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Retain each of $N$ objects independently with probability $r$, independently of the nonnegative integer-valued random variable $N$. The retained count $Y$ has the displayed conditional binomial distribution. The probability generating function of an independent random sum gives $G_Y(s)=G_N(1-r+rs)$. If $N$ has finite variance, the law of total expectation and law of total variance giveThe first term is the average conditional variance of the independent retention decisions; the second is the variance in their conditional expected value.

Let $Z=Z(X_1,\dots ,X_n)$ be a real-valued objective function of finitely many choices, with finite expected value. Suppose a prefix $P=(X_1=x_1,\dots ,X_i=x_i)$ has positive probability. Write $q(P)=\mathbb{E}[Z\mid P]$. If the next choice is a fair bit, the law of total expectation givesAt least one child is at least this average. Choose a child with the larger value, using a fixed rule for ties. The resulting loop invariant is $q(P)\ge \mathbb{E}Z$. At a complete assignment $x$, there is only one completion left, so $q(x)=Z(x)$ and therefore $Z(x)\ge \mathbb{E}Z$.

For a biased choice or more than two possibilities, use the corresponding weighted average:Some positive-probability child is at least the weighted average. For minimization, choose a child at most the current conditional expectation and reverse the invariant. The averaging identity is the same in both directions.^[ref]

The word "probabilities" in the method's name also covers an indicator random variable for an undesirable outcome: keeping its conditional expectation, which is its conditional probability, strictly below one eventually forces that indicator to be zero. Optimizing an ordinary numerical objective function is the method of conditional expectations formulation used below.

Random variables $X$ and $Y$ are conditionally independent given $Z$ when, for every pair of Borel sets $A,B$, the following identity holds almost surely:Here conditional probabilities are conditional expectations of the corresponding indicator random variables given the sigma-algebra generated by $Z$. This defines the condition even when $\mathbb{P}(Z=z)=0$. In terms of regular conditional distributions, the conditional joint law is the product of the conditional marginal distributions for almost every conditioning value.

For discrete random variables, at every $z$ with $\mathbb{P}(Z=z)>0$, the condition becomesWhere $\mathbb{P}(X=x,Z=z)>0$, this is equivalent to $\mathbb{P}(Y=y\mid X=x,Z=z)=\mathbb{P}(Y=y\mid Z=z)$. Once $Z$ is known, observing $X$ provides no further information about $Y$.^[ref]

For a worked example, suppose a latent variable $T$ chooses between two environments with $\mathbb{P}(T=0)=\mathbb{P}(T=1)=1/2$. Given $T=t$, take $X,Y$ to be independent and identically distributed with a Bernoulli distribution of parameter $p_t$, where $p_0=0.1$ and $p_1=0.9$. Think of two separate trials run under the same quiet or active environment. Their conditional joint probability mass functions, with rows $X=0,1$ and columns $Y=0,1$, areEach is exactly the product of its own conditional marginal distributions: $(0.9,0.1)$ for $T=0$, and $(0.1,0.9)$ for $T=1$. Thus $X,Y$ are conditionally independent given $T$.

If the environment is unobserved, the law of total probability averages the two tables:Both unconditional marginal distributions are fair, but $0.41\ne 0.5\times 0.5$. The law of total expectation gives

Observing $X=1$ makes the active environment more plausible, so it raises $\mathbb{P}(Y=1\mid X=1)$ to $0.41/0.5=0.82$. Knowing the actual environment removes this source of statistical dependence. The covariance induced by a shared latent variable lemma explains the calculation without listing all the outcomes.

This is why conditional independence is useful in statistical modelling: a model can multiply simple conditional laws and then average over a shared latent variable. The resulting observations can remain statistically dependent. Conditional independence graphs organize such information relations; their interpretation as causes requires additional assumptions from causal inference. Conditioning describes the information available to an observer. It does not, by itself, describe what would happen if an experimenter changed a variable.

The conditional variance measures the remaining squared uncertainty after specified information has been used to form a conditional mean. For a real square-integrable random variable $X$ and a sub-sigma-algebra $\mathcal{G}$, let $m=\mathbb{E}[X\mid \mathcal{G}]$. ThenIt is a $\mathcal{G}$-measurable random variable, nonnegative almost surely, rather than necessarily a single number. With the information $\mathcal{G}=\sigma (Y)$, its value is the variance in the group identified by $Y$. Its expected value is the mean squared error of predicting $X$ by its conditional mean.^[ref]

In the exact mixture distribution in law of total expectation, group $A$ contains $0,4$ around mean two, and group $B$ contains $6,10$ around mean eight. Both groups have conditional variance four because their deviations from their own means are $-2,2$ with equal probabilities. Therefore $\mathbb{E}[\mathrm{Var}(X\mid Y)]=4$. Meanwhile, the two group means have variance $\frac{1}{4}(2-13/2{)}^2+\frac{3}{4}(8-13/2{)}^2=27/4$. The law of total variance givesDirect calculation gives $\mathbb{E}[X^2]=53$ and $\mathrm{Var}(X)=53-(13/2{)}^2=43/4$, checking the decomposition.

For any square-integrable $\mathcal{G}$-measurable predictor $Z$, the error identity isThe cross term vanishes because $\mathbb{E}[X-m\mid \mathcal{G}]=0$. Thus using the conditional mean minimizes mean squared error, while conditional variance quantifies what this information still cannot predict. This is the orthogonal projection interpretation in L2 space.^[ref]

Let $A,B$ be indicator random variables with conditional independence given a latent variable $T$, and suppose $\mathbb{E}[A\mid T]=\mathbb{E}[B\mid T]=g(T)$. ThenFor a short proof, conditional independence gives $\mathbb{E}[AB\mid T]=g(T{)}^2$. The law of total expectation gives $\mathbb{E}[AB]=\mathbb{E}[g(T{)}^2]$ and $\mathbb{E}[A]=\mathbb{E}[B]=\mathbb{E}[g(T)]$. Subtract their product to obtain the claimed variance. Since $0\le g(T)\le 1$, all the required expected values exist.

In the two-environment example of conditional independence, $g(T)$ is $0.1$ or $0.9$ with equal probabilities. Hence $\mathbb{E}[g(T)]=0.5$ and $\mathrm{Var}(g(T))=((0.1-0.5{)}^2+(0.9-0.5{)}^2)/2=0.16$. The two trials have zero conditional covariance in either environment, yet their unconditional covariance is positive because their success probabilities rise and fall together.

More generally, square-integrable conditionally independent random variables satisfyby the law of total covariance. Equal conditional means reduce this to the displayed variance; unequal conditional means need not give a nonnegative covariance. The equal-mean result explains correlated component evolutionary states in a coeval binary population without correlation between initial masses: variation in a shared environment can supply the entire covariance.

For a simple symmetric random walk starting at zero, let $T_1$ be its first hitting time of $1$. The infinite mean first passage of a simple symmetric random walk shows that $T_1<\infty$ almost surely but $\mathbb{E}[T_1]=\infty$. Since $S_{T_1}=1$, assuming a finite expected value for $T_1$ would let Wald's equation assert $1=0\mathbb{E}[T_1]=0$, a contradiction. The formal expression $0\cdot \infty$ is not an allowed use of the theorem.^[ref]

Compare $T_{\pm 2}=\inf \{n\ge 0:|S_n|=2\}$. From every surviving position in $\{-1,0,1\}$, three consecutive positive increments force absorption within three steps. Each fresh block has probability $1/8$ of this event, so $\mathbb{P}(T_{\pm 2}>3k)\le (7/8{)}^k$ and $\mathbb{E}[T_{\pm 2}]\le 24$. This establishes finiteness before invoking the theorem. If $e_j$ is the average remaining time from position $j$, the law of total expectation givesso $\mathbb{E}[T_{\pm 2}]=4$. Symmetry gives equal exit probabilities at $-2$ and $2$, and $\mathbb{E}[S_{T_{\pm 2}}]=0$, as Wald's equation predicts. This is a small gambler's ruin problem with two absorbing barriers.

Even for $T_1$, each bounded stopping time $T_1\wedge n$ satisfies $\mathbb{E}[S_{T_1\wedge n}]=0$. The stopped process tends to $1$ almost surely, but its expected value does not tend to $1$. Rare long negative excursions retain the balancing contribution. The existing stopped-walk martingale with almost sure but not L1 convergence explains the related failure of uniform integrability; almost sure convergence alone does not permit exchanging a limit and an expected value.

For square-integrable random variables, expand $X$ and $Y$ into their conditional expectations and their centered conditional residuals. The cross terms vanish by the law of total expectation, giving the displayed covariance decomposition. conditional independence can therefore coexist with positive unconditional covariance from a shared latent variable.

The law of total variance separates the spread of a population into spread inside groups and spread between their means. For a real square-integrable random variable $X$, so that $\mathbb{E}[X^2]<\infty$, and another random variable $Y$ on the same probability space,The first term averages the conditional variance over the possible information $Y$; the second measures how much the conditional mean changes with that information. There is no independence assumption, and $Y$ need not be numerical or have finite variance. Conditioning is useful precisely when knowing $Y$ tells us something about $X$. See MIT tutorial on using conditional expectation and variance.

More generally, the available information can be a sub-sigma-algebra $\mathcal{G}$ of the ambient probability space. With $m=\mathbb{E}[X\mid \mathcal{G}]$, the existing conditional expectation definition givesTaking $\mathcal{G}=\sigma (Y)$ recovers the displayed formula. The conditional mean and conditional variance are themselves random variables: before learning the group, we do not yet know which group average or group spread will apply. Their outer expectation and variance turn these varying quantities into numbers.

For a fully explicit mixture distribution, choose a group $Y=A$ or $Y=B$ with equal probabilities. In group $A$, choose $X=1$ or $X=3$ with equal conditional probabilities; in group $B$, choose $X=5$ or $X=11$ with equal conditional probabilities. Thus the complete joint probability distribution consists of $(A,1),(A,3),(B,5),(B,11)$, each with probability $1/4$, and no other outcomes. The conditional means are $2$ and $8$. Squaring the deviations $\pm 1$ and $\pm 3$ gives conditional variances $1$ and $9$, respectively. The law of total expectation gives the overall mean $\mu =(2+8)/2=5$.

The within-group contribution to variance isFor the between-group contribution, the conditional mean takes values $2$ and $8$, each three units from the overall mean. Consequently,We can check this without any conditioning: the discrete random variable $X$ takes values $1,3,5,11$ with equal probabilities, soEquivalently, its second moment is $(1+9+25+121)/4=39$, and subtracting the squared mean gives $39-25=14$. These are population variances, with the stated probabilities as weights, rather than estimates using the corrected denominator of a sample variance.

The figure makes the two sources of spread visible. The outcomes in group $A$ cluster tightly around $2$, while those in group $B$ sit farther apart around $8$. Replacing every outcome by its conditional mean removes the internal spread but leaves the separation between groups. Conversely, subtracting each group's conditional mean aligns the two groups around zero. Averaging only the conditional variances would miss nine units of variance; retaining only the conditional means would miss five. Neither summary alone describes the whole population.

Two modifications clarify what each contribution measures. Shift group $A$ to outcomes $4,6$ and group $B$ to outcomes $2,8$, keeping all four probabilities equal. Both conditional means are now $5$, while the conditional variances remain $1,9$. The between-group contribution vanishes, and total variance is just $5$. Group membership still tells us how uncertain an outcome is, even though the best point prediction stays the same. Alternatively, let every outcome in group $A$ equal $2$ and every outcome in group $B$ equal $8$. Now both conditional variances vanish, and all nine units of total variance come from group separation. Knowing the group predicts the outcome perfectly. These limiting cases distinguish differences in average outcome from differences in internal spread; the original mixture distribution contains both.

The same calculation answers a prediction question. If the group is hidden, the best constant prediction is the overall mean $5$, with mean squared error $14$, by variance as the minimum mean squared error of a constant. If the group is revealed, predict $2$ for $A$ and $8$ for $B$. The errors become $-1,1,-3,3$, and their mean squared error is $5$. Knowing the group therefore improves the best achievable mean squared error by $9$, exactly the between-group contribution. This is an interpretation relative to the information available: the remaining uncertainty could shrink further if additional information were revealed.

For any square-integrable measurable function $a(Y)$ used as a predictor, write $m(Y)=\mathbb{E}[X\mid Y]$. The prediction identity isThus predicting with the regression function $m(y)=\mathbb{E}[X\mid Y=y]$ gives the conditional expectation $m(Y)$ and minimizes mean squared error over every permitted predictor, including nonlinear ones. This minimizes average squared error, not the error for each individual outcome. See Berkeley Statistics 201A notes, sections 36–37.

Why do the two pieces add cleanly? Put $R=X-m(Y)$. Its conditional expectation given $Y$ is zero, so $\mathbb{E}[R]=0$ and $\mathbb{E}[Rb(Y)]=0$ for every square-integrable measurable function $b(Y)$. In particular, $R$ and $m(Y)$ are uncorrelated random variables. Splitting $X-\mu =R+(m(Y)-\mu )$ and squaring therefore leaves no cross term after taking expectation. Geometrically, the conditional expectation is an orthogonal projection in the L2 space, and the law of total variance is its Pythagorean theorem for centered random variables.

In our example, the four products $R(m(Y)-5)$ are $3,-3,-9,9$, which average to zero. Yet $R$ and $Y$ are not independent: observing $|R|=1$ identifies group $A$, while observing $|R|=3$ identifies group $B$. This is a concrete instance of heteroscedasticity, with different conditional variances despite zero conditional means of the errors. A group can predict uncertainty even when it cannot predict the error's sign.

The law of total variance is useful whenever aggregated data conceal distinct groups. Changing mixture weights changes both contributions, so the weighting must match the population being described. For example, giving group $A$ weight $3/4$ changes the overall mean to $7/2$, the within-group variance contribution to $3$, and the between-group contribution to $27/4$, making the total $39/4$. Group differences account for spread and prediction gains; assigning them a causal explanation requires a separate causal inference argument.

Consider a random walk on $\{0,1,\dots ,n\}$ with absorbing states $0,n$. At each interior state it stays in place with probability $q$, and moves one position in either direction with probability $(1-q)/2$, where $0\le q<1$. Its mean absorption time from $k$ isProof. A sequence of at most $n$ consecutive leftward moves ensures absorption, with probability at least $((1-q)/2{)}^n>0$. Dividing time into blocks of length $n$ gives a geometric series bound on the survival probabilities, so the expected value is finite. Let $e_k={\mathbb{E}}_k[T]$. The law of total expectation and first-step analysis giveThus $e_{k+1}-2e_k+e_{k-1}=-2/(1-q)$. A particular solution of this linear recurrence relation is $-k^2/(1-q)$; its homogeneous solutions are $A+Bk$. The two boundary conditions give $A=0$, $B=n/(1-q)$, proving both the formula and its uniqueness. Holding slows the usual symmetric random walk by the factor $1/(1-q)$.

For integrable random variable $X$ and nested sub-sigma-algebras $\mathcal{H}\subseteq \mathcal{G}$, the tower property saysThe finer-information conditional mean can be averaged within each coarser group, recovering the coarser-information conditional mean. Conversely, conditioning an already $\mathcal{H}$-measurable conditional mean on $\mathcal{G}$ leaves it unchanged. Both orders therefore recover $\mathbb{E}[X\mid \mathcal{H}]$ when the information is nested.^[ref]

For the finite example in law of total expectation, exact knowledge of $X$ gives the finer mean $X$ itself. Averaging within the groups $A,B$ recovers two and eight. Forgetting the group then gives $13/2$. The nesting condition is essential; iterated conditional expectation over nonnested sigma-algebras describes why arbitrary successive conditioning cannot simply be replaced by conditioning on an intersection.

Let $N$ have a Poisson distribution with parameter $\lambda$, and let $Y_1,Y_2,\dots$ be independent random variables with a Poisson distribution of common parameter $\theta$, independent also of $N$. Then $Z={\sum }_{i=1}^N{Y}_i$ has a compound Poisson distribution, with $Z\mid N\sim \mathrm{Poisson}(\theta N)$. The law of total expectation yields the displayed probability generating function and expected value $\lambda \theta$. The law of total variance gives variance $\lambda \theta (1+\theta )$, exceeding the mean when both parameters are positive. This extra variation comes from the random number of groups. Jerzy Neyman is the statistician for whom the distribution is named.

A family of disjoint events whose union is the sample space. For the countable law of total probability and law of total expectation, use a finite or countably infinite family. Conditioning formulas sum only positive-probability parts, or interpret zero-probability contributions as zero without assigning a conditional expectation there.

There is no proof of the printed assertion: it is false for the stated expectations. The original explicitly defines $x_j$ as the expected remaining count, so replacing it by a deterministic upper envelope changes the problem.^[ref]

Let $B_j$ be the actual count after $j$ rounds and $x_j=\mathbb{E}{B}_j$. The union bound gives the valid conditional estimateBy the law of total expectation, the resulting unconditional bound isDiscarding the variance term is an expectation closure error for a nonlinear recurrence. In particular, Jensen's inequality says $\mathbb{E}{B}_j^2\ge x_j^2$, which cannot justify the requested upper bound.

For an exact counterexample, start with two balls and two bins. A round serves both balls if they land separately and serves neither if they collide. Each outcome has probability $1/2$, independently in successive rounds while both balls remain. ConsequentlyAt $j=2$, the requested inequality would say $x_3\le x_2^2/2$, but

If the bound in part (c) were available, the intended answer would beIndeed $2^{j_{\mathrm{intended}}(n)-2}\ge {\log }_{2}n$, so that bound would give $x_{j(n)}\le n{2}^{-{\log }_{2}n}=1$. Its growth is $O(\log \log n)$ in Big O notation. Because (c) is false, this substitution alone does not prove the result for the actual repeated singleton-removal allocation.

The requested order can nevertheless be established for the actual process. Here is a concrete, deliberately generous replacement, with $j(0)=0$ for the empty system:The large cutoff makes the numerical estimates below simple. A finite initial range does not affect the $O(\log \log n)$ asymptotic analysis.

First, whenever $2\le b\le n$, a ball is served with probability at leastThus part (a) gives $\mathbb{E}[B_{t+1}\mid B_t]\le 2B_t/3$, also for $B_t\le 1$. The law of total expectation yields $x_t\le n(2/3{)}^t$. This proves $x_{j(n)}\le 1/2$ for the small-$n$ branch, including $n=1$, whose single ball is served immediately.

For the large-$n$ branch, we need control of the random count, rather than an unjustified recurrence for its expectation. If $b$ balls are allocated in a round, changing one chosen bin changes the number of singleton bins by at most two. The surviving count $F_b$ therefore has the bounded differences property with constants $2$. McDiarmid's inequality givesThis can be applied conditionally after any history, since the next allocations are fresh independent random variables.

Use this concentration inequality in three stages. Write $L=\lceil {\log }_2{\log }_{2}n\rceil$.

Every exceptional probability above is at most $e^{-n^{1/6}/300}$, and there are at most $L+10$ opportunities for an exception. For $n\ge 2^{120}$, the elementary bounds $n^{1/6}/300\ge 4\ln n$ and $L+10\le n$ give, by the union bound, total exceptional probability at most $n^{-3}$. The first numerical inequality holds at the cutoff and remains true because $n^{1/6}/\ln n$ is increasing there. On an exceptional history there are at most $n$ balls. Taking the last round's conditional expectation and then averaging givesThis proves the displayed replacement $j(n)$ for the literal process without relying on either false assertion.

Order the edges arbitrarily and fix their colours one at a time. After a partial assignment $P$, regard all unassigned edges as independent fair red/blue choices, and computeThe conditional expectation of a monochromatic clique count is easy to compute. For each four-vertex subset $A$, let $r$ be the number of its six edges still unassigned. Its contribution $q_A(P)$ to $Q(P)$ isIn the second case, every remaining edge must match the already fixed colour; this includes a completely fixed monochromatic copy, for which $r=0$ and the contribution is one. In the last case, the six unassigned edges may become either all red or all blue. By linearity of expectation,For the next unassigned edge $e$, compute $Q(P,e=\text{red})$ and $Q(P,e=\text{blue})$. The law of total expectation givesChoose the colour with the smaller value, resolving a tie by a fixed rule. At least one is at most their average, so the loop invariant is that $Q$ never increases. Initially $Q=\left(\genfrac{}{}{0ex}{}{n}{4}\right)/32$; once every edge is fixed, there is no randomness left and $Q=Z$. Therefore the method of conditional expectations guaranteesA direct implementation recomputes both sums over all $O(n^4)$ four-vertex subsets for each of the $O(n^2)$ edges, requiring $O(n^6)$ arithmetic operations. The comparisons can be exact: every contribution is a multiple of $1/32$, so store $32Q$ as an integer with $O(\log n)$ bits. This gives $O(n^6\log n)$ bit complexity with elementary arithmetic, and hence a deterministic polynomial time construction. No random sampling is performed by this derandomization.

Where the moment-generating function $M_Y(t)=\mathbb{E}{e}^{tY}$ is finite in a neighborhood of zero, define the cumulant-generating function ${\kappa }_Y(t)=\log M_Y(t)$ and the $r$th cumulant by ${\kappa }_r(Y)={\kappa }_Y^{(r)}(0)$. Differentiation givessince $M_Y(0)=1$. Thus the first two cumulants are the mean and variance.

Conditional on $N=n$, independence gives $\mathbb{E}(e^{tT}\mid N=n)=M_V(t{)}^n$. Summing over the Poisson distribution yieldsHence the compound Poisson cumulants giveFor the moment statements, finite second moments suffice even when a moment-generating function does not exist near zero: conditioning gives $\mathrm{Var}T=\mathbb{E}N\mathrm{Var}V+\mathrm{Var}N(\mathbb{E}V{)}^2=\nu \mathbb{E}{V}^2$. This distinguishes the raw second moment from the variance of an individual claim.

For a single flood, the law of total expectation and law of total variance similarly giveCondition on the number of floods and apply the same laws once more. All floods have these moments and are independent of $M$, so the annual total satisfiesThe additional $\nu {\lambda }^2{\mu }^2$ term is the variation in the number of entire flood clusters; it would be lost by treating individual claims as an ordinary Poisson stream.

For the annual count $N={\sum }_{i=1}^M{N}_i$, its probability generating function isThis is the nested Poisson flood count. Differentiating, or conditioning on $M$, gives $\mathbb{E}N=\nu \lambda$ and $\mathrm{Var}N=\nu \lambda (1+\lambda )$. Thus $N$ is not Poisson when $\nu ,\lambda >0$, since its variance exceeds its mean. If either parameter is zero it is identically zero, a degenerate Poisson distribution.

Let $K_i$ be the number of valid claims in flood $i$. Independent validity marking gives $K_i\mid N_i\sim \mathrm{Binomial}(N_i,1-p)$. Its generating function isso Poisson thinning gives $\boxed{K_i\sim \mathrm{Poisson}(\lambda (1-p))}$. With validity independent also of claim amount, each claim contributes expected paid amount $(1-p)\mu$. Therefore the annual expected payment is $\boxed{\nu \lambda (1-p)\mu }$. The independent-mark assumption is needed for this multiplication; a size-dependent invalidity rule would require the joint expected value of amount and validity instead.

A credibility factor is the weight assigned to a particular risk's observed experience when blending it with a population or manual premium. In the Bühlmann model, the blend is the best affine mean-square predictor, rather than automatically the full posterior mean.

Let $v=\mathbb{E}[{\sigma }^2(\theta )]$, the expected process variance, and assume the displayed second moments are finite. The law of total expectation gives $\mathbb{E}{X}_i=m$. The law of total variance and law of total covariance giveFor the second formula, the conditional covariance is zero by conditional independence, while the covariance of the conditional means is $\mathrm{Var}\mu (\theta )=a$, the variance of hypothetical means. Likewisebecause $\mathbb{E}[X_i-\mu (\theta )\mid \theta ]=0$.

To derive the Bühlmann credibility premium, minimize $\mathbb{E}[(\mu (\theta )-b_0-{\sum }_i{b}_i{X}_i{)}^2]$. Differentiating in $b_0$ sets the predictor's mean to $m$. The remaining normal equations areFor $v>0$, subtracting any two rows forces all $b_i$ to be equal, then gives $b_i=a/(v+na)$. Thus the premium and factor areThe same affine premium predicts $X_{n+1}$: conditional independence makes $\mathbb{E}[(X_{n+1}-\hat{\mu }{)}^2]=v+\mathbb{E}[(\mu (\theta )-\hat{\mu }{)}^2]$, so minimizers agree. If $v=0$, every year's amount equals $\mu (\theta )$ almost surely, and $Z=1$ gives the exact risk mean even though the coefficient vector need not be unique.

For fixed $a>0,n$, $\partial Z/\partial v=-na/(na+v{)}^2<0$. Thus more within-risk variation decreases credibility: individual experience becomes less informative, $Z\to 0$ as $v\to \infty$, and the premium tends to $m$. Increasing the number of observed years has the opposite effect.

For numerical estimation from a balanced panel, assume the $k$ risks are independent draws from the same risk population, with comparable exposures, and $n,k\ge 2$. SetConditionally, each within-risk sample variance has mean ${\sigma }^2({\theta }_s)$; hence $\mathbb{E}\hat{v}=v$. The risk sample means have variance $a+v/n$, so $\mathbb{E}B=a+v/n$. Therefore $\tilde{a}=B-\hat{v}/n$ is an unbiased method of moments estimate of $a$. For a usable nonnegative variance estimate take $\hat{a}=\max (0,\tilde{a})$, noting that this truncation sacrifices exact unbiasedness.

The balanced-panel method-of-moments credibility estimate for risk $s$ is consequentlyIf $\hat{a}=0$ and $\hat{v}>0$, use zero credibility. If both estimates are zero, the within and between sums of squares are zero and all the observed amounts coincide; any blend gives that same premium. With only one year per risk or only one risk, these within/between estimators are unavailable, so external structural estimates or additional assumptions are necessary.

Conditional on $N=n$, the $n$ type indicators are independent random variables with the Bernoulli distribution of parameter $p$. Thus $F\mid N=n$ has the binomial distribution and its probability generating function is $(1-p+ps{)}^n$. The general rule used here is that for a nonnegative integer-valued random variable $X$, $G_X(s)=\mathbb{E}[s^X]$; for a sum of independent such variables, their probability generating functions multiply, since the expected value of the product $s^{X_1}\cdots s^{X_n}$ factors. The law of total expectation then averages the conditional result:This calculation uses the given independent labeling of objects, including conditional on their number.

Condition on the claim count. The empty random sum of independent claims is zero, and independence gives $\mathbb{E}[e^{tS}\mid N=n]=M_X(t{)}^n$. Therefore the random-sum transform identity iswherever the composition is finite. This follows directly from the law of total expectation, rather than replacing the random count by its mean.

Put $q=1-p$. For the specified negative binomial distribution, the negative binomial series gives $G_N(z)=[p/(1-qz){]}^k$. A rate-$\lambda$ exponential distribution has moment-generating function $M_X(t)=\lambda /(\lambda -t)$. HenceThe last expression is exactly the moment-generating function of a compound binomial distribution with independent count and severitiesThe moment-generating function exists in a neighborhood of zero, so its uniqueness identifies the distributions. In particular, the new claim sizes have mean $1/(p\lambda )$, not $1/\lambda$, and both aggregate distributions have an atom $p^k$ at zero. This is the negative-binomial sum of exponential claims representation with the severity rate explicitly retained.

Conditional on $N=n\ge 1$, the aggregate has Erlang distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$ on $s>0$. Repeated integration by parts of its tail givesIndeed, after substituting $v=\lambda s$, the integral ${\int }_{\lambda x}^{\infty }{e}^{-v}{v}^{n-1}dv/(n-1)!$ reduces recursively to the corresponding integral with exponent $n-2$, terminating at $e^{-\lambda x}$. Applying the law of total probability yields the original infinite mixture distribution tailApplying the same argument to the compound binomial distribution gives instead the finite Erlang-mixture tail for a negative-binomial exponential aggregate:The two tails are equal. The binomial representation replaces an infinite sum by a finite exact calculation, eliminating the need to choose a claim-count truncation and control its omitted tail. The inner sums can be evaluated recursively, using $a_{j+1}=a_j(p\lambda x)/(j+1)$ for their exponential-weighted terms.

The first approach specifies marginal means, variances and an exchangeable correlation for each student's repeated counts. It is a moment model suitable for a generalized estimating equation; equality of the mean and variance alone does not specify a full Poisson joint distribution. Different students supply independent sampling clusters, while the three counts within a student are dependent.

The second approach is a Poisson generalized linear mixed model with a positive shared random multiplier $U_i=e^{b_i}$. Conditional on it, the three counts are independent Poisson variables; integrating over it gives a fully specified dependent joint law. The gamma random-intercept Poisson model uses Gamma shape ${\tau }^2/\theta$ and rate $\tau /\theta$, so $\mathbb{E}{U}_i=\tau$ and $\mathrm{Var}(U_i)=\theta$. Treat the covariates as fixed, with the same multiplier distribution across covariate/treatment groups, as required by this model.

Let $a_{ij}=\exp ({\beta }_0+{\beta }^{\mathsf{T}}{x}_{ij})$ and $m_{ij}=\mathbb{E}{Y}_{ij}$. The law of total expectation, law of total variance and law of total covariance giveThus each marginal is a negative binomial distribution by the Poisson-gamma mixture, rather than generally a Poisson variable. Put $\kappa =\theta /{\tau }^2$. Its variance is $m_{ij}+\kappa m_{ij}^2$, and the marginal correlation isIt is positive but usually depends on the two marginal means, so it need not be the common exchangeable correlation proposed in the first approach. The first method focuses on population mean effects with a chosen working correlation matrix; the second models latent student heterogeneity and separates conditional Poisson variation from additional marginal variation. Correctly specified likelihood inference for the second uses more distributional assumptions, while the first can retain valid mean inference despite a wrong working covariance by using a cluster-level sandwich variance.