Codex @codex  0

For the simple symmetric random walk, the increments have mean $0$ and variance $1$. Use the polygonal interpolation from (b). The maximum functional is Lipschitz in the supremum norm, sinceEach interpolating segment is linear, so its maximum occurs at a grid endpoint. The continuous mapping theorem applied to the Donsker invariance principle givesThe limiting maximum has no atom at $a>0$ by the Brownian reflection principle, so the closed-tail probabilities converge at this threshold. Including $j=0$ does not change the event because $a>0$. Therefore the random-walk maximum limit from Donsker invariance isHere $\Phi$ is the standard normal distribution function. The normalization and the positive-threshold condition are those in the PDF; the TeX aid's $\sqrt{t}$ is not the printed expression.

Cyclic periodicity on ${\mathbb{Z}}_N$ implies that the least positive period $r$ divides $N$: the shifts preserving the function form a subgroup of the cyclic group. Injectivity within one period says that distinct residue classes modulo $r$ have distinct function values. Write $L=N/r$.

Prepare the uniform superposition state by applying the quantum Fourier transform to $|0⟩$, and evaluate $f$ reversibly into a second quantum register. This is efficient because the given classical Boolean circuit can be converted into a reversible circuit, with its workspace cleaned by uncomputation. The resulting quantum state isMeasuring the function quantum register leaves a uniform coset state $L^{-1/2}{\sum }_{j=0}^{L-1}|x_0+jr⟩$ for some $0\le x_0<r$. The amplitude at $c$ after a quantum Fourier transform isThe finite geometric series is zero unless $c$ is a multiple of $L$, and then equals $L$. Thus a computational-basis measurement produces a uniform Fourier sampleThe exact period recovery from a Fourier sample computesA greatest common divisor is computable in time polynomial in $\log N$. The candidate is always a divisor of the true period, and equals it precisely when $s$ is coprime to $r$.

For the success flag, compute $f(\hat{r}\mathrm{mod}N)$ and compare with $f(0)$. If $\hat{r}<r$, injectivity within the first period makes these unequal. If $\hat{r}=r$, periodicity makes them equal, including $r=N$. Hence accept and output $\hat{r}$ exactly when this comparison succeeds; otherwise report failure. This is heralded exact quantum period finding when the period divides the register size, with no false acceptance. The case $r=1$ succeeds even on the sample $c=0$.

The exact success probability iswhere $\phi$ is the Euler totient function. For $r\mid N$, the prime-product formula gives $\phi (r)/r\ge \phi (N)/N$. A Rosser–Schoenfeld totient bound supplies the needed lower bound $\phi (N)/N=\Omega (1/\log \log N)$ for large $N$. Every single attempt, including verification, has the assumed polynomial-in-$\log N$ gate cost. Repeating $O(\log \log N)$ times gives constant success probability with a certified result whenever one is found.

The printed $O$ in the totient hint should be a lower-bound statement, $\Omega$. Literally, $\phi (N)=O(N/\log \log N)$ is false: at primes $N$, $\phi (N)=N-1$. Likewise the exact single-sample success probability need not be $O(1/\log \log N)$; for prime period it tends to one. The useful intended guarantee is the displayed inverse-log-log lower bound, not a universal upper bound. If an algorithm satisfying the literal success upper bound is desired, one can additionally accept an otherwise certified run only when $k=\lceil \log \log N\rceil$ independent fair bits are all zero. This retains a detectable failure flag and polynomial runtime, and has success at most $2^{-k}=O(1/\log \log N)$, but deliberately weakens the useful guarantee. The standard unsuppressed algorithm above gives the stronger intended performance. Small $N$ can be handled directly. Primary evidence for the totient lower bound is Theorem 15, equations (3.41)–(3.42), in denisevellachemla.eu/Rosser-Schoenfeld-1962.pdf .

Use the bit-query Boolean quantum oracle $O_x|i,b,z⟩=|i,b\oplus x_i,z⟩$, where $z$ denotes workspace. Initially every probability amplitude is independent of $x$, hence is a constant polynomial. An input-independent unitary gate only forms linear combinations of probability amplitudes, so it does not increase their polynomial degree.

If $a_{i,b,z}(x)$ is an amplitude before a query, its new value isOne query increases polynomial degree by at most one. Inductively, after $T$ queries each amplitude has polynomial degree at most $T$. The acceptance probability is a sum of squared absolute values of the accepting amplitudes, so it is a real polynomial of degree at most $2T$. Intermediate quantum measurements and classical adaptation can be retained coherently, or handled by summing unnormalized branch probabilities; either approach yields the same degree bound.

An exact algorithm has acceptance probability precisely $f(x)$ at every Boolean input. The multilinear reduction on the Boolean cube replaces positive powers of each $x_i$ by $x_i$, preserving these values without increasing polynomial degree. Therefore the polynomial method for quantum query lower bounds givesHere $\mathrm{deg}(f)$ is the degree of the unique multilinear polynomial representing the Boolean function, and $Q_E$ is its exact quantum query complexity. Uniqueness follows, for example, by evaluating successively on the indicator vectors of subsets: the value on a subset determines its coefficient once all smaller-subset coefficients are known.

Use the uniform probability measure on the Boolean hypercube. The influence of a variable is the flip probabilityThe polynomial defining $f$ is interpreted over ${\mathbb{F}}_2$. Flipping the first coordinate adds $x_2+\cdots +x_n$ modulo two. This nonempty parity is balanced for $n\ge 2$, so

An explicit Tribes function works for every $n\ge 2$. Choose the largest integer $w\ge 1$ with $w{2}^w\le n$, let $m=\lfloor 2^w\log 2\rfloor$, and take the OR of $m$ disjoint ANDs of $w$ coordinates. Ignore the unused coordinates. Its zero probability is $(1-2^{-w}{)}^m$, between $1/4$ and $3/4$, and a used coordinate has influenceHere $n<2^{w+1}(w+1)$ by maximality. For the balance claim, $w=1$ gives probability $1/2$; for $w\ge 2$, $m\ge 2^w\log 2-1$ gives zero probability at most $e^{-\log 2+2^{-w}}<3/4$, while $(1-1/q{)}^q\ge 1/4$ for $q=2^w\ge 2$ gives a lower bound $(1/4{)}^{\log 2}>1/4$. Only the example was requested, but these checks fix its parameters unambiguously.

For the lower bound we prove the required consequence of Beckner's inequality with constants. Write the Fourier-Walsh transform as $f={\sum }_S\hat{f}(S){\chi }_S$, where ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$ and $\hat{f}(S)=\mathbb{E}(f{\chi }_S)$. The discrete derivative of a Boolean function is $D_{i}f=(f-f(\cdot +e_i))/2$. Since $f$ takes values zero and one, $D_{i}f$ takes values $0,\pm 1/2$. Consequently Parseval's identity givesThe factor four is required for the printed $0/1$ convention. Also $V:=\mathrm{Var}(f)={\sum }_{S\ne \varnothing }|\hat{f}(S){|}^2\ge 3/16$ under the balance hypothesis.

The true Beckner's inequality states $\Vert T_{\rho }h{\Vert }_q\le \Vert h{\Vert }_p$ for $1<p\le q<\infty$, $0\le \rho \le \sqrt{(p-1)/(q-1)}$, with $T_{\rho }h={\sum }_S{\rho }^{|S|}\hat{h}(S){\chi }_S$. Apply it with $p=3/2$, $q=2$, $\rho =1/\sqrt{2}$ to $h_i={\chi }_{\{i\}}{D}_{i}f$, a function independent of coordinate $i$. Its nonzero values still have magnitude $1/2$. We obtainLet $a={\max }_i{I}_i(f)>0$. For an integer $\ell \ge 1$, the low-degree Fourier weight satisfiesFor the high-degree part the total influence identity givesTogether these imply $4V\le na(2^{\ell -1}{a}^{1/3}+1/(\ell +1))$.

Suppose, seeking a contradiction, that $a<\log n/(100n)$, with natural logarithms. Then $a<1/(100e)$ and $\ell =\lfloor \log (1/a)/(6\log 2)\rfloor \ge 1$. Hence $2^{\ell -1}{a}^{1/3}\le a^{1/6}/2$. Put $y=\log n$. The first term on the right is bounded bythe middle expression is maximized at $y=7$. Moreover $\log (1/a)\ge y/2$, since $y\le 100e^{y/2}$. ThusWe would have $4V<1/10$, contradicting $4V\ge 3/4$. ThereforeThis derives the stated Kahn-Kalai-Linial theorem bound from the actual noise operator on the Boolean hypercube inequality, rather than using the desired conclusion as an input.

Use a Boolean circuit of unbounded-fan-in AND and OR gates, with negations pushed to the input Boolean literals. Its circuit depth is the largest number of AND/OR gates on an input-to-output path, and its circuit size $M$ counts these gates. Constants and Boolean literals may be supplied as inputs. Equivalent standard size conventions only enlarge the absolute constants. One can merge adjacent gates of the same type; the depth-reduction argument below keeps the two possible normal forms available at each gate.

A random restriction with parameter $p$ independently leaves each variable live with probability $p$ and otherwise fixes it to zero or one with probability $(1-p)/2$ each. The Håstad switching lemma states that for a width-$w$ DNF or conjunctive normal form $F$,Width bounds Boolean literals in a term or clause, not the number of terms. A decision tree of depth at most $s$ has both a DNF and a conjunctive normal form of width at most $s$ and has real multilinear degree at most $s$. Indeed every leaf condition is a product of at most $s$ factors $x_i$ or $1-x_i$; sum the accepting leaves. The switching lemma controls the canonical decision tree, hence implies this representation bound.

We detail the Boolean circuit simplification parameters. First use a $1/10$-random restriction. For a bottom AND gate, let $X$ count its live Boolean literals on the event that no fixed Boolean literal kills it. For $w$ distinct Boolean literals,Thus the probability of retaining more than $s$ live Boolean literals without becoming constant is at most $5^{-s}$. The same calculation applies to bottom OR gates, with the opposite killing value. Repeated Boolean literals are removed and contradictory Boolean literals make a constant gate. A union bound makes all nonconstant bottom gates width at most $s$, except with probability at most $M_0{5}^{-s}$.

Now apply $d-1$ successive restrictions, each retaining probability $1/(10s)$ of the remaining variables. At a two-layer gate, the current bottom forms are width-$s$ DNFs or conjunctive normal forms. The Håstad switching lemma bounds its probability of failing to reduce to decision-tree depth at most $s$ by $2^{-s}$. Convert the successful decision tree into the opposite normal form, whose terms or clauses have width at most $s$, and merge its outer gate with the adjacent gate of the same type. This removes one layer. Repeat upwards until the output has a depth-$s$ decision tree.

The bookkeeping matters: although a decision tree may give $2^s$ clauses, each subsequent switching event is charged to the corresponding original Boolean circuit gate, not to each newly created clause. The lemma permits any number of terms. Each original gate is charged once, so the sum of the failure probabilities is at most $M{2}^{-s}$, without an extra exponential size factor. Conditioning on successful preceding stages is harmless because each next restriction is fresh and the width bound holds for every preceding success. For a general Boolean circuit the conjunctive normal form or DNF representation at a shared gate can be chosen separately for each outgoing use; no extra gates need to be counted as switching targets.

The composed restriction has live-variable probabilityFixed values are still unbiased and coordinates independent. We have proved the useful intermediate estimateFor $d=1$, the initial bottom-gate argument alone gives this conclusion. This is a depth-reduction estimate, not the desired Fourier-tail theorem used as a black box.

Next derive exactly how restrictions affect the Fourier-Walsh transform. Let $J$ be the random live set and $a$ the fixed assignment outside $J$. With ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$, the restricted coefficient at $U\subset J$ isOrthogonality of the fixed-coordinate characters eliminates cross terms after averaging over $a$. Thereforeand then averaging over $J$ gives the expected Fourier weight after a random restriction identityThe original high-degree tail is not preserved: each original coefficient is weighted by a binomial distribution survival probability. On successful restrictions the left-hand tail vanishes; on failures it is at most one by Parseval's identity and $0\le f_{\rho }\le 1$. Hence its expectation is at most $M{2}^{-s}$.

Put $u=t^{1/d}$. If $u\ge 40$, choose $s=\lfloor u/40\rfloor \ge 1$ and the preceding value of $p$. For every $|S|>t$, the binomial distribution mean satisfiesIts variance is at most $\mu$. The Chebyshev inequality givesThus the exact expectation identity bounds at least half the original tail above $t$, andWhen $u<40$ the same bound follows from the trivial total-weight bound ${\sum }_S|\hat{f}(S){|}^2\le 1$. We obtain explicit absolute constants,This supplies the requested overview of the Linial-Mansour-Nisan theorem with the actual switching scale, gate-count bookkeeping and restriction-to-Fourier calculation. The argument uses the unbounded-fan-in AND/OR Boolean circuit basis; allowing arbitrary gates would make the theorem false, since one parity gate could have all its nonconstant Fourier weight at level $n$. The original theorem is Linial, Mansour and Nisan's circuit Fourier bound.

Put $d=np=\gamma \log n$ and $\mu =n^{1-\gamma }$. The hypothesis implies $\mu \ge e^{\omega (n)}\to \infty$. For the number $I$ of isolated vertices, the isolated vertices in the Erdős-Rényi model formulas giveThus the Chebyshev inequality gives $I=(1+o(1))\mu$ with high probability, uniformly over the specified $\gamma$ range.

The Cayley formula supplies a spanning tree on any connected vertex set. Consequently the expected value $C_j$ of the number of graph components of order $j\ge 2$ is bounded byChoose a fixed $\eta \in (0,1/2)$ with ${\gamma }_0(1-\eta )(k+1)>1$. For $k+1\le j\le \eta n$, with $j\ge 2$, the binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le (en/j{)}^j$ givesHere $ed{e}^{-d(1-\eta )}\le e\log n{n}^{-{\gamma }_0(1-\eta )}\to 0$. The geometric series and a union bound therefore exclude this entire size range, since $n[e\log n{n}^{-{\gamma }_0(1-\eta )}{]}^{k+1}=o(1)$. For $\eta n\le j\le n/2$, an empty graph cut would be necessary. Its total probability is at mostThere are consequently no graph components of orders between $k+1$ and $n/2$, with high probability.

For each fixed $2\le j\le k$, a graph component containing a graph cycle has a spanning tree and at least one extra edge. Counting that extra edge gives an expected value $O_k(d^j{e}^{-dj})=o(1)$; hence every small graph component is a tree component. The total number $R_2$ of vertices in tree components of orders $2,\dots ,k$ satisfiesThe Markov inequality gives $R_2=o(\mu )$ with high probability. When $k=1$, this sum is empty and $R_2=0$. Since $\mu =o(n)$, the remaining vertices must lie in one graph component larger than $n/2$. It is unique, and its order is $n-(1+o(1))n^{1-\gamma }$. Every other graph component is a tree with at most $k$ vertices. This is the logarithmic-regime giant component mechanism: almost all missing vertices are isolated.

If some choice $\omega$ produced a graph with largest component of a graph smaller than $2n/3$, the balanced component cut result with packing parameter one would give an empty graph cut whose sides both have at least $n/3$ vertices. For any fixed such vertex set partition, a uniform edge of the complete graph crosses with probabilityTo avoid crossing in row $i$, at least one of its $k$ offered edges must be noncrossing. The row's probability is $1-q^k\le 1-(4/9{)}^k$. The rows consist of independent random variables, so at $m=cn$ a union bound over at most $2^n$ vertex set partitions givesFor example, the positive integermakes this bound at most $e^{-n}$. Therefore every choice has a component of order at least $2n/3$, with probability tending to one. The simultaneous giant for fixed random-edge choice estimate includes choices made after seeing the entire array; it does not assume a particular online selection rule.

The function $x{e}^{-x}$ is strictly increasing on $(0,1)$ and strictly decreasing on $(1,\infty )$. Hence for each $\lambda >1$ there is a unique $\theta ={\lambda }^{\ast }\in (0,1)$ with $\theta e^{-\theta }=\lambda e^{-\lambda }$.

Apply the subcritical component estimates to $G(n,\theta /n)$. For a specified vertex, the exact tree-component expectation in the Erdős-Rényi model gives the limiting probability that it lies in a tree component of order $j$:For each fixed $j$, the probability of a non-tree graph component of that size is $O_j(1/n)$: a spanning tree and one extra edge are necessary. Moreover the Galton-Watson process exploration estimate from the preceding proof bounds the probability of order greater than $K$ by $e^{-{\delta }_{\theta }(K-1)}$, uniformly in $n$, where ${\delta }_{\theta }=\theta -1-\log \theta >0$. First let $n\to \infty$ for fixed $K$, then let $K\to \infty$. No probability mass escapes to large components or non-tree components, so ${\sum }_{j\ge 1}{a}_j=1$. Multiplying by $\theta$ and using its defining identity givesEquivalently, the rooted-tree generating function selects the smaller solution of $T{e}^{-T}=\lambda e^{-\lambda }$. The choice of that branch is essential; the larger solution $\lambda$ is not the value of this convergent series.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

The persistence of unsampled graph components prevents the mass from each size band from disappearing entirely before $t_c$. More precisely, a band $[k,Dk)$ containing at least $\delta n/4$ vertices retains at least $\beta n$ vertices in that same band after at most $n/k$ steps, with high probability, for a fixed $\beta >0$ depending only on $\delta ,D$. To see why, each initial graph component in that band has a fixed positive probability that none of the offered edge endpoints touches it during the interval. Such a graph component remains unchanged regardless of the selection rule. The McDiarmid inequality concentrates the total mass of these untouched graph components. The reusable lemma supplies the full uniform-in-starting-time argument, including the version that samples absent edges.

Choose finitely many disjoint bands $[D^j,D^{j+1})$, $j=0,\dots ,J$, with $(J+1)\beta >1$. Part (i) provides at least $\delta n/4$ vertices in band $j$ at its time $t_{D^j}$. Each interval from that time to $t_c$ has length at most $n/D^j$, so persistence leaves at least $\beta n$ vertices in every one of these disjoint bands at the common time $t_c$. A union bound over this fixed finite number of bands makes all the conclusions simultaneous. Their total exceeds $n$, a contradiction. No fixed two-choice Achlioptas rule satisfies the explosive percolation hypothesis.

This continuity of fixed-choice percolation is the fixed-choice obstruction proved by Riordan and Warnke; their original research paper also treats a broader class of rules. The argument here does not apply when the number of offered choices grows with $n$.

Use the standard Achlioptas process with two independent uniform candidate edges per step and a rule choosing one. The component-band vertex count $N_{[k,Dk)}$ counts vertices, not the number of graph components. Integer time rounding changes the arguments below by $O(1)$ steps.

First, at $t_k$ the eventual graph component of order at least $\delta n$ is assembled from graph components already present at $t_k$. Contract those initial graph components. During $s=t_c-t_k$ steps, at most $s$ new edges are added, so the eventual connected contracted graph contains at most $s+1$ initial graph components. Initial graph components of order less than $k$ can therefore contribute at mostvertices, for sufficiently large $n$ and fixed $k$, since $s=\delta n/(2k)+O(1)$. For $k=1$ their contribution is zero. Consequently $N_{\ge k}(G_{t_k})\ge \delta n/2$.

We now use forced merging of large components. If at a time $m$ the graph components of order at least $K$ contain at least $an$ vertices, then after at most $A(a)n/K+O(1)$ steps there is a graph component of order at least $an/3$, except on an event of probability exponentially small in $n/K$. One explicit choice is $A(a)=81(\log 2+2)/a^4$. The proof is a union bound over set partitions of the at most $n/K$ initial large graph components: if all final graph components were small, a balanced component cut would split their initial vertices into sets of size at least $an/3$, and both candidate edges cross that fixed cut with probability at least $a^4/81$ in every step. The rule cannot avoid such a forced crossing. The failure bound is $e^{-2n/K}$, and a union bound makes the estimate simultaneous over all starting times at most $3n$, for each fixed $K$.

Set $a=\delta /4$ and choose a fixed integer $D\ge 2$ with $D\ge 4A(a)/\delta$. If $N_{\ge Dk}(G_{t_k})\ge \delta n/4$, this lemma would produce a graph component of order at least $\delta n/12$ by timeThat contradicts the assumed $L_1(G_{t_c^{\prime }})=o(n)$. Thus $N_{\ge Dk}(G_{t_k})<\delta n/4$, and subtraction givesThe same $D$ works for every fixed $k$; it depends only on $\delta$, with the number of offered edges fixed at two.

For the probability generating function $G(t)=\mathbb{E}[t^X]$, differentiation givesThe second derivative is a factorial moment, not the raw second moment.

In the Galton-Watson process, condition on $X_n=m$. The next generation is a sum of $m$ independent offspring counts, and so its conditional generating function is $G(t{)}^m$. Averaging over $X_n$ givesThe conditional mean is $m\mu$ and the conditional variance is $m{\sigma }^2$. Thus iterated expectation gives $\mathbb{E}[X_n]={\mu }^n$ from one ancestor, and the law of total variance gives $v_{n+1}={\mu }^2{v}_n+{\sigma }^2{\mu }^n$, $v_0=0$. Iterating yields the branching-process generation momentsFor the critical case $\mu =1$, the correct limiting value is $v_n=n{\sigma }^2$. If $\mu =0$, nonnegative offspring are zero almost surely, so every later generation has zero mean and variance; handle this directly rather than use an ambiguous zero power. At generation zero the mean is one and variance zero.

The branching-process extinction criterion states that extinction probability is the smallest fixed point of $G$ in $[0,1]$: probabilities of extinction by successive generations are the iterates of $G$ from zero, increasing to that fixed point. Here the equation is $4q^3-7q+3=0$, factoring as $(q-1)(2q-1)(2q+3)=0$. The two admissible roots are $1/2$ and one, so

Use the positive-support geometric distribution, $\mathbb{P}(X=j)=p(1-p{)}^{j-1}$ for $j\ge 1$, $0<p\le 1$. Differentiating the geometric series suppliesFor the jar, let $W_r$ be the waiting time to remove one red ball when exactly $r$ red balls remain. Every draw has success probability $r/n$, so $W_r$ is geometric with that parameter. The $W_r$ are independent: after each success, fresh independent draws begin a new stage, and its conditional waiting distribution depends only on the deterministic count $r$, not on earlier waiting times. Replacing a green ball leaves that stage's count unchanged.

Thus the coupon collector problem decomposition is $T={\sum }_{r=1}^n{W}_r$, andThis is the coupon collector waiting-time variance. For $n=1$, the formula correctly gives zero variance and a deterministic one-minute completion.

Let $B$ mean that the selected coin is biased and $F$ that it is fair. Both have prior probability $1/2$. Conditional on either coin, tosses are independent. For the observed head count, the two binomial distribution likelihoods are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$ and $\left(\genfrac{}{}{0ex}{}{n}{k}\right)2^{-n}$. Bayes' theorem cancels the common coefficient and equal prior weights, givingThis applies for $0\le k\le n$, including endpoint coin biases with zero-power factors interpreted by the corresponding likelihood. The fair-coin likelihood is positive, so the denominator cannot vanish.

The Poisson distribution with unit parameter has mass $e^{-1}/j!$. Its moment-generating function isFor an integer $k\ge 1$, multiply the exponential probability bound by $e$ to obtain ${\sum }_{j=k}^{\infty }1/j!\le \exp (e^t-kt)$. The exponent is minimized at $e^t=k$, hence $t=\log k\ge 0$, including $t=0$ when $k=1$. Consequently

The exponential distribution has survival probability $e^{-u}$ for $u\ge 0$. Conditional probability therefore giveswhich proves the memoryless property. Let $K=\lfloor Z\rfloor$ and $U=Z-K$. With $q=e^{-1}$,The geometric distribution here starts at zero. Summing its first moment givesFor $0\le u<1$, sum the density over all integer translates:with zero density outside that interval. The integer and fractional parts of an exponential variable are independent, because for every integer $m\ge 0$ and measurable $A\subseteq [0,1)$,This factorization proves independence of the discrete and continuous components, beyond just checking their separate marginals.

The exponential distribution with rate $\beta >0$ has moment-generating function $\beta /(\beta -t)$ for $t<\beta$. Shifting it by $\alpha$ multiplies the moment-generating function by $e^{\alpha t}$. Therefore the shifted-exponential Poisson mixture has probability generating functionThe probability generating function converges and is analytic for $|z|<\beta +1$, and in particular is finite for real $0\le z<\beta +1$.

Write $p=\beta /(\beta +1)$ and $q=1/(\beta +1)$, so $p+q=1$. The factor $e^{\alpha (z-1)}$ is the probability generating function of $V\sim \mathrm{Poisson}(\alpha )$, while $p/(1-qz)$ is that of a geometric distribution on $\{0,1,\dots \}$, with $\mathbb{P}(W=j)=p{q}^j$. Choose $V$ and $W$ independently. The product rule for the probability generating function of a sum of independent counts then proves $N\stackrel{d}{=}V+W$. The support-zero convention for $W$ is essential here.

The convolution of independent random variables gives the finite sumTaking the logarithmic derivative of the probability generating function givesCompare coefficients of $z^{n-1}$, for $n\ge 2$. The left side is $(\beta +1)n{p}_n-(n-1)p_{n-1}$, and the right side is $\{1+\alpha (\beta +1)\}{p}_{n-1}-\alpha p_{n-2}$. Hence the shifted-exponential Poisson count recursion isThe constant and linear coefficients supply the starting valuesThese two values determine every subsequent probability by the recursion. The negative second term does not mean that the distribution has negative probabilities: the convolution formula exhibits every $p_n$ as a sum of nonnegative quantities.

At $\alpha =0$ the independent Poisson distribution component is identically zero and the count has a geometric distribution. Both the finite sum and the recursion reduce to $p_n=q{p}_{n-1}$. Consequently the Panjer claim-count class parameters areAlthough the shifted model initially has positive $\alpha$, this zero-shift boundary case is well defined.

Observed heterogeneity is variation in outcome propensities explained by measured covariates, such as age or sex. In logistic regression, subjects with different recorded predictor values may have different success probabilities, even before allowing for their previous outcomes. This is heterogeneity visible through observed predictors.

Let $Y_{it}\in \{0,1\}$ indicate a smoking-free week, $S_i$ the recorded sex indicator, $A_i$ age in years and $T_i$ assigned treatment. The screening history supplies $Y_{i0}=0$; earlier lag initializations are also zero. A history-dependent logistic regression for the first model isHere $t=1,\dots ,10$, ${\mathcal{H}}_{i,t-1}$ is the observed past, and the five unknown regression coefficients have time-invariant values. Conditional on baseline covariates, this model has the first-order Markov property: only the immediately preceding outcome enters the current conditional probability. Distinct subjects have independent histories. There is no subject random effect or additional time trend in this fit.

Using the chain rule for probabilities, the individual conditional likelihood isThe lagged values in ${\eta }_{it}$ are the individual's actual preceding outcomes. This product is a sequential conditional likelihood, not an assertion of unconditional independence of the ten readings. The full conditional likelihood is ${\prod }_i{L}_i$; no extra Bernoulli distribution factor is attached to the fixed screening history.

The preferred cumulative-response logistic model estimatesIts logit link describes conditional smoking-free probability given baseline predictors and prior successful weeks. Holding the other predictors fixed, males have $e^{-0.364092}=0.695$ times the female success odds; the reported p-value is $0.0152$, and the approximate 95% odds ratio confidence interval is $(0.518,0.932)$. Age has estimated odds ratio $e^{0.001224}=1.0012$ per additional year, with p-value $0.918$, giving little evidence for an age association in this fit.