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Let $Z$ be a real random variable with finite expectation $m$ and finite variance ${\sigma }^2$. The Chebyshev inequality isThe proof is the pointwise comparisonTaking expectations and using the definition of variance givesDivision by the positive number $t^2$ proves the Chebyshev inequality, including the case ${\sigma }^2=0$. Equivalently, this is the Markov inequality applied to the nonnegative random variable $(Z-m{)}^2$.

For the exponential distribution with expected value $\mu$,The Poisson distribution has both expected value and variance equal to $\lambda$. Substitution into the random sum of independent claims formulas gives the portfolio A moments

To distinguish the random intensity from its possible values, write it as $\Lambda$. Its law is a gamma distribution with shape $2$ and rate $p/q$. HenceFor the Poisson mixture, conditional expectation and conditional variance both equal $\Lambda$. The law of total expectation and the law of total variance yieldwhere $p+q=1$. Applying the random sum of independent claims formulas with the exponential distribution of the claim sizes gives the portfolio B momentsAt the matched intensity ${\lambda }_0$, the expected value for portfolio A is also $2q\mu /p$, whereas its variance is $4q{\mu }^2/p$. Thus the expected totals agree, but mixing increases the variance:The extra term is precisely ${\mu }^2\mathrm{Var}(\Lambda )$.

Write $m_X=\mathbb{E}{X}_1$ and $v_X=\mathrm{Var}(X_1)$. For the random sum of independent claims, conditioning on $N$ givesThe law of total expectation and the law of total variance therefore give the aggregate momentsThe first term in the variance measures variation of the individual claims at a fixed count; the second measures variation of the count itself. These formulas require the indicated moments to be finite.

For the moment-generating function, independent random variables givewhere $G_N(z)=\mathbb{E}[z^N]$ is the probability generating function. This identity holds wherever the expectations are finite; in particular a moment-generating function need not exist for positive $t$ for an arbitrary positive claim distribution. The empty sum for $N=0$ is zero.

The expected value and variance under quota share reinsurance follow by scaling the exponential distribution:The retained stop loss moments for an exponential aggregate follow from the payout $Y=\min (S,M)$, its survival function equals $e^{-x/{\mu }_S}$ for $0\le x<M$ and zero for $x\ge M$. The tail integral formula for moments gives, with $m=M/{\mu }_S$ and $r=e^{-m}$,Consequently the retained moments areMatching the two expected values forces $\alpha =1-r$, which lies strictly between zero and one. The difference of the variances simplifies toIndeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h^{\prime }(m)=1-e^{-m}\ge 0$ for $m\ge 0$. Because $M>0$, the difference is actually positive. At equal retained expected value, aggregate stop loss reinsurance reduces the variance more than quota share reinsurance.

Write $Y=g(S)$ and $Z=\min (S,M)$. Their common expected value is $c$. Expanding about the retention gives the requested variance identityThe stop loss variance minimization principle follows from a pointwise comparison. For $0\le x\le M$, the constraint $0\le g(x)\le x$ impliesFor $x>M$, the squared distance of $\min (x,M)=M$ from $M$ is zero, so the same squared-distance comparison is immediate. ThereforeSubtracting the same $(M-c{)}^2$ proves optimality of the retained stop loss payout:Since $Z$ is bounded, its variance is finite; if $\mathbb{E}[Y^2]=\infty$, the inequality remains valid with infinite variance on the left. When it is finite, equality requires $g(S)=\min (S,M)$ almost surely, because the pointwise squared-distance inequality is strict whenever the two payouts differ.

In the Bühlmann model, a latent risk parameter $\Theta$ is drawn from a population distribution. Conditional on $\Theta$, the yearly observations are independent and identically distributed random variables, with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. Define the structural parametersHere $v$ is the expected process variance, while $a$ is the variance of hypothetical means. The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ from the observed claims, under mean squared error. Predicting the next claim gives the same affine estimate: the extra conditional observation noise contributes the constant $v$ to the prediction error.

The law of total variance and conditional independence giveAn affine estimate can be written as $\hat{m}=m+{\sum }_{j=1}^n{b}_j(X_j-m)$: for any chosen $b_j$, optimizing the constant makes its expected value equal to $m$. The normal equations for the linear least-squares projection areFor $v>0$ they force all $b_i$ to agree, with $b_i=a/(v+na)$. Thus the credibility factor and premium areThe credibility factor increases with the observation count and between-risk variance, and decreases with within-risk variance. If $a=0$, the risk mean is known and $Z=0$; if $v=0$ and $a>0$, one observation reveals it and $Z=1$. If both vanish, the premium is the fixed value $m$ and the factor is immaterial.

In the specified model, the conditional law is a gamma distribution with shape $\alpha$ and scale $\theta$. ThereforeThe prior is an inverse-gamma distribution with shape $k$ and scale $\lambda$. To obtain its moments directly, substitute $y=\lambda /\theta$ in the defining integral, obtainingThe Gamma function recurrence yieldsThe assumption $k>2$ makes both structural variances finite. HenceIt follows that the model-specific credibility estimate is

For the Bayes estimator under squared error loss, the quantity to estimate is $\alpha \Theta$, so the optimum is its posterior mean. The likelihood function, viewed as a function of $\theta$, is proportional toMultiplication by the prior shows gamma scale inverse-gamma conjugacy:Although the printed hint only mentions integer shapes, the same substitution and Gamma integral normalize this posterior for every positive real shape, so no integrality of $\alpha$ is needed. Its posterior mean gives the Bayesian estimate and comparisonThis exact Bühlmann credibility for gamma claims holds for every observed sample, not merely on average. Here the posterior mean is affine in the sample mean, so the best affine Bühlmann credibility premium is also the unrestricted Bayes estimator under squared error loss.

Multiple imputation by chained equations specifies one conditional distribution for each incomplete variable. In general, let $V_1,\dots ,V_q$ be the incomplete variables, $Z$ the fully observed predictors, and ${\varphi }_j$ the parameters of a model $p_j(V_j\mid V_{-j},Z;{\varphi }_j)$. The notation $V_{-j}$ means all variables except $V_j$. The models should respect each variable's support and contain the terms needed by the planned analysis.

First choose suitable models and initialise only the missing entries with admissible provisional values. One sweep updates $j=1,\dots ,q$ in turn. At update $j$, fit its model to records whose $V_j$ is observed, using the latest filled-in predictors. DrawThe first draw represents parameter uncertainty through a Bayesian posterior; the second represents residual uncertainty through a posterior predictive distribution. Replace the missing entries with these draws, leaving every genuinely observed entry fixed. Repeat whole sweeps until the chain has settled, assessing traces, mixing and plausibility. Obtain $M$ completed datasets from independently initialised chains, or sufficiently separated draws after convergence. Both parameter and missing-value uncertainty must be propagated; repeated deterministic prediction is not multiple imputation.

Here use a logistic regression for $C$ conditional on $P$, age and sex. For $P$, use an appropriate discrete count model conditional on $C$, age and sex, such as a suitably checked negative binomial regression, allowing enough flexibility for skewness and nonlinear effects. Include interactions or transformations needed by the analysis in the imputation models. If condom use is structurally undefined for a record, preserve that applicability rule rather than treating it as ordinary missing data.

Impute the eligible survey information before discarding records solely because the observed $P$ is absent. In each completed dataset select its own $P>1$ records and fit the requested logistic regression. This allows a record with missing $P$ to contribute to the target group in some imputations and not others. Models must respect any known eligibility restrictions; if eligibility was independently known, use that information.

For a scalar regression coefficient, let ${\hat{Q}}_m$ be its estimate and $U_m$ its estimated variance in completed dataset $m$. Rubin's rules giveThe standard error is $\sqrt{T}$. Use the corresponding multiple-imputation confidence interval, rather than treating one completed dataset as fully observed. For vector coefficients replace the squared differences by outer products and $U_m$ by the covariance matrix. This combines within-imputation and between-imputation uncertainty.

Treat the extra measurements as auxiliary variables for multiple imputation. Include alcohol consumption and disease history as predictors in both the $C$ and $P$ conditional distributions, together with age, sex and the other incomplete response. If these auxiliary variables themselves have missing entries, give them appropriate models and update them within the same multiple imputation by chained equations sweeps. Preserve observed entries throughout.

Useful auxiliary measurements can improve predictions, precision and the plausibility of missing at random. Their association with the missing responses reduces uncertainty in the imputed values; association with nonresponse may explain some selection that otherwise depends on unobserved values. This can reduce selection bias under the enlarged missing at random model and reduce between-imputation variance. It does not guarantee elimination of missing not at random dependence. They need not be added to the final substantive logistic regression merely because they help imputation: the final analysis should still estimate its intended association.

Use $w_i=1/v_i$ for the inverse within-study variances, and $m=21$. For Cochran's Q statistic, the DerSimonian–Laird estimator of the between-study variance isConsequently the estimated heterogeneity variance and standard deviation areThese describe dispersion of underlying study log odds ratios, not ordinary sampling error of a single trial. The between-study heterogeneity is additional to the within-study variances.

The I-squared statistic isIt estimates the share of variation beyond that expected from sampling error in this collection of studies. Approximately half the variation is attributed to heterogeneity on that scale; this is neither the percentage of trials with different effects nor a percentage change in mortality. The PDF omits the symbol naming its heterogeneity estimate, so both ${\hat{\tau }}^2$ and $\hat{\tau }$ are reported explicitly.

A fixed-effect meta-analysis assigns inverse-variance weights. Using the unrounded within-study variance from part (b), this trial has $w=8$, so its percentage weight isUsing the printed rounded standard error of $0.35$ instead gives approximately $2.04\%$; the raw cell calculation yields the more precise answer.

For a random-effects meta-analysis, the corresponding weight and percentage weight areWith ${\hat{\tau }}^2=0.05$, its unnormalised weight is $40/7\approx 5.714$. The supplied sums of fixed-effect weights do not determine the denominator of the random-effects percentage; the individual $v_i$ are needed. Positive heterogeneity makes the weight distribution more even. Percentage weight measures the trial's contribution to a pooled mean with fixed weights; total influence can also involve re-estimation of heterogeneity.

For an independent subgroup contrast in meta-analysis, the difference of disjoint independent subgroup estimates has estimated variance equal to the sum of the two variances. The subgroup contrast and standard error areThe Wald test of equal subgroup mean log odds ratios therefore givesUnder the null, $Z$ is approximately standard normal and $Z^2$ has a chi-squared distribution with one degree of freedom. A quantile-2 confidence interval for the difference is $(0.29-0.8,0.29+0.8)=(-0.51,1.09)$. There is no statistically persuasive evidence of a subgroup difference. The result does not establish equal effects. Nor does a between-trial subgroup comparison, by itself, identify a causal advantage of one technique: other study characteristics may differ.