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The logarithm of the stretch after $N$ independent uniform renovations is a sum of independent identically distributed bounded increments. For singular stretching $e^{\gamma }$, $\gamma =\mathrm{arsinh}(|a|/2)$, the increment has Fourier expansion $r+{\sum }_{n\ge 1}(-1{)}^{n+1}{q}^n\cos (2n\theta )/n$, with $q=\tanh \gamma$ and $r=\frac{1}{2}\log (1+a^2/4)$. Orthogonality proves the displayed variance. The central limit theorem gives an approximate central log-normal distribution of ${\lambda }_N$, not an exact description of extreme tails. For $|a|\ll 1$, both the mean and variance of $\log {\lambda }_N$ are $N{a}^2/8$ to leading order. Direct expansion of fixed-power moments gives $⟨{\lambda }_N^p⟩=\exp [Np(p+2)a^2/16+O(N{a}^4)]$; in particular mean stretch grows faster than typical stretch.

Let $d\sigma =-\sigma dt/\tau +\sqrt{\kappa }dW$, ${\sigma }_0=0$, and $\dot{B}=\sigma B$, $B_0>0$. The logarithm $\log (B_t/B_0)={\int }_0^t{\sigma }_{s}ds$ is a centered Gaussian random variable of variance $V(t)=\kappa {\tau }^2[t-2\tau (1-e^{-t/\tau })+\tau (1-e^{-2t/\tau })/2]$. Squaring the stochastic kernel $\sqrt{\kappa }\tau [1-e^{-(t-u)/\tau }]$ and integrating proves this formula. The exponential moment of a Gaussian linear functional gives $\mathbb{E}{B}_t^n=B_0^n{e}^{n^{2}V(t)/2}$ and the displayed long-time growth exponent. Positive moment growth with zero mean strain reflects the breadth of the log-normal distribution; it does not imply positive growth of the mean logarithm.

Write $a=\Lambda T$. The deformation gradient of one interval of shear flow is $S(a)=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An initial unit material line at angle $\theta$ therefore becomes $(\cos \theta +a\sin \theta ,\sin \theta )$. Its orientation, including the correct quadrant, isFor an unoriented line the angle is taken modulo $\pi$. If $\Lambda =0$, there is no shear flow: every ${\lambda }_N=1$ and $\mu =0$. The fluctuation and maximizing calculations below concern nonzero shear.

To quantify the renovating shear flow, use the intended isotropic model: each shear orientation is independent and uniform modulo $\pi$. Independence by itself does not specify a probability law and cannot fix the answer. Under the uniform law, conditional on every earlier deformation, the relative angle to the next shear is again uniform. Thus the logarithmic increments ${\xi }_n=\log \ell ({\theta }_n)$ are independent and identically distributed, even though the absolute orientation of the stretched material line need not be uniform. Since $\log {\lambda }_N={\sum }_{n=1}^N{\xi }_n$, only the one-step angular mean is needed.

Write ${\ell }^2=A+B\cos (2\theta -\delta )$, where $A=1+a^2/2$, $B=|a|\sqrt{1+a^2/4}$ and $A^2-B^2=1$. The supplied angular logarithmic integral, with parameter $B/A$, givesTherefore the mean logarithmic stretching rate isThis is the mean logarithmic stretching in an isotropic renovating shear, and the strong law of large numbers gives the same almost-sure Lyapunov exponent. Taking $\Lambda >0$ without loss for this statistic puts it in the requested form $\Lambda F(\Lambda T)$.

For $b=|\Lambda |T$, $F(b)=\log (1+b^2/4)/(2b)$. At small $b$, $\mu ={\Lambda }^{2}T/8+O({\Lambda }^4{T}^3)$; at large $b$, $\mu \sim \log (b/2)/T$, tending to zero when $T$ increases at fixed shear rate. The maximizing nonzero $b$ solvesThe difference $2(1-z^{-1})-\log z$ rises from zero up to $z=2$ and then strictly decreases to $-\infty$, so there is exactly one root beyond $2$. Numerically,Very rapid renovation gives little deformation per interval and cancels the first-order mean stretch. Long intervals provide strong deformation but progressively align a material line with the shear direction, so one interval stretches it only algebraically in $T$. Intermediate persistence combines substantial deformation with frequent reorientation, producing the finite maximum.

For the distribution, let $\gamma =\mathrm{arsinh}(|a|/2)$, so the singular stretches of $S(a)$ are $e^{\pm \gamma }$. In the right-singular-vector basis, ${\ell }^2=e^{2\gamma }{\cos }^2\phi +e^{-2\gamma }{\sin }^2\phi$. Put $q=\tanh \gamma =|a|/\sqrt{a^2+4}$. Factoring this expression and expanding its logarithm givesThe variance follows from angular orthogonality. These bounded independent increments implyAccordingly the central part of the large-$N$ stretch distribution is approximately described by a log-normal distribution, with density $[\lambda \sqrt{2\pi Nv}{]}^{-1}\exp [-(\log \lambda -Nr{)}^2/(2Nv)]$. This central limit theorem is not an assertion about arbitrarily extreme tails: the exact stretch has support inside $e^{-N\gamma }\le {\lambda }_N\le e^{N\gamma }$.

For $|a|\ll 1$, the log-stretch fluctuations in an isotropic renovating shear haveTo leading order the mean and variance of $\log {\lambda }_N$ both equal $N{a}^2/8$. Typical stretch is $\exp (N{a}^2/8)$, while its mean grows as $\exp (3N{a}^2/16)$. These moment statements can be justified directly rather than extrapolating the central Gaussian approximation: expansion of ${\ell }^p$ givesfor fixed $p$. In particular $⟨{\lambda }_N^2⟩=(1+a^2/2{)}^N$ exactly. The leading small-$a$ log-normal approximation is especially precise in the many-small-renovations scaling with $N{a}^2$ fixed and $N{a}^4\to 0$.

If the orientation law is not isotropic these formulas need not hold. For example, repeating a fixed shear direction gives $S(a{)}^N=S(Na)$, only algebraic stretch, and zero asymptotic logarithmic rate. The associated degenerate direction variables are independent, illustrating why the stated independence condition alone is insufficient to select the isotropic result.

The hierarchical Bayesian model separates variability between individuals from variability between repeated measurements on an individual. In WinBUGS, dnorm takes a precision parameter, so inverse variance is the appropriate second argument.

Line (1) gives $\log {\sigma }_j^2\sim N(\varphi ,{\psi }^2)$, hence a positive log-normal distribution for each within-individual variance. Estimating the shared parameters supplies partial pooling of these variances without forcing them equal. Line (2) assigns a proper bounded flat prior to the between-individual standard deviation $\tau$, and line (3) does the same for the standard deviation $\psi$ on the log-variance scale. The broad bounded priors on $\mu$ and $\varphi$ are also proper.

This is a reasonable structure for heterogeneous repeated measurements, but flat does not mean uninformative on all transformed scales. The bounds must be wide enough for plausible values, while such enormous ranges can put substantial mass on implausible predictions. Use prior predictive checks and sensitivity analysis, and check the Markov chain Monte Carlo convergence diagnostics. Serial dependence or systematic trends would require an extension of the independent-measurement likelihood.

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.