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Condition on the count and use the independence of random variables of the claim sizes. For $n\ge 1$ their joint exponential transform factors, while for $n=0$ the empty sum contributes $1$. Consequently the law of total expectation givesThis random sum of independent claims transform uses both independence assumptions: the count must be independent of the entire claim-size sequence, and the sizes must be mutually independent with the same probability distribution. The probability generating function is interpreted through its defining nonnegative series. The identity holds as a finite moment-generating function wherever that series is finite; outside that domain the expectation and series can agree at $+\infty$. In particular a positive argument may take $M_X(r)$ beyond $1$, so finiteness does not follow merely from the usual unit-disk domain of a probability generating function.

Conditional on $\Theta =\theta$, summing the individual independent Poisson counts gives $Y_j\sim \mathrm{Poisson}(m_j\theta )$. The between-year conditional independence makes the likelihood functionMultiplying by the gamma distribution prior density, proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$, gives the Poisson-gamma conjugacy with unequal exposures:For an action $t$, the posterior squared-error loss decomposes asHence the Bayes estimator under squared error loss is the posterior mean, givingFuture counts are independent of the observed years conditional on $\Theta$, so the law of total expectation then gives the posterior predictive expected countThe Bayesian and credibility estimates coincide exactly. This exact Bühlmann–Straub credibility for Poisson-gamma counts occurs because the posterior mean is already affine in the exposure-weighted experience, and therefore belongs to the class over which the credibility estimate minimizes mean squared error.

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 covariance is zero. Let ${\mathcal{F}}_t=\sigma ({\epsilon }_s:s\le t)$. The stationary noise-series representation makes $f(X_t)$ measurable with respect to ${\mathcal{F}}_t$, whereas the martingale difference sequence property gives $\mathbb{E}[X_{t+h}\mid {\mathcal{F}}_{t+h-1}]=0$. The law of total expectation therefore givesSince $\mathbb{E}{X}_{t+h}=0$, this is the required covariance. All products are integrable by the Cauchy-Schwarz inequality, using finite second moments of $X$ and the stipulated square integrability of $f(X_t)$.

The printed identity needs an expectation on its right-hand side. For fixed $r\ge 1$, condition on $X_2,\dots ,X_r$ and use independence of $X_1$:The expression inside this expectation is generally random and cannot equal the unconditional probability by itself.

For each independent Monte Carlo method replicate, draw $R_j$ from its given distribution. If $R_j=r\ge 1$, draw $r-1$ independent $F$-variables, form $W_j={\sum }_{i=2}^r{X}_{j,i}$ and set $H_j=F(x-W_j)$. For $r=1$ the empty sum is $0$. If $R$ may equal zero, use $H_j=1_{\{x\ge 0\}}$ when $R_j=0$, since $S_0=0$. The law of total expectation givesSince $0\le H_j\le 1$, the estimator has statistical consistency by the strong law of large numbers. It is the conditional Monte Carlo estimator obtained by Rao-Blackwellization of the direct indicator $1_{\{S_R\le x\}}$, and the law of total variance gives $\mathrm{Var}(H_j)\le l(1-l)$.

This version assumes that the cumulative distribution function $F$ can be evaluated. An easy sampler alone does not automatically supply an easy cumulative distribution function evaluation. If only sampling is available, use the direct indicator estimator, or replace each conditional cumulative distribution function by the average of several independent indicators $1_{\{X_{j,1,k}\le x-W_j\}}$; the nested version remains an unbiased estimator. With $L$ such inner draws its variance per outer replicate is $\mathrm{Var}(H_j)+L^{-1}E[H_j(1-H_j)]$.

Use the standard risk-neutral quasilinear utility model: utility is value received minus payment. A bidder may abstain for utility zero, payments to the seller are nonnegative, and the zero-value type has utility zero. The last normalization is necessary for the requested revenue formula; if arbitrary subsidies were allowed, an additive payment constant would remain undetermined. The independent private values model is symmetric, and each bidder has a unit-demand valuation in the two-item part. Participation satisfies individual rationality.

Let $G(\theta )$ be the interim probability of receiving an item and $m(\theta )$ the interim expected payment of a type $\theta$. Write $u(\theta )=\theta G(\theta )-m(\theta )$. By mimicking the Bayesian Nash equilibrium bid of a type $z$, a type $\theta$ could obtain utility $\theta G(z)-m(z)$. Equilibrium therefore gives, for $\theta >z$,The lower inequality uses type $\theta$'s incentive constraint, and the upper uses type $z$'s. Since $G$ is continuous in these auctions, the inequalities give $u^{\prime }(\theta )=G(\theta )$. With $u(0)=0$,This is the interim payment identity underlying revenue equivalence.

For one item, independence and the uniform distribution give $G(\theta )={\theta }^{n-1}$: all other values must be below $\theta$. Hence the expected payment from a bidder conditional on its value isThis is an unconditional-in-winning interim payment, not the amount paid conditional on winning. In the first-price sealed-bid auction, $m(\theta )=b(\theta )G(\theta )$, so the Bayesian Nash equilibrium bid iswith $b(0)=0$. For $n=1$, the seller's revenue is zero and a zero bid suffices under the stipulated allocation rule.

For two items and three unit-demand bidders, a type $\theta$ wins if at most one of the other two values exceeds it. ThereforeThe interim payment identity gives $m(\theta )={\theta }^2-2{\theta }^3/3$. Dividing by the winning probability gives the symmetric Bayesian Nash equilibrium bidTo check that this is an Bayesian Nash equilibrium rather than just a necessary formula, its derivative is $b^{\prime }(\theta )=2(1-\theta )(3-\theta )/[3(2-\theta {)}^2]$, positive for $0\le \theta <1$. A type $\theta$ mimicking a type $z$ receives utility $\theta G(z)-m(z)$, whose derivative in $z$ is $(\theta -z)G^{\prime }(z)$. It is positive before $z=\theta$ and negative after it, so the truthful type-matching bid is globally optimal. Bids above the highest Bayesian Nash equilibrium bid can only increase payment without increasing the winning probability; the usual nonnegative bid range covers the lower boundary.

By symmetry and the law of total expectation, the seller's expected two-item revenue isFor one item and the same three bidders,More generally the single-item revenue with $n$ bidders is $(n-1)/(n+1)$. Allocating a second item lowers competition enough that it adds no expected revenue in this particular three-bidder uniform model.

The final mechanism is a direct revelation mechanism because the message submitted by bidder $i$ is its valuation report, in the same type space $[0,1]$, and allocation and payment are explicit functions of those reports. It is the Clarke pivot mechanism for selecting two unit-demand winners. Truthful reporting is a dominant strategy: bidder $i$'s utility equals its true allocation value plus the other bidders' reported allocation values, minus a term depending only on the others' reports. Reporting its true value makes the efficient allocation maximize the first two terms, while the last is unaffected by its report.

Let the ordered values be ${\theta }_{(1)}\le {\theta }_{(2)}\le {\theta }_{(3)}$. For a winner $i$, the maximum welfare achievable by the other two bidders is their combined value; in the actual allocation only the other winner receives an item. Thus the difference defining its payment is the excluded bidder's value ${\theta }_{(1)}$. For the loser, both other bidders receive items already, so its payment is zero. Consequently both winners pay the lowest valuation, andThis equals the first-price sealed-bid auction revenue above. A conditional check gives the same interim payment identity: the minimum of the other two values has density $2(1-z)$, and bidder $i$ wins exactly when that minimum is below $\theta$. Its conditional expected pivot payment is ${\int }_0^{\theta }2z(1-z)dz={\theta }^2-2{\theta }^3/3=m(\theta )$. The mechanisms have equal expected revenue, even though their realized payments need not coincide.

Immediately after departure $n$, either $Q_n>0$ and one of those customers starts the next service, or $Q_n=0$ and the next arrival both initiates service and is not counted among those waiting after it. During that service $A_n$ further customers arrive. Thus the embedded departure chain of an M-G-1 queue obeys

Conditional on the next service time $S_{n+1}=s$, the Poisson process property givesThe law of total expectation and law of total variance therefore giveand

In equilibrium let $Q$ have the common distribution of $Q_n,Q_{n+1}$ and put $H=1_{\{Q>0\}}=h(Q)$. The fresh pair $(A_n,S_{n+1})$ is independent of $Q_n$. Taking expectations in the recursion givesBecause $QH=Q$ and $H^2=H$, stationarity of the second moment givesSolving,

In stationarity, level crossing together with Poisson arrivals see time averages identifies this as the mean number $L$ in the shop. The Little law gives $L=\lambda (W_q+\mathbb{E}S)$, where $W_q$ is waiting time before service. Since $\rho =\lambda \mathbb{E}S$,This is the Pollaczek-Khinchine mean formula.

Under $H_0$, conditional independence givesMultiplication by the measurable sign $s(z_1)$ and the law of total expectation prove the first identity.

WriteThenConditional orthogonality under $H_0$ and the variance bounds in (ii) giveand similarly for the term containing ${\xi }_i{a}_i$. The Cauchy-Schwarz inequality bounds the last term bywhich is $o_p(n^{-1/2})$ by $n{\mathrm{MSPE}}_f{\mathrm{MSPE}}_g\to 0$.

The leading summands ${ϵ}_i{\xi }_{i}s(z_i)$ are independent, centered, and have variance $\mathrm{Var}({ϵ}_1{\xi }_1)$ because $s^2=1$. The central limit theorem and consistency of ${\tau }_D$ therefore give, by the Slutsky theorem,This is the generalized covariance measure statistic.

Without the null, condition first on $(x_1,z_1)$. Sincethe term involving $\mathbb{E}(x_1\mid z_1)$ vanishes, givingFor the specified alternative, independence and $\mathbb{E}{x}_1=0$ make the right sideThe constant choice $s=1$ gives zero because $\mathbb{E}h(z_1)=0$, so no first-order power is expected. Takinginstead gives $\mathbb{E}(x_1^2)\mathbb{E}|h(z_1)|>0$, producing asymptotic power.

For a fixed observed colour ${\tilde{c}}_s$, Bayes' theorem givesCompleting the square shows that this is a truncated normal distribution with variance ${\sigma }_c^2$, lower endpoint zero, and untruncated locationHenceConditioning first on $E_s$ and using the law of total expectation yields the dusty colour-magnitude relationAs ${\tilde{c}}_s\to -\infty$, the inverse Mills ratio implies ${\overline{E}}_s\to 0$ with vanishing derivative, so the asymptotic slope is $\beta$. As ${\tilde{c}}_s\to +\infty$, ${\overline{E}}_s\sim {\tilde{c}}_s-c_0-{\sigma }_c^2/\tau$, so the asymptotic slope is $R$.

The no unmeasured confounding assumption, or conditional exchangeability, isAlso assume consistency of potential outcomes, no interference in causal inference, positivity in causal inference, and finite expectations. For $a\in \{0,1\}$, the law of total expectation, exchangeability, and consistency givePositivity ensures that the observed conditional means exist on the covariate support being averaged. Subtracting the two cases identifies the average treatment effect as

Under the null, ${\epsilon }_1$ and ${\xi }_1$ have conditional independence given $z_1$. The law of total expectation gives

For an integrable random variable $X$, the law of total expectation for the partition of a sample space isParts of zero probability contribute zero and need no definition of $\mathbb{E}[X\mid B_i]$. To see the identity, write $X={\sum }_{i}X{1}_{B_i}$ and use linearity of expectation; integrability justifies the countable sum since ${\sum }_i\mathbb{E}[|X|1_{B_i}]=\mathbb{E}[|X|]<\infty$.

First the expected value of the absorption time is finite. From any interior state, consecutive leftward moves lead to absorption within at most $n$ steps, with probability at least $p=3^{-n}$. By independence of the subsequent steps, the probability of surviving $jn$ steps is at most $(1-p{)}^j$. Summing this geometric series bounds the expected value of the absorption time by $n/p$.

Now apply first-step analysis. Starting from $k$ in the interior, one step is always used. Conditional on that step, the remaining expected values are $e_{k+1}$, $e_k$, and $e_{k-1}$, each with probability $1/3$. The law of total expectation therefore givesThe term $1$ remains even when the particle stays at $k$, because a step has elapsed. Rearranging gives

A partition of a sample space is a finite or countable family $(A_j)$ of pairwise disjoint events whose union is the whole sample space. If $X$ is integrable, the law of total expectation statesThis follows by summing $X{1}_{A_j}$ over the partition of a sample space; absolute integrability justifies the countable sum. For nonnegative $X$, the identity also holds with infinite expectations. Zero-probability parts contribute zero without requiring a conditional value on them.

Apply the law of total expectation to the partition of a sample space given by the possible values of $R_k$. Since the conditional mean is affine,Here $r_0=n$. This uses linearity of expectation and does not assert independence of successive urn counts.

Let $R$ denote the red die. Conditional on $R=r$, each white roll succeeds independently with probability $p_r=(7-r)/6$. Consequently $T\mid R=r$ has the geometric distribution on $\{1,2,\dots \}$ and expected value $1/p_r=6/(7-r)$. The law of total expectation givesThis is $2.45$ rolls; the averaging must be over conditional expected values, not over success probabilities before taking a reciprocal.

After the first child, $X_2$ is uniform on $\{0,\dots ,N-1\}$. Conditional on $X_2=k\ge 1$, the mean number removed is $(k+1)/2$; conditional on $X_2=0$, it is zero. The law of total expectation therefore givesFor $N=1$, the sum is empty and the same formula gives zero.

For $|s|\le 1$, all the generating-function expectations are absolutely bounded. Conditional on $X=k$, independence givesAt $k=0$ the product is one, consistent with the empty-family case. The law of total expectation now yieldsThis is the total progeny generating function of a branching process. The given almost-sure finiteness of $T$ ensures $G_T(1)=1$. The conditional-expectation proof avoids assuming a finite expected total before it has been established.

If $B$ and $U$ are independent random variables, conditioning on the Bernoulli random variable $B$ giveswith the continuous value $M_Z(0)=1$. Indeed $Z=0$ when $B=0$, while the conditional probability distribution of $Z$ given $B=1$ is the original uniform distribution.

Under the literal assumptions alone, the law of total expectation gives onlyThis is not determined by $M_U$ and $p$. For an explicit counterexample, put $p=1/2$ and $B=1_{\{U>0\}}$. Then $B$ has the required Bernoulli distribution, butwhich differs from the independent-case formula.