Apply Itô formula to $V(t,S_t)$. Its Brownian coefficient isThe self-financing portfolio's Brownian coefficient is ${\pi }_t\sigma S_t$. Since $S_t\sigma >0$, equality of the two value processes forcesThus the stock holding is the claim's option delta.

For constant coefficients, the density process $Y_t{B}_t/B_0$ is a true exponential martingale and defines the risk-neutral measure $Q$. Under $Q$,The minimal value process is thereforeThe Markov property and the lognormal transition law make this a deterministic function of $(t,s)$, and

SetThe boundedness of ${\xi }_T$ and positivity of the deflator make $M$ a nonnegative true martingale. By the Brownian martingale representation theorem, $d{M}_t={\eta }_{t}d{W}_t$. A self-financing wealth process with stock holding ${\pi }_t$ satisfiesThe product $YV$ then has diffusion coefficientChooseThen $Y_t{V}_t=M_t$, so $V_T={\xi }_T$ and the strategy replicates the claim. It is admissible because $V=M/Y$ is nonnegative.

For any other admissible replicating wealth $\tilde{V}$, the nonnegative local martingale $Y\tilde{V}$ is a supermartingale. HenceThe constructed strategy hasso this is the minimal replication cost.

A local martingale deflator makes both $YB$ and $YS$ local martingales. Since the filtration is generated by $W$, the martingale representation theorem and the finite-variation drift forced by $YB$ givefor a continuous adapted $\lambda$, where $\eta =-Y\lambda$. Applying the Itô product rule to $YS$ gives driftIt vanishes exactly when

Condition on $S$ and use the characteristic function of a standard normal distribution:The integrand has modulus one, so Fubini's theorem is immediate. Taking expectation over $S$ yields

By the bound in part (a), Fubini's theorem applies. Conditional on $S$,where the Characteristic function of the Cauchy distribution was used. Since $\mathbb{E}S=1$,The formula expresses a European call option value through complex moments of $S$. In an affine stochastic-volatility model such as the Heston model, those moments are available from an explicit transform, so call prices reduce to a one-dimensional Fourier expectation or integral.

HereThe Laplace transform of an exponential distribution is $\mathbb{E}[e^{-sX}]=(1+s{)}^{-1}$ for $\mathrm{Re}s>-1$. Since $\mathrm{Re}(t\theta )\ge 0$,

For $\theta =p+iq$ with $0\le p\le 1$,The elementary inequality $x^p\le 1+x$ for $x>0$ givesTherefore the Mellin transform $M(\theta )=\mathbb{E}[S^{\theta }]$ is absolutely well-defined throughout the strip and

For $c\in \{0,1\}$ setThe quadratic negative terms makeeverywhere finite and smooth. Existence of the assumed $\rho$ rules out the arbitrage direction in part (c) by part (a). Hence each $F_c$ has a bounded minimizing sequence, and part (b) supplieswith $\mathbb{E}{\rho }_c=1$ and $\mathbb{E}({\rho }_{c}X)=0$. Uniqueness forces ${\rho }_0={\rho }_1$. Taking logarithms and cancelling the common quadratic terms givesThuswith $b=h_1-h_0$ and $a=\log F_1(h_1)-\log F_0(h_0)$. Since $Y$ was arbitrary, the one-period market is complete.

We prove the contrapositive. LetThe function $F$ is constant along $N$, so minimize it on $N^{\perp }$. Suppose there is no $H$ with $H\cdot X\ge 0$ almost surely and strict inequality with positive probability. If a sequence $h_k\in N^{\perp }$ satisfies $\Vert h_k\Vert \to \infty$, pass to a subsequence withBecause $H\notin N$ and there is no arbitrage direction, $\mathbb{P}(H\cdot X<0)>0$. On that event, $e^{-h_k\cdot X}\zeta \to \infty$, and Fatou lemma gives ${\mathrm{liminf}}_{k}F(h_k)=\infty$. Hence every finite sublevel set of $F$ in $N^{\perp }$ is bounded. It is also closed, so $F$ attains its infimum there and has a bounded minimizing sequence. This contradicts the assumption. Therefore there is a unit vector $H$ satisfying

Let $(h_k)$ be a bounded minimizing sequence. A subsequence converges to some $h_{\ast }$, and continuity gives $F(h_{\ast })=f$. Since $F$ is smooth and $h_{\ast }$ is an unconstrained minimizer,DefineThen $\rho >0$, $\mathbb{E}\rho =1$, andThus the normalized exponential tilt is the required state-price density.

If such $\rho$ existed, then integrability and $H\cdot X\ge 0$ would giveBut $\rho >0$ almost surely and $H\cdot X$ is nonnegative and strictly positive with positive probability. Hence $\rho (H\cdot X)$ is nonnegative and nonzero with positive probability, so its expectation is strictly positive. This contradiction proves that no such positive state-price density exists.

Letand use constant baseline hazard $h_0=m$. Define the two-point frailtyThen $\mathbb{E}U=1$, and the conditional rates are exactly $\gamma$ and $\beta$. The experimental-treatment population hasandSince standard treatment has hazard $\beta$, the population hazard ratio isAt zero,whereas for $\pi >0$,The treatment effect is therefore non-proportional and strengthens among later survivors as the high-rate subgroup is depleted. A trial should allow adequate follow-up, avoid relying only on a constant-hazard-ratio Cox model, and prespecify survival-curve, milestone-risk, restricted-mean-survival, or time-varying-effect analyses. Its power and interpretation will depend materially on follow-up duration.

A frailty random variable is an unobserved positive multiplicative risk factor. A proportional frailty model hasThe scale of $U$ is not separately identifiable from $h_0$: multiplying $U$ by a constant and dividing $h_0$ by it leaves the model unchanged. We may therefore normalize $\mathbb{E}U=1$, which lets $h_0$ represent the mean initial hazard multiplier and makes relative frailty interpretable.

If $U\sim \mathrm{Exponential}(1)$ and $h_0(t)=\lambda$, then the Laplace transform of $U$ givesThus $h(0)=\lambda$, while $h(t)\to 0$ as $t\to \infty$. Survivors become increasingly enriched for low-frailty individuals.

Among the 12 pairs with control time first, the eight in which that first time is an event contribute $(1+\beta {)}^{-1}$ each. Among the eight pairs with treated time first, the four in which that first time is an event contribute $\beta /(1+\beta )$ each. Later events in one-eye risk sets contribute one. HenceDifferentiating the log likelihood gives $4/\beta -12/(1+\beta )=0$, and therefore

Without randomization, treatment side can be associated with prognosis. Always treating the left eye confounds treatment with systematic left-right differences; choosing the worse eye creates severe confounding by indication, baseline imbalance, and possible regression toward the mean. The within-patient comparison then no longer identifies a treatment effect without stronger adjustment assumptions.

In a proportional hazards model, individual $i$ hasThe exponential term is the hazard multiplier relative to the baseline hazard $h_0$. With $h_0$ unspecified, the Cox partial likelihood multiplies, over event times, the failing subject's multiplier divided by the sum of multipliers in the current risk set. After estimating $\beta$, the Breslow estimator is

A Stratified Cox model uses a separate baseline hazard $h_{0s}$ for each stratum but a common $\beta$. Its partial likelihood is the product of within-stratum partial likelihoods, and a separate integrated baseline hazard is estimated in each stratum.

In a matched pair, the only informative comparison occurs while both members remain at risk. If the exposed member fails first, the pair contributes $e^{\beta }/(1+e^{\beta })$; if the unexposed member fails first, it contributes $1/(1+e^{\beta })$. A censoring as the first recorded time gives no informative failure comparison, and any later one-person risk set contributes one. With only two observations per stratum, each stratum supplies almost no information about its arbitrary baseline hazard, so a useful common integrated baseline-hazard estimate is generally unavailable.