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Let $F(y)$ be the manager's utility function as a function of nonnegative terminal fund wealth. WriteThenBoth pieces are increasing and strictly concave functions. However, the derivative jumps upward from $a{e}^{-a{w}_0}$ to $b{e}^{-a{w}_0}$ at $w_0$. The payoff utility is increasing but not concave. A concave function must have nonincreasing one-sided slopes.

For concavification of incentive utility, the common tangent for exponential incentive utility replaces the upward kink by its common tangent. There are contact points $\ell <w_0<h$ and a common slope $s>0$ satisfyingSince $F(\ell )=-s/a$ and $F(h)=-s/b$, the chord condition gives $h-\ell =1/a-1/b$. Solving the derivative matching yields the contact pointsThe assumption that $w_0$ is large enough means, explicitly, thatso $\ell >0$. The inequalities $\ell <w_0<h$ follow from $\log q<q-1$ and $\log q>1-1/q$ for $q>1$.

The least concave majorant on $[0,\infty )$ isIt is increasing and concave: the derivative is decreasing outside the interval and equals $s$ inside. The tangent line lies above each original branch. Any concave majorant must lie above the chord joining the two contacts, so this one is the least.

The state-price budget constraint now reduces the problem to pointwise maximization of $\bar{U}(y)-qy$, where $q=\eta {\zeta }_T$ and $\eta >0$ is chosen so that $\mathbb{E}[{\zeta }_T{Y}^{\ast }]=w_0$. Because ${\bar{U}}^{\prime }(0)=a$ is finite, the nonnegative wealth constraint must be included. The optimal terminal wealth isAt $q=s$, any point of $[\ell ,h]$ maximizes the concavified objective. If $\kappa \ne 0$, the state-price density has a continuous log-normal distribution, so this event has probability zero. The optimizer then avoids $(\ell ,h)$ almost surely and satisfies $\bar{U}(Y^{\ast })=F(Y^{\ast })$. The supporting-line proof from part (i), with supergradients at the contacts, proves optimality for the concavified problem; the pointwise equality proves the same optimizer and value solve the original manager's problem. Nonnegative replication is available in the complete market.

If $\kappa =0$, the state-price density is deterministic. If the required deterministic mean terminal wealth lies in the linear segment, randomize between $\ell$ and $h$ at the common slope instead of choosing an interior wealth. This maintains the budget and attains the same concavified value. The Brownian market can replicate that bounded lottery even when its market price of risk is zero.

In the special case $r=\mu =0$, ${\zeta }_T=1$ and the fund has a fair-game wealth process. Since $\ell <w_0<h$, the manager uses fair-game gambling induced by an incentive fee:For example choose a threshold event in $W_T$ with the displayed probability and replicate its payoff. Then $w_t=\mathbb{E}[Y^{\ast }\mid {\mathcal{F}}_t]$ remains between $\ell$ and $h$, so the strategy respects nonnegative wealth. It earns no risk premium but raises expected incentive utility function above the value at the unrandomized $w_0$, because the common tangent lies strictly above the kink.

Use the Radon-Nikodym derivative at $T_1$ and the discounted bond price martingale from part (b):For the variance, put $b_t^i={\int }_t^{T_i}\sigma (t,u)du$ and $d_t=b_t^2-b_t^1={\int }_{T_1}^{T_2}\sigma (t,u)du$, for $t\le T_1$. The ratio $R_t=P(t,T_2)/P(t,T_1)$ has, by the Itô formula,Under the T-forward measure ${\mathbb{Q}}_{T_1}$, $d{W}_t=d{W}_t^{T_1}-b_t^{1}dt$, so its drift becomes $-d_t^2/2$. Since $R_{T_1}=P(T_1,T_2)$,The first two terms are deterministic; the Itô isometry supplies the variance of the last. The two requested answers areIn fact the terminal zero-coupon bond price has a log-normal distribution under this forward measure, with the deterministic negative half-variance correction ensuring the displayed mean.

By the invariance property of maximum likelihood estimation,The sampling distribution is $\hat{\theta }\sim N(\theta ,{\sigma }_{\hat{\theta }}^2)$, where$\hat{h}$ has a log-normal distribution withWriting $v={\sigma }_{\hat{\theta }}^2/{\alpha }^2$, its exact fractional bias and variance areHence, to leading order,

Set $Y_n=\log Z_n$ andDefine a probability measure ${\mathbb{P}}^{\ast }$ by the positive densityThe density has expectation one and is strictly positive, so ${\mathbb{P}}^{\ast }$ is equivalent to $\mathbb{P}$. By exponential tilting, the variables $Y_k$ remain independent under ${\mathbb{P}}^{\ast }$ and have distribution $N({\mu }_{\ast },{\sigma }^2)$. Thus $Z_k$ has a log-normal distribution andFor the discounted stock ${\tilde{S}}_n=S_n/S_n^0$,Hence ${\mathbb{P}}^{\ast }$ is an equivalent martingale measure.

No. A standard counterexample comes from the log-normal distribution. Letand, for a fixed $0<|\epsilon |\le 1$, letThis is a nonnegative density distinct from $f$. For every nonnegative integer $n$, substituting $z=\log x$ makes the difference of the $n$th moments proportional toIt is the imaginary part ofwhich vanishes because its phase is $2\pi n$. The case $n=0$ also proves that $f_{\epsilon }$ is normalized. Thus the two distributions have every finite moment equal but are different. Unbounded random variables need not be determined by their moments.

The invariance property of maximum likelihood estimation givesSince $\hat{\theta }-\theta \sim N(0,{\sigma }_{\theta }^2)$, the ratio $\hat{h}/h$ has a log-normal distribution. Its exact variance isand the lowest-order delta method approximation isFor fixed ${\sigma }_{\mathrm{tot}}$, case (i) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2$, whereas case (ii) has $A={\sigma }_{\mu }^2+{\sigma }_{\mathrm{tot}}^2/K$. Because $K\ge 2$ and $B$ is the same in both cases, independent supernova-level variation in case (ii) gives the smaller fractional variance: averaging reduces it, while a shared galaxy fluctuation does not average away.

By invariance of maximum likelihood,Because $\hat{\theta }-\theta \sim N(0,{\sigma }_{\theta }^2)$, the ratio $\hat{h}/h$ is log-normal. Its exact variance isAs $N_{\mathrm{SN}},G,N_{\mathrm{LMC}}\to \infty$, all sampling terms vanish but $V_1$ retains ${\sigma }_{\mathrm{LMC}}^2$. The dominant remaining uncertainty is therefore the parallax calibration of the LMC distance modulus.