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Let $v=\mathrm{Var}(S)>0$ and $v_I=\mathrm{Var}(S_I)$. The variance-matching quota share reinsurance in this part has the nonnegative retained fraction ${\alpha }^{\ast }=\sqrt{v_I/v}$. For claimwise retentions $0\le h(x)\le x$, its existence within $[0,1]$ follows from $v_I=\lambda \mathbb{E}[h(X{)}^2]\le \lambda \mathbb{E}[X^2]=v$. The comparison below only needs the matching contract specified in the question.

Because $S_R=S-S_I$, the variance and covariance identities giveThe Cauchy-Schwarz inequality bounds $\mathrm{Cov}(S,S_I)\le \sqrt{v{v}_I}={\alpha }^{\ast }v$. Therefore the quota share comparison at matched retained variance isThe last expression is the sum of variances under the matching quota share reinsurance. HenceIf $v=0$, the given matching contract forces $v_I=0$, so all relevant totals are constant and equality holds. For $v>0$, equality in the bound is equivalent to $S_I-\mathbb{E}[S_I]={\alpha }^{\ast }(S-\mathbb{E}[S])$ almost surely.

For a single claim $x$, quota share reinsurance with retained fraction $\alpha \in [0,1]$ makes the direct insurer pay $\alpha x$ and the reinsurer pay $(1-\alpha )x$. Under excess of loss reinsurance with retention $M\ge 0$, the direct insurer pays $\min (x,M)$ and the reinsurer pays the positive part $(x-M{)}_{+}$. Thus the concise payout pairs areThe cap in excess of loss reinsurance applies separately to every claim; it is not a cap on the entire annual aggregate.

For the following variance calculations take $\lambda >0$ and $\mathbb{E}[X^2]<\infty$, so the displayed variances are finite. For any per-claim payout $Y$, the law of total variance in a compound Poisson distribution givesThe final term is the raw second moment, not the single-claim variance. Both parties' totals are retained compound Poisson aggregates, with different payout functions of the same claims; they are generally dependent.

A credibility estimate blends the risk's own sample mean with a population or prior expected value. In the basic equal-exposure form it is $Z\overline{X}+(1-Z)\mu$, where the credibility factor $Z\in [0,1]$ is the weight attached to the individual experience. More data ordinarily increase that weight; greater within-risk variation relative to between-risk variation reduces it.

Write $m(\theta )=\mu (\theta )$ for the conditional claim expected value, keeping $\mu$ for the prior-level parameter. Normalization of the probability density function givesAt interior points of its finite domain, differentiation under the integral sign yieldsThis is an instance of the exponential-family derivative identities for natural parameter $-\theta$. The minus sign is present in the PDF and is lost in the converted TeX. The conditional variance is likewise $(\log q{)}^{\prime \prime }(\theta )$.

Differentiate the log of the natural conjugate prior:Integrate on a compact subinterval and then let its endpoints approach those of the parameter interval. The prescribed vanishing of the prior density eliminates the boundary term. Since $m(\theta )\ge 0$, monotone convergence theorem justifies the limit of its expectation, and the identity givesIn particular the assumed proper prior and boundary conditions themselves establish this first-moment integrability.

For $T={\sum }_{i=1}^n{x}_i$ and $\overline{x}=T/n$, conditional independence makes the likelihood function proportional to $q(\theta {)}^{-n}{e}^{-\theta T}$. Multiplication by the prior distribution gives the Bayesian posteriorThus this is a conjugate prior family, with updated parameters

The integration-by-parts calculation for the posterior mean also needs vanishing posterior boundary values. Here that property can be checked, rather than assumed. Let $a,b$ be the essential lower and upper endpoints of the claim support. Every nondegenerate tilted claim law has $a<m(\theta )<b$, so the established prior mean satisfies $a<\mu <b$. Supported observations have $\overline{x}\in [a,b]$; hence $a<{\mu }_n<b$. At a finite parameter endpoint the prior's vanishing forces $q(\theta )\to \infty$, because its exponential factor has a finite positive limit; the increased power $q^{-k_n}$ makes the posterior vanish there too. At $\theta \to +\infty$, choose $a<d<{\mu }_n$ with positive base-measure mass below $d$. Then $q(\theta )\ge C{e}^{-d\theta }$ and the unnormalized posterior is at most $C^{-k_n}{e}^{-k_n({\mu }_n-d)\theta }\to 0$. At $\theta \to -\infty$, choose ${\mu }_n<d<b$ with positive mass above $d$ to obtain the corresponding bound $C^{-k_n}{e}^{k_n(d-{\mu }_n)\theta }\to 0$. These bounds also ensure posterior propriety at infinite endpoints. This is the endpoint control for a Laplace-family conjugate posterior.

Apply the previous score integration to the Bayesian posterior. The natural conjugate credibility identity isThus $Z=n/(n+k)$ is the credibility factor. This is an exact posterior mean, rather than an affine approximation to one.

For the specific shape-two gamma distribution, takeThen $m(\theta )=2/\theta$ and $\mathrm{Var}(X\mid \theta )=2/{\theta }^2$. The gamma rate gamma conjugacy prior isthat is, $\Theta \sim \mathrm{Gamma}(2k+1,\text{rate }k\mu )$. Its normalizing constant in the notation of the question is $c(\mu ,k)=\Gamma (2k+1)/(k\mu {)}^{2k+1}$. The gamma distribution density vanishes at both zero and infinity for $k>0$. The posterior has shape $2k+2n+1$ and rate $k\mu +T$, so its conditional-mean posterior mean is again $(k\mu +T)/(k+n)$.

For the last calculation, gamma distribution integration givesThe second identity requires the printed $k>1/2$. Therefore the expected process variance and variance of hypothetical means, denoted by $a,b$ in this question, areConsequently $a/b=k$, and the Bühlmann credibility factor agrees with the exact Bayesian weight:This is the exact Bühlmann credibility for gamma claims, expressed using the reciprocal rate as the random scale. The symbols $a,b$ here are the process and between-risk variances, respectively; their roles should not be interchanged when comparing with other notation for the Bühlmann model.

Use a baseline-adjusted linear regression for an intention-to-treat analysis:Condition on the observed baseline covariates and assignment. The parameter of primary interest is $\beta$, the mean effect of assignment to the experimental intervention, under this constant-effect conditional mean model. Independent normal distributions with common variance give the usual Gaussian likelihood and regression inference. Prognostic baseline covariates improve precision by explaining outcome variability.

Do not include the actual intervention-use variable $D_i$ as an ordinary adjustment variable for this estimand: it is measured after assignment and can mediate the effect being estimated. Post-randomization adjustment changes a treatment estimand; the resulting coefficient on $Z_i$ would generally represent a different comparison. Participants remain in their assigned groups even when adherence differs.

Only sets 4, 5, 7 and 8 have an earlier failure with both members at risk. Sets 4 and 8 have B fail first and favor A, giving $6+3=9$ informative pairs. Sets 5 and 7 have A fail first and favor B, giving $18+9=27$. Both-censored pairs contribute zero; in sets 3 and 6 the earlier observation is censoring, so the later failure occurs with no remaining comparator and also contributes zero.

There are thus 36 informative independent strata. The stratified log-rank statistic, its null variance, and its standardized value areEquivalently the squared statistic is nine. The normal approximation gives a two-sided p-value $2[1-\Phi (3)]<0.0027$ using the stated bound. The exact conditional sign test instead gives $2{\sum }_{k=0}^9\left(\genfrac{}{}{0ex}{}{36}{k}\right)2^{-36}\approx 0.00393$; the slight difference is discreteness rather than a different direction of effect. There is strong evidence favoring Method B, which is associated with the later failure in three quarters of the informative pairs, under the test assumptions.

Yes: different adapted volatilities can give perfectly correlated terminal prices. The counterexample in part ii already establishes this with one nonconstant volatility. Both can be nonconstant as well. Put $h(x)=1+\frac{1}{2}\tanh x$ and takeThe Doléans-Dade exponential has volatility $h(B_t)$, bounded between $1/2$ and $3/2$. The second stock has volatility $h(B_t)S_t/(1+S_t)$, also bounded and nonconstant. Their squared volatility difference integrates toThe Novikov condition and boundedness give true square-integrable martingales, and the Itô isometry gives positive variance for each $t>0$. Yet $S_t^{\prime }=(1+S_t)/2$ gives correlation one. The mechanism is positive affine dependence, not equality of the multiplicative volatility coefficients; it also explains the failure of the literal part-ii premise.

The principle terminal correlation does not identify adapted stock volatility exposes a false printed assertion for the correlation coefficient defined in the introduction. Take $r=0$, $S_0=S_0^{\prime }=1$, andThen $d{S}_t=S_{t}d{B}_t$ andHence ${\sigma }_t=1$ and ${\sigma }_t^{\prime }=S_t/(1+S_t)$ are bounded. Both terminal prices have positive finite variance and are perfectly correlated, because one is a positive affine function of the other. Nevertheless $\boxed{{\int }_0^1({\sigma }_t-{\sigma }_t^{\prime }{)}^{2}dt={\int }_0^1\frac{dt}{(1+S_t{)}^2}>0}$ on every continuous finite stock path. Equal initial prices fix the means, not the affine slope.

Here is the exact general conclusion and the intended qualified proof. Bounded volatilities make these zero-rate stochastic exponentials true square-integrable martingales. Writing $s_0=S_0=S_0^{\prime }$, perfect terminal correlation gives $S_1-s_0=c(S_1^{\prime }-s_0)$ for some $c>0$. Conditional expectation therefore gives $S_u-s_0=c(S_u^{\prime }-s_0)$ for every $0\le u\le 1$. Comparing stochastic integrals, the Itô isometry yieldsThis relation need not imply equality of the volatilities.

The terminal proportionality identifies bounded stock volatility criterion repairs the statement. If one additionally assumes equal terminal variances, then $c=1$ and the whole stock paths coincide, so positivity gives $\sigma ={\sigma }^{\prime }$ almost everywhere and the requested integral is zero. The same repair follows if terminal proportionality $S_1=c{S}_1^{\prime }$ is assumed: equal martingale means force $c=1$, after which the conditional-expectation and Itô isometry argument applies. This proves the intended conclusion under a sufficient additional hypothesis without treating centered correlation as uncentered proportionality.

Assume positive deterministic initial stock prices and finite, positive terminal variances, as required to define the correlation coefficient. The explicit stochastic-exponential solutions with constant volatilities giveThe common, possibly random, accumulated interest rate cancels. If $\sigma ={\sigma }^{\prime }$, this ratio is a positive constant, so the prices are perfectly correlated.

Conversely, perfect correlation would give $S_t=c{S}_t^{\prime }+d$ almost surely for constants $c>0,d$. If $\sigma \ne {\sigma }^{\prime }$, the displayed ratio is a nonconstant variable with a lognormal distribution with full support $(0,\infty )$. If $d>0$, the ratio would always exceed $c$, and if $d<0$ it would always be below $c$, both impossible. If $d=0$, the ratio would be constant, also impossible. Therefore $\boxed{\rho (S_t,S_t^{\prime })=1⟺\sigma ={\sigma }^{\prime }}$, whenever the stated terminal variances exist and are positive. No deterministic-interest assumption is needed. Degenerate terminal prices, for example zero volatilities and deterministic rates, have undefined correlation and must be excluded.

For centered square-integrable variables with positive variances, the correlation coefficient isPut $v=\sqrt{\mathbb{E}[X^2]}$ and $v^{\prime }=\sqrt{\mathbb{E}[(X^{\prime }{)}^2]}$. ThenThus $\boxed{\rho =1⟺X=(v/v^{\prime })X^{\prime }\text{ almost surely}}$, with positive proportionality factor. Conversely positive proportionality immediately gives correlation one. For variables with nonzero means, this criterion applies to their centered versions and yields an affine, not necessarily proportional, relationship. That distinction is essential for the positive stock prices below.

For a predictable position $\pi$, the current net wealth increment $\pi \Delta S_t+Y_t-y$ is conditionally Gaussian with mean $\pi \mu +a-y$ and variance ${\pi }^2{\sigma }^2+2\pi \rho b\sigma +b^2$. Its exponential-loss multiplier isDifferentiate its quadratic exponent. The unique minimizing position isThe first term is the speculative demand; the second offsets the part of the income correlated with the tradable price increment. Independent period vectors and backward induction justify using this same position at every date. Completing the square gives minimal exponent $-\kappa -\gamma d(y)$, whereTherefore hedging a Gaussian income stream with exponential utility has valueThe residual variance is $b^2(1-{\rho }^2)$; it disappears for perfectly correlated income.

LetPart (c) has $d(y)=\overline{y}-y$. Comparing the optimized values with and without the contract, $V_0^{\mathrm{swap}}=-e^{-\gamma x-T\kappa -\gamma T(\overline{y}-y)}$ and $V_0=-e^{-\gamma x-T\kappa }$, shows that $y>\overline{y}$ makes the swap value strictly more negative. Thus The investor prefers not to enter when $y>\overline{y}$. He is indifferent at equality and prefers entry for a smaller payment.

The periodic utility indifference payment for Gaussian income equals the mean income less the cost of hedging its correlated component and the exponential-utility penalty for the residual unhedgeable variance. Comparing unoptimized strategies would not justify this price threshold.

For a statistically homogeneous isotropic dimensionless field, write its two-point autocorrelation function of a random field as $C(r)$ andThe formal real-space scale invariance condition is $C(\lambda \mathbf{r})=C(\mathbf{r})$ for all $\lambda >0$. Changing variable $\mathbf{q}=\lambda \mathbf{k}$ in the first expression and equating the nonzero-mode Fourier transforms gives ${\lambda }^{-3}P(q/\lambda )=P(q)$, equivalently $P(\lambda k)={\lambda }^{-3}P(k)$. Taking $\lambda =k/k_0$ yieldsThus a scale-invariant inflationary power spectrum has equal variance per logarithmic wavenumber interval, rather than equal power per Fourier volume.

There is an important mathematical qualification to the literal covariance condition. A nonzero $k^{-3}$ power spectrum over all scales has $C(0)=A\int dk/k/(2{\pi }^2)$, divergent at both endpoints; it is not the covariance of a finite-variance ordinary field. Moreover, exact dilation invariance of a continuous isotropic $C(r)$ makes it constant for $r>0$ and, by continuity, at zero: its spectrum can then consist only of a zero-mode delta measure. The nontrivial cosmological statement consequently concerns nonzero modes, with cutoffs or subtraction of an unobservable constant. For example the finite subtracted covarianceis invariant under simultaneous rescaling of $r,r_0$. This is the precise infrared qualification of a scale-invariant covariance; a regulated unsubtracted $C(r)$ generally shifts by an additive constant under dilation. The formal derivation above gives the intended nonzero-mode scaling, with this qualification rather than an impossible finite-variance premise.

Let $G={\Phi }_G$ and $f=f_{NL}^{\mathrm{local}}$. The Gaussian random field covariance in Fourier space isThe Fourier transform of the centered quadratic term isThe Gaussian three-point function vanishes. At first order in $f$, the only terms are $f⟨Q_1{G}_2{G}_3⟩$ and its two cyclic placements. For the first, Wick theorem pairs four Gaussian factors in three ways. Pairing the two factors inside $Q_1$ together is canceled by the mean subtraction. Pairing one of those factors with $G_2$ and the other with $G_3$ gives $P(k_2)P(k_3)$ and the overall momentum delta; exchanging the two factors gives the same contribution. ThusAdding the three placements givesThere is no second-order term because it would contain five centered Gaussian factors. The two-point function receives its first correction at $f^2$, whereas the displayed local-type primordial non-Gaussianity gives a three-point correlation already at $f$. For $P\propto k^{-3}$ the squeezed limit is $B_{\Phi }(q,k,k)\simeq 4fP(q)P(k)$; this explains the strong correlation between a long potential and short-scale power. A finite variance or regulator is needed to define the subtracted $⟨G^2⟩$, as noted in part (b).