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Integrating the uniform density gives $\mathbb{E}{X}^k={\int }_0^1{x}^{k}dx=1/(k+1)$. By independence of random variables, $\mathbb{E}(XY{)}^k=\mathbb{E}{X}^k\mathbb{E}{Y}^k=1/(k+1{)}^2$. For the remaining expected value, integrate first in $x$:The value at $y=0$ is supplied by continuity. Therefore, with the harmonic number $H_m={\sum }_{j=1}^{m}1/j$,

Conditional on one sample point being $(x,y)$, it is a maximal external point precisely when none of the other $n-1$ points falls in the upper-right rectangle $(x,1]\times (y,1]$. Its area is $(1-x)(1-y)$, so its conditional probability of being maximal is $[1-(1-x)(1-y){]}^{n-1}$. Both $1-X$ and $1-Y$ are independent uniform random variables, hence a given point is maximal with probability $H_n/n$. Summing their indicator random variables and applying linearity of expectation gives the expected number of coordinatewise maxima:This grows only logarithmically with the sample size, even though all $n$ points are potential candidates.

Use the usual assumption that the two die scores are independent random variables. A fair score has expected value $7/2$ and variance $35/12$. Linearity of expectation and the variance of a sum therefore giveCount the ordered pairs of scores to obtain the probability mass functionsOn these attainable probability supports, the maxima occur at $x=7$ and $y=0$, each with probability $1/6$; the minima occur at $x=2,12$ and $y=-5,5$, each with probability $1/36$. Outside the respective probability support, the mass is zero.

They are not independent: $Y=0$ implies equal die scores and hence $X$ is an even number, so $P(X=7,Y=0)=0$, whereas $P(X=7)P(Y=0)=1/36>0$. In fact $\mathrm{Cov}(X,Y)=\mathrm{Var}(S_1)-\mathrm{Var}(S_2)=0$: this is an example of uncorrelated random variables that are dependent.

We first specify the standard hypotheses of percolation uniqueness on an amenable quasi-transitive graph. A locally finite graph has finite degree at every vertex; a quasi-transitive graph, or graph of finite type, has finitely many vertex orbits under the acting group $\Gamma$ of graph automorphisms. Here the site percolation law is invariant under $\Gamma$ and independent, with one parameter $p_j$ for each orbit. Initially suppose $0<p_j<1$. This gives the finite-energy property of Bernoulli percolation: conditional on all states outside any finite set, every assignment inside that set has positive probability. An amenable graph has finite sets $W_n$ with $|{\partial }^{+}{W}_n|/|W_n|\to 0$, where ${\partial }^{+}{W}_n$ denotes the external vertex boundary. Finite type bounds the degrees uniformly, so internal and external boundary conventions are equivalent here.

Let $N$ be the number of infinite percolation clusters, taking values in $\{0,1,2,\dots ,\infty \}$. Each event $\{N=j\}$ is invariant. Part (i), together with countability of these possibilities, shows that $N$ is an almost-sure constant.

Suppose first that this constant is a finite integer $j\ge 2$. Some finite connected set $F$ meets all $j$ infinite percolation clusters with positive probability, since increasing finite sets exhaust the graph. Make every vertex of $F$ open. This joins all the infinite clusters. Opening a finite set cannot produce a new infinite cluster from finite clusters: only finitely many clusters meet its finite boundary, and their finite union is finite. By the finite-energy property the modified event has positive probability, contradicting the almost-sure value $j$. Thus a finite $N$ must be $0$ or $1$.

Suppose instead that $N=\infty$ almost surely. Some finite connected set $F$ meets three distinct infinite percolation clusters with positive probability. Open all of $F$. Its vertices belong to one open cluster, and deleting $F$ from that cluster leaves at least three infinite components. This is an encounter box in percolation. We use boxes rather than assume that the local geometry permits a single trifurcation vertex in percolation.

Here is the boundary-counting contradiction in detail. Translates of $F$ have the same positive encounter probability. Finite type and bounded degree allow a deterministic packing of mutually disjoint such translates whose centers have positive lower density in the interiors of the Følner sets $W_n$: every vertex is within a fixed distance of one chosen orbit, and a bounded-radius greedy packing discards only a bounded number of possible centers per chosen center. Boundary layers of any fixed thickness have size $o(|W_n|)$. The expected number of encounter boxes wholly inside $W_n$ is consequently at least $c|W_n|$ for a fixed $c>0$ and all large $n$.

On the other hand, contract the open encounter boxes in each cluster to vertices and retain the part inside $W_n$. Each contracted encounter box separates at least three branches that run to infinity, and each branch must leave $W_n$. Choose a minimal forest connecting all these boundary exits. Every encounter junction has degree at least three in that forest: its three infinite branches remain in different components after deleting the junction, so none can be omitted. The leaves are boundary exits. For each nontrivial tree, the degree-sum identity gives at most the number of leaves minus two vertices of degree at least three. Thus the number of encounter boxes is at most a constant times $|{\partial }^{+}{W}_n|$, where the constant accounts for the bounded boundary degrees. Taking expected values contradicts amenability. This excludes $N=\infty$ and provesThere is a genuine endpoint qualification if different orbit parameters are allowed to be $0$ or $1$. Subdivide each horizontal edge of the square lattice once. Open every original vertex deterministically and close every new midpoint deterministically. This is an invariant independent law on an amenable graph that is also a quasi-transitive graph, but its infinite open clusters are the distinct vertical columns. Thus the finite-energy property must be included when interpreting a general vector $\mathbf{p}$. The homogeneous endpoint laws themselves cause no problem: they give zero or one infinite cluster.

The conclusion is false for $k$-independent laws, even with all site marginals equal to $1/2$. Here is a fully specified monochromatic line-graph percolation counterexample. Form a graph $B$ by replacing each vertex of ${\mathbb{Z}}^2$ by a complete graph on ten vertices and putting every possible edge between the blocks of two neighboring lattice vertices. Every vertex of $B$ has degree $9+4\cdot 10=49$. The graph is connected, periodic and amenable: a square of blocks has volume of order $n^2$ and boundary of order $n$. Assign independent fair red/blue colors to the vertices of $B$.

Call a block good when it contains both colors. Good-block indicators are independent, withThey percolate. To verify this without presupposing a site threshold, put $q=1-g=1/512$. A finite nearest-neighbor good-block cluster around a good origin is surrounded by a circuit of bad blocks in the matching square graph, where horizontal, vertical and diagonal neighbors are allowed. A surrounding circuit of length $\ell$ has an anchor within $[-\ell ,\ell {]}^2$. Overcounting its choices by $9{\ell }^2{8}^{\ell }$ gives the Peierls argumentThe sum bound follows directly from ${\sum }_{\ell \ge 1}{\ell }^2{r}^{\ell }=r(1+r)/(1-r{)}^3$, with $r=1/64$. An infinite good-block cluster therefore exists with positive probability, and by part (i) it exists almost surely. Inside each of its blocks the red vertices are connected to one another, as are the blue vertices; between adjacent good blocks the complete bipartite joining edges connect each color. Hence $B$ has both an infinite red component and an infinite blue component almost surely.

Now use site percolation on $\Gamma =L(B)$, the line graph of $B$: its vertices are the edges $\{u,v\}$ of $B$, two adjacent when they share an endpoint. Declare such a site open exactly when $u,v$ have the same color. Each site has open probability $1/2$. Sets of sites at graph distance greater than one involve disjoint sets of underlying color variables, so this is a translation-invariant $1$-independent law. The line graph is again connected, locally finite, amenable and of finite type. The infinite red and blue components of $B$ yield infinite open clusters in $L(B)$. An open path in $L(B)$ cannot change color at a shared endpoint, so those two clusters are distinct. Thus $N\ge 2$ almost surely. Local color constraints destroy the finite-energy property, precisely the ingredient missing from the uniqueness proof.

For $a<b$, let $U_N[a,b]$ be the number of completed upcrossings by the observations $X_1,\dots ,X_N$: buy when the value is at most $a$, sell at the first subsequent value at least $b$, and repeat. With $z^{-}=\max \{-z,0\}$, one form of the Doob upcrossing inequality for a martingale isFor completeness, hold either zero or one unit according to this rule. The holding at each step is measurable before the next martingale increment, so the total gain $G_N$ has expected value zero. Completed trades earn at least $b-a$ each; an unfinished trade loses at most $(X_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(X_N-a{)}^{-}$, proving the inequality.

Suppose $C={\sup }_n\mathbb{E}|X_n|<\infty$. For each rational pair $a<b$ the inequality and monotone convergence giveEvery rational interval therefore has only finitely many upcrossings, simultaneously outside a countable union of null sets. If $\mathrm{liminf}{X}_n<\mathrm{limsup}{X}_n$, there is a rational interval strictly between them and infinitely many upcrossings, a contradiction. Thus the limit $X_{\infty }$ exists in the extended real line almost surely. Fatou's lemma now givesThe limit is finite almost surely and integrable. This proves the martingale convergence theorem rather than using it as the deduction.

The exact extra condition for convergence in L1 is uniform integrability:To see sufficiency, first discard the tails of $X_n$ uniformly and the tail of the integrable $X_{\infty }$. On the set where both absolute values are at most $K$, their difference is bounded by $2K$ and tends to zero almost surely; dominated convergence finishes the proof. Conversely, if $X_n\to X_{\infty }$ in $L^1$, thenMake the first term uniformly small for all sufficiently large $n$, then control the finitely many remaining integrable variables by increasing $K$. Hence uniform integrability is necessary too. In this case the conditional expectation identity $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ follows by taking the $L^1$ limit in $X_n=\mathbb{E}[X_m\mid {\mathcal{F}}_n]$ for $m\ge n$.

The coin-doubling martingale shows why an $L^1$ bound is not enough. Let ${\xi }_1,{\xi }_2,\dots$ be independent fair coin indicators and setConditional on the first $n$ coins, the next value is either $2X_n$ or zero with equal probabilities, so this is a nonnegative martingale. Its expected value is always $1$, while a first failure occurs almost surely, giving $X_n\to 0$ almost surely. Yet $\mathbb{E}|X_n-0|=1$ for every $n$. Indeed, whenever $2^n>K$, its tail expected value above $K$ is still $1$. Thus convergence in L1 fails because uniform integrability fails.

A martingale describes an observable after its predictable drift has been removed. For a Markov chain the drift is the change produced by its transition operator; for Brownian motion it is one half of the Laplacian. Requiring the resulting processes to be martingales for a sufficiently rich class of observables determines the transition law itself. The same identities, combined with optional stopping, determine hitting probabilities, expected exit times and recurrence or transience. The duplicated Q5 paragraph in the PDF is a repeated printing of this one essay request.

For a finite or countable-state Markov chain with transition matrix $P$, define $(Pf)(x)={\sum }_{y}P(x,y)f(y)$ for a bounded test function $f$. The Markov property says $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=(Pf)(X_n)$. Consequentlyis a martingale, since each increment has conditional expected value zero. Conversely, if these processes are martingales for every bounded $f$, taking the conditional expectation of an increment recovers that identity. With $f=1_{\{y\}}$ it gives $\mathbb{P}(X_{n+1}=y\mid {\mathcal{F}}_n)=P(X_n,y)$ for every state $y$. Thus, given the initial law, these identities characterize the Markov chain, not merely a few of its moments. This is the discrete-time Markov chain martingale characterization.

The drift operator is $P-I$. If $(P-I)h=0$ away from a target set, $h(X_{n\wedge T})$ is a stopped martingale, where $T$ is the first hit of that set. Under boundedness or suitable uniform integrability, the optional stopping theorem expresses $h$ in terms of boundary hitting probabilities. Similarly, if $(P-I)v=-1$ away from the target and $v=0$ on it, then $v(X_{n\wedge T})+(n\wedge T)$ is a martingale, giving $v(x)={\mathbb{E}}_{x}T$ when the terminal limit is justified.

For a concrete calculation, take a simple symmetric random walk on $\{0,1,\dots ,N\}$ stopped at $T$, its first hit of $0$ or $N$. In the interior, $h(i)=i$ has zero drift, while $v(i)=i(N-i)$ satisfiesThe stopped identity gives ${\mathbb{E}}_i(T\wedge n)=v(i)-{\mathbb{E}}_{i}v(X_{T\wedge n})\le v(i)$, proving $T$ finite with finite mean. Bounded convergence in both stopped identities now yields the gambler's ruin answersThis illustrates why the integrability justification is part of using a martingale, rather than an automatic consequence of writing an equation.

For a finite-state continuous-time Markov chain with generator matrix $Q=(q_{ij})$, the corresponding martingale problem for a continuous-time Markov chain usesThese are martingales. Conversely, suppose an adapted càdlàg finite-state process satisfies these identities. Let $P_t=e^{tQ}$, fix $T$, and set $g(s,i)=(P_{T-s}f)(i)$. Then ${\partial }_{s}g+Qg=0$. Apply deterministic integration by parts to the finite collection of indicator martingales to obtain the time-dependent Dynkin formula: $g(s,X_s)-g(0,X_0)-{\int }_0^s({\partial }_{r}g+Qg)(r,X_r)dr$ is a martingale. ThereforeThis recovers the Markov property and the entire semigroup, proving the characterization. On countable spaces one uses nonexplosion and localization to justify the analogous unbounded-rate formulas.

For $d$-dimensional Brownian motion, the Itô formula shows that, for $f\in C_c^{\infty }({\mathbb{R}}^d)$,is a true martingale, since its bounded derivatives make the integral square-integrable on each bounded horizon. This is the martingale problem for Brownian motion with generator $L=\frac{1}{2}\Delta$.

Here is why it also characterizes Brownian motion. Suppose a continuous adapted process $X$, with $X_0=x$, satisfies that martingale problem. Use smooth cutoff test functions equal to $x_i$ and $x_i{x}_j$ inside successively larger balls, and stop before leaving those balls. The coordinate tests show that $X^i-x_i$ are continuous local martingales; the product tests show that $X^i{X}^j-x_i{x}_j-{\delta }_{ij}t$ are local martingales. The Itô product rule consequently givesthe difference between the two proposed compensators is a continuous finite-variation local martingale, hence constant and zero at time zero. For each $u\in {\mathbb{R}}^d$, another application of the Itô formula gives the exponential local martingaleIts modulus is the deterministic quantity $e^{|u{|}^{2}t/2}$, so it is a true martingale on every bounded time interval. Taking its conditional expectation and dividing by its known value at time $s$ givesThe conditional characteristic function identifies a centered multivariate normal distribution of covariance $(t-s)I$, independent of the past. Together with continuity, this is the defining law of Brownian motion. This proves the Lévy characterization of multidimensional Brownian motion in the form needed for the generator characterization.

The analytic use of these martingales parallels the chain case. A harmonic function $h$ satisfies $\Delta h=0$, so it yields a stopped local martingale until the process leaves its domain of harmonicity. Bounded stopped observables permit optional stopping and solve boundary hitting problems. A solution of $\frac{1}{2}\Delta v=-1$ with zero boundary values gives an expected exit time. For example, in the ball of radius $R$, $v(x)=(R^2-|x{|}^2)/d$ gives ${\mathbb{E}}_x{T}_R=(R^2-|x{|}^2)/d$; stopping $|B_t{|}^2-dt$ first at $T_R\wedge n$ proves finiteness and justifies the limiting calculation. The Brownian local time construction in Q4 is another example: the predictable compensators of approximations to $|B_t|$ recover its increasing occupation term.

We now prove the required three-dimensional transience of Brownian motion in dimension at least three, in the strong sense $|B_t|\to \infty$ almost surely. Start first at a point $x$ with $r<|x|<R$, and let $T$ be the first exit from this annulus. The radial Laplacian in dimension three givesThus $|B_{t\wedge T}{|}^{-1}$ is a bounded martingale. The exit time is finite: stopping $|B_t{|}^2-3t$ gives $3\mathbb{E}(t\wedge T)\le R^2-|x{|}^2$. At exit, continuity puts $|B_T|$ at either $r$ or $R$. The optional stopping theorem therefore yieldsand hence the Brownian sphere-hitting probability in dimension threeAs $R\to \infty$, the events $\{{\tau }_r<{\tau }_R\}$ increase to $\{{\tau }_r<\infty \}$: a continuous path is bounded on the finite interval preceding any finite inner hit. This argument assumes no transience. ConsequentlyTo pass from a positive escape probability to almost-sure eventual escape, take Brownian motion started at zero. Let ${\sigma }_R$ be its first hit of the sphere of radius $R$. Stopping the squared-norm martingale proves $\mathbb{E}{\sigma }_R=R^2/3<\infty$. At ${\sigma }_R$, the Strong Markov property and the preceding hitting formula giveThe Strong Markov property here follows from independent Brownian increments by approximating a stopping time from above with dyadic-valued stopping times and using path continuity. Choose $R_m=2^{m}r$. The exit times ${\sigma }_{R_m}$ increase to infinity, since a continuous path is bounded on every compact time interval. The events of a return to the closed ball of radius $r$ after ${\sigma }_{R_m}$ decrease and have probabilities $2^{-m}$. Their intersection has probability zero. Thus almost surely there is some finite exit time after which the path never again enters that ball. Apply this simultaneously to all positive integer $r$. This proves the last visit to a bounded ball for three-dimensional Brownian motion conclusion and therefore

The command uses the model without covariates, starts in mild fatigue at time zero, and calculates expected accumulated years in each state between times $0$ and $10$. Let $Z(t)$ be the fatigue state and $Q$ its fitted generator matrix. For the occupation time of a continuous-time Markov chain,the expected finite-horizon occupation time isThe first equality follows by interchanging a bounded time integral and expected value, and the second uses the homogeneous continuous-time Markov chain transition matrix. The reported expectations are 7.014800 years mild, 1.724686 years moderate, and 1.260514 years severe. They sum to ten because the person occupies exactly one state at every time. All repeated entries are counted: these are not the lengths of single holding times, the durations until first reaching each state, or the probabilities of occupying the states at year ten. The interpretation of the time limits is also given in the package documentation.

For $m\ge 2$, $f^{\prime \prime }(x)=m(m-1)x^{m-2}\ge 0$ on $[0,\infty )$, so $f$ is a convex function; for $m=1$ it is linear and also convex. The equally weighted random variable has expected value $EX=N^{-1}{\sum }_i{x}_i=1/N$. Applying Jensen's inequality givesMultiplying by $N$ provesFor $m>1$ equality requires all $x_i=1/N$, by strict convexity; for $m=1$ equality holds for every admissible vector.

Use the same first-sex conditioning as in part (b). The conditional variances are $p/q^2$ after a male first capture and $q/p^2$ after a female first capture. ThusThe conditional expected values are $1+1/q$ and $1+1/p$, with probabilities $p,q$. A variable taking two values $a,b$ with probabilities $p,q$ has variance $pq(a-b{)}^2$, soThe law of total variance now givesAt $p=q=1/2$ this gives variance $2$, consistent with $N$ being one plus a geometric waiting time of parameter $1/2$.

For $q>p$, let $M={\sup }_{n\ge 0}{S}_n$, including $S_0=0$. It is a nonnegative integer-valued random variable, possibly infinite before the following estimate. The tail-sum formula for expectation and part (i) giveThus $\boxed{\mathbb{E}\underset{n\ge 0}{\sup }{S}_n\le p/(q-p)}$, and the supremum is finite almost surely. The sum identity follows by writing $M={\sum }_{k\ge 1}{1}_{\{M\ge k\}}$ and applying monotone convergence theorem. If the printed supremum is read literally over $n\ge 1$, its positive part has exactly the same positive-level tails and its negative part is bounded by one. The same upper bound on its expected value remains valid.

Put $\rho =p/q\in (0,1)$ and $M={\sup }_{n\ge 0}{S}_n$. For $k\ge 1$, $\{M\ge k\}=\{H_k<\infty \}$, so part (ii) gives $\mathbb{P}(M\ge k)={\rho }^k$. Since $\mathbb{P}(M\ge 0)=1$, taking differences givesThis is the geometric distribution of parameter $1-p/q$ on the failures-before-first-success convention, with support $\{0,1,\dots \}$. Its expected value is $\rho /(1-\rho )=p/(q-p)$, so the bound in part (b)(ii) is attained. This proves the maximum of a downward-biased random walk formula.

The time-zero convention is necessary here. If $M_1={\sup }_{n\ge 1}{S}_n$, then the first step can be $-1$ and the walk may never return to zero. Using the same hitting probability calculation after that step gives $\mathbb{P}(M_1=-1)=q(1-\rho )=q-p>0$. Also $\mathbb{P}(M_1\ge 0)=p+q\rho =2p$, while its tails at positive $k$ remain ${\rho }^k$. Its masses are thereforeThus the printed geometric conclusion is correct for the standard maximum that includes $S_0=0$, and false for a supremum restricted to the positive times.

The increments have expected value $p-q<0$. The strong law of large numbers for independent and identically distributed random variables with finite absolute mean givesHence $S_n\to -\infty$. Since $q/p>1$,Its expected value nevertheless stays equal to one. Thus this martingale is not uniformly integrable, and it does not converge to zero in $L^1$. This distinction will be avoided in part (ii) by stopping at an upper level, which makes the stopped process bounded.

For each fixed $t\in [0,1]$, the variables $1_{\{X_k\le t\}}$ are independent and identically distributed random variables with Bernoulli distribution of parameter $t$. Their expected value is $t$, and the strong law of large numbers gives $F_n(t)\to t$ almost surely.

For a finite set $F$, intersect these probability-one events over its finitely many elements. On the resulting single event, every one of the finitely many errors tends to zero, so their maximum tends to zero:If $F$ is empty, take the displayed maximum to be zero. The important point is simultaneous convergence on one event; pointwise convergence at an uncountable collection of arguments would not alone imply the uniform claim in part (ii).

Under the null hypothesis, $Y\sim \mathrm{Bin}(n,1/2)$, with expected value $n/2$ and variance $n/4$. Approximating its probability at an integer by the normal distribution density over a unit-width cell givesDividing by the prior predictive distribution from the preceding part gives the uniform-alternative binomial Bayes factorThe second approximation replaces $n+1$ by $n$. Both are large-sample approximations for counts in the central range, rather than exact identities at extreme counts.

Condition first on the count in the random sum of independent claims. For $N=k$, independence of the summands and independence from $N$ giveThe empty product is one when $k=0$. Taking expected values proves the random-sum transform identity:For positive claim sizes this composition is always finite for $u\le 0$, and is an identity of Laplace transforms of nonnegative random variables after writing $u=-s$. For positive $u$, it holds wherever the composed series is finite; no positive exponential moment is implied merely by positivity of the summands.

Let $P(z)$, $F(z)$ and $G(z)$ be the probability generating functions of $N$, one severity, and $S$, respectively. Positivity of the severities gives $F(0)=0$ and $g_0=p_0$; the random-sum transform identity gives $G=P\circ F$. The count recurrence in the Panjer claim-count class impliesUse the chain rule and substitute $F(z)$ to obtainEquating the coefficient of $z^{r-1}$ givesThe factor multiplying $f_j{g}_{r-j}$ after rearrangement is $a(r-j)+(a+b)j=ar+bj$. Therefore the Panjer recursion isThese power-series calculations are valid inside the convergence discs of the probability generating functions, and hence justify the coefficient identities without assumptions about positive exponential moments.

For a Poisson distribution count, $p_n/p_{n-1}=\lambda /n$, so $a=0$ and $b=\lambda$. The aggregate recursion becomesTo obtain a moment recursion, write ${\mu }_j=\mathbb{E}{X}^j$, $m_j=\mathbb{E}{S}^j$, and $m_0=1$. For nonnegative measurable $h$, conditioning on $N=n$ and using exchangeability of the severities giveswhere $X$ on the right is independent of $S$. The second equality uses $n{p}_n=\lambda p_{n-1}$, the Poisson size-bias identity. Choosing $h(s)=s^{k-1}$, applying the binomial theorem, and using independence proves the raw moment recursion for a compound Poisson distribution:For any particular $k$, a finite severity $k$th moment suffices: $({\sum }_{i=1}^n{X}_i{)}^k\le n^{k-1}{\sum }_{i=1}^n{X}_i^k$, and the Poisson distribution has finite moments of every order. No positive moment-generating function domain is required.

The first three iterations giveSubtracting the appropriate powers of the expected value yieldsIn particular, the aggregate variance and third central moment involve the raw second and third severity moments, rather than their centered counterparts.

The severity is an equal-weight mixture distribution of rate-$2$ and rate-$2/3$ exponential distributions. Its expected value isThe moment-generating function, finite exactly for $r<2/3$, isCanceling the nonzero $r$ in the adjustment coefficient equation and multiplying by the positive denominator on $0<r<2/3$ givesThus the required polynomial isSince $P(0)=4\theta >0$, $P(2/3)=-2<0$, and the leading coefficient is positive, one root lies in $(0,2/3)$ and the other exceeds $2/3$. Only the smaller root is in the moment-generating function domain. ConsequentlyThe rationalized expression avoids subtraction of nearly equal numbers when $\theta$ is small.

The erroneously chosen same-mean exponential distribution gives $R_{\exp }=\theta /(1+\theta )$. Direct substitution shows $P(R_{\exp })=-\theta <0$, so $R_{\exp }$ lies strictly between the two polynomial roots, and in particular $R_{\mathrm{mix}}<R_{\exp }$. Thereforewhile both bounds equal one at $u=0$. The exponential misspecification produces a smaller claimed upper bound. The Lundberg inequality certifies $e^{-R_{\mathrm{mix}}u}$ for the actual mixture; it does not certify the smaller expression computed from the wrong severity law.

Let $C_t$ be the cumulative claim amount and $U_t=u+ct-C_t$ the surplus in the classical risk model. The supplied adjustment coefficient satisfies $\lambda (M_X(R)-1)=cR$. The Compound Poisson process has independent increments, and the random-sum transform identity givesIt follows that $Z_t=\exp (R(C_t-ct))$ is a nonnegative martingale with $Z_0=1$: conditional on the past, the multiplicative increment has expected value one. This is the exponential surplus martingale.

Let $\tau =\inf \{t\ge 0:U_t<0\}$ be the ruin stopping time. Surplus decreases only at claim arrivals, so on $\{\tau \le t\}$ the deficit at ruin gives $C_{\tau }-c\tau >u$ and $Z_{\tau }>e^{Ru}$. The optional stopping theorem at the bounded stopping time $\tau \wedge t$ yieldsAs $t$ increases without bound, the events $\{\tau \le t\}$ increase to eventual ruin. Hence the ultimate ruin probability satisfies the Lundberg inequalityThe proof uses only bounded-time stopping; it does not presume that the martingale can be stopped with equality of expected values at an unbounded ruin time.

The ratio of expected process variance to variance of hypothetical means is $K=s^2/v=(\alpha +1)/(k-2)$, and the credibility factor is $Z=n/(n+K)$. Holding $\alpha ,\lambda$ fixed, an increase in $k$ decreases $K$. Hence the credibility factor increases, approaching one as $k$ tends to infinity. Extending the formula to real $k>2$ makes the sign explicit:For the integer values in the model this establishes the same strict monotonicity. The collective claim expected value $m=\alpha /[\lambda (k-1)]$ also changes, but the conclusion concerns the relative weight of experience: conditional noise declines faster than the variation of the risk means.

Set $h=\mathbb{E}\theta >0$ and $w=\mathrm{Var}(\theta )>0$. Keeping $h$ fixed in the gamma distribution family gives $\alpha =h^2/w$ and $\lambda =h/w$. The ratio of expected process variance to variance of hypothetical means becomesAs the prior variance $w$ decreases, $K$ increases, so the credibility factor decreases. In the limit $w\downarrow 0$, $Z\downarrow 0$: the risk parameter is already essentially known from the prior, leaving little need to use the noisy sample mean. The population claim expected value $h/(k-1)$ remains fixed throughout this comparison.

The PDF gives an inverse-gamma distribution with shape $k$ and scale $\theta$, and a gamma distribution prior with shape $\alpha$ and rate $\lambda$. Write $\mu (\theta )=\theta /(k-1)$. The supplied conditional moments, together with $\mathbb{E}\theta =\alpha /\lambda$ and $\mathbb{E}{\theta }^2=\alpha (\alpha +1)/{\lambda }^2$, give the three structural quantities of the Bühlmann model:Here $v$ is the variance of hypothetical means and $s^2$ the expected process variance. The law of total covariance and conditional independence of different years implyAll these quantities are finite because $k>2$.

For any coefficients $c_i$, minimizing the mean squared error over the intercept first gives $c_0=m(1-{\sum }_i{c}_i)$. Let $Y=\mu (\theta )-m$ and $W_i=X_i-m$. Expanding the remaining mean squared error givesIts normal equations are $s^2{c}_i+v{\sum }_j{c}_j=v$ for every $i$. Since $s^2>0$, subtracting any two equations shows that all $c_i$ agree. The quadratic part is strictly positive for every nonzero coefficient change, so the solution is the unique global minimizer. Solving givesA credibility estimate combines individual experience with the collective expected value, using an experience weight $Z\in [0,1]$. In this equal-exposure setting, the Bühlmann credibility premium isThus the computed coefficients give exactly a credibility estimate, with a fixed credibility factor determined by known parameters and the number of observed years. This is Bühlmann credibility for inverse-gamma observations.