso ${\psi }_A(0)=0$ andUsing the restriction exponent $5/8$ gives

Write the semimartingale decomposition as $U=M+A$, where $M$ is a continuous local martingale and $A$ has finite variation. Applying Itô formula to $\log Z_t(z)$ shows that its finite-variation part isIt vanishes for every $z$. Multiplying by $Z_t(z{)}^2$ givesSubtract this identity for two points with distinct $Z_t$ to obtain $d{A}_t=0$; then $d⟨M{⟩}_t=4dt$. Since the curve starts at zero, $U_0=0$. The Lévy characterization of Brownian motion now gives $U_t=2B_t$. Hence the Loewner chain is

Fix $z$ and writeBy assumption, $L(z)$ is a continuous local martingale, so $Z_t(z)=e^{L_t(z)}$ is a semimartingale. The Chordal Loewner equation giveswhich has finite variation. Thereforeis a semimartingale. Thus the Loewner driver $U$ is a continuous semimartingale.

The imaginary partis a bounded local martingale and hence a martingale. As the simple transient trace passes $z$, this angle converges to $\pi$ if the trace passes to the right of $z$ and to $0$ if it passes to the left. Bounded convergence therefore givesso the SLE4 left-passage probability is

Let $Z_t=g_t(z)-U_t$. For ${\mathrm{SLE}}_4$, $d{U}_t=2d{B}_t$, and the Chordal Loewner equation givesThe complex Itô formula yieldsTherefore

For compact $K\subset \mathbb{H}$, letBy the Tonelli theorem and part (ii),Every point in the SLE range has conformal radius tending to zero and therefore belongs to every $E_{ϵ}$. Hence the range inside $K$ has zero expected Lebesgue measure, and so has zero measure almost surely. Exhausting $\mathbb{H}$ by countably many compact sets proves

At time zero, ${{\rm Y}}_0(z)=\mathrm{Im}z$. Compactness of $K\subset \mathbb{H}$ givesAlso $M_0(z)$ is uniformly bounded above on $K$. Onone has $M_{{\tau }_{ϵ}}\ge {ϵ}^{-p}{2}^{-q}$. Optional stopping, Fatou lemma, and the assumed conditional angular estimate giveThusThe exponent $(8-\kappa )/\kappa$ requested in the question does not follow and is false as written. The SLE Green-function estimate gives probability comparable to ${ϵ}^{1-\kappa /8}$, confirming that the denominator in the requested exponent should be $8$.

Putso $M_t={{\rm Y}}_t^{-p}{S}_t^q$. Before ${\tau }_{ϵ}$, one has ${{\rm Y}}_t\ge ϵ$ and $0<S_t\le 1$. ThereforeThe supplied continuous local martingale is thus bounded after stopping, and a bounded local martingale is a true martingale. Hence

It has the Loewner local growth property when, after mapping out the old hull, each short new increment is small: for every $T,ϵ>0$ there is $\delta >0$ such thatEquivalent formulations use a crosscut of diameter below $ϵ$ separating the new increment from infinity.

It has the half-plane-capacity parameterization under the standard chordal convention when

The family is a nondecreasing family of sets when

The statement is true. In the notation of part (ii), equality of the capacities forces $\mathrm{hcap}(D)=0$. Every nonempty compact H-hull has strictly positive half-plane capacity: by the Brownian representation of half-plane capacity, Brownian motion started sufficiently high has positive harmonic measure of a boundary portion of positive height. Hence $D=\varnothing$, so $C\setminus A=\varnothing$ and

Map out $A$ first. The imagewith its bounded filling is a compact H-hull, and uniqueness of hydrodynamic normalization givesComparing the coefficients of $1/z$ at infinity yields the half-plane-capacity composition ruleThus half-plane capacity is monotone under inclusion.