Use one particle for each telephone line, including unused lines. Let $f$ be the number of free lines, $h$ the number of callers waiting for or being checked by an operator, and $a$ the number using automated service. Then $f+h+a=N$. These are the three colonies of a closed migration process, with directed cycle $f\to h\to a\to f$ and departure ratesRejected attempted calls leave the state unchanged. They must not be represented as extra customers in the waiting stage.

Equivalently the state is $(h,a)$ with $h+a\le N$. Its transitions areThe directed cycle has equal traffic weights; choosing each weight to be $\nu$ gives the unnormalized product-form stationary distribution of a closed migration processWe can verify global balance for a continuous-time Markov chain directly. Weighted incoming transitions from $(h-1,a)$, $(h+1,a-1)$ and $(h,a+1)$ contribute, respectively,Their sum equals the weight times the outgoing rate. This proves stationarity after normalization, without falsely asserting reversibility of the directed migration cycle.

For $n=h+a$ occupied lines, summing over the automated population givesThe normalizing constant is $G_N={\sum }_{n=0}^{N}H(n)$. The attempted calls form an independent Poisson process, so Poisson arrivals see time averages: the rate of attempted arrivals seeing a full state is $\nu$ times its stationary probability, and division by the total attempted-arrival rate gives the loss proportion. ConsequentlyThe numerator is essential; the PDF contains it even though the converted TeX loses it.

With two independent operators, the total check-completion rate is $\lambda \min (h,2)$. In particular it is only $\lambda$ when one caller is present. DefineThe human-stage weight becomes $(\nu /\lambda {)}^h/D_2(h)$, while the free and automated weights are unchanged. The same global-balance calculation applies using these departure rates. ThusThis is the blocking probability in a finite-line sequential-service network with two operators. It remains valid for $N=0$, when every attempted call is lost; for $N=1$ it coincides with the single-operator answer.

Let $(A_t,B_t)$ be a Markov jump process on integer populations, with rateswhen $a,b>0$, where $\kappa >0$. Make both axes absorbing, so the fight ends when one population reaches zero. Start at $(N,b_N)$, with $b_N=\lfloor \alpha N\rfloor$; when $\alpha N$ is integral this is the printed initial condition exactly. The harmless rounding convention makes the model defined for every integer $N$.

Use the embedded jump chain $(A_j,B_j)$ rather than physical time. Its probabilities of the two losses are respectively $b/(a+b)$ and $a/(a+b)$, independently of $\kappa$. It reaches an axis after at most $N+b_N\le 2N$ steps; hold it constant thereafter. PutBefore absorption its two increments are $-2a+1$ and $2b-1$, soAfter absorption set ${\mu }_j=0$. Consequentlyis a discrete-time martingale. All populations are at most $N$, so each martingale difference is bounded in absolute value by $2N+2\le 4N$. The total drift up to step $2N$ is at most $2N$. This is the squared-population near-invariant for stochastic attrition.

The two-sided Azuma-Hoeffding inequality states that a martingale with $m$ differences bounded by $C$ satisfies $\mathbb{P}(|M_m|\ge z)\le 2\exp (-z^2/(2m{C}^2))$. Its exponential estimate follows by applying the conditional Hoeffding lemma to each centered difference, iterating, and optimizing the exponential Markov bound. With $m=2N$ and $C=4N$ it gives

Let $c=\sqrt{1-{\alpha }^2}>0$, and let $X^N=A_{2N}/N$ be the final proportion. If the first population survives, then the second is zero and $D_{2N}/N^2=(X^N{)}^2$. Hence $|X^N-c|>\delta$ impliesIf the first population is extinguished, $X^N=0$ and $D_{2N}\le 0$. This can satisfy the indicated deviation event only when $\delta <c$, and then $|D_{2N}-c^2{N}^2|\ge c^2{N}^2>c\delta N^2$. Thus the same implication holds in both cases.

The initial rounding error obeys $|D_0-c^2{N}^2|=|{\alpha }^2{N}^2-b_N^2|\le 2N$. Combining it with the drift bound shows that the deviation event forcesfor all sufficiently large $N$. Set $\eta =c\delta /2$ in the preceding concentration inequality. We concludefor sufficiently large $N$, and thereforeThe symmetric stochastic attrition process thus has exponential concentration of terminal stochastic attrition survivors. The deterministic equations conserve $a^2-b^2$ and predict the same surviving fraction, but the martingale argument supplies the stronger exponential probability estimate required here.

The original PDF specifies a bounded solution class $C_b^2$, rather than the $C_0^2$ transcribed in the TeX. Write $\varphi (x)=\min (1,\max (0,-x))$ and fix a time $t>0$. For Brownian motion starting at $x$, considerThe Itô formula and ordinary product rule give a drift proportional to $-u_t+\frac{1}{2}{u}_{xx}-\lambda \varphi u$, which vanishes by the equation. Therefore $F$ is a local martingale. It is bounded on this horizon because $u$ is bounded and $\varphi \ge 0$, so the bounded local martingale criterion makes it a true martingale. Its endpoint values are $F_0=u(t,x)$ and $F_t=e^{-\lambda {\int }_0^t\varphi (B_r)dr}$, using $u(0,\cdot )=1$. ConsequentlyThis proves the required Feynman-Kac formula with a bounded potential directly, and also gives uniqueness within the stated bounded solution class.

Let $I={\int }_0^t\varphi (B_s)ds$. By dominated convergence, $e^{-\lambda I}\to 1_{\{I=0\}}$ in expectation. Continuity of $B$ and $\varphi$ shows that $I=0$ exactly when $B_s\ge 0$ throughout $[0,t]$: a negative value at any time produces a nontrivial interval with positive integrand. For $x<0$ such an interval occurs immediately, so the limiting probability is zero. For $x>0$, the Brownian reflection principle and the atom-free distribution of the running minimum giveFor $x=0$ this probability is zero: the reflection formula for levels $-\epsilon$, followed by $\epsilon \downarrow 0$, shows that Brownian motion goes below zero before $t$ almost surely. ThusThis is the soft killing limit for Brownian nonnegative survival. The assertion is pointwise for positive time, as requested; no claim of uniform convergence across the initial-time boundary is needed.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

Use the physical dynamics of the Black-Scholes model, $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with bank rate $\rho$ and $\sigma >0$. Write $\lambda =(\mu -\rho )/\sigma$ for the market price of risk. If $w_t$ is portfolio wealth, ${\pi }_t$ the dollar amount in stock, and $c_t\ge 0$ the consumption rate, the budget dynamics areAdmissible controls are progressively measurable, locally integrable in the required drift and diffusion norms, and keep $w_t\ge 0$. Such a solvency restriction excludes doubling-type strategies. In a complete Brownian market, the investment-consumption problem is to maximizewhere the utility functions are increasing and concave. A terminal-only problem omits the integral. Strict concavity gives uniqueness of optimal consumption and terminal wealth when an optimum exists.

The state-price density isThe Itô formula givesThus $H_t{w}_t+{\int }_0^t{H}_s{c}_{s}ds$ is a nonnegative local martingale, hence a supermartingale, yielding the state-price budget constraintConversely, a nonnegative consumption plan and terminal payoff $\xi$ satisfying budget equality and the necessary integrability can be financed in the Brownian complete market. DefineThe Brownian martingale representation theorem for the total future budget supplies a stochastic-integral coefficient; comparing it with $H_t(\sigma {\pi }_t-\lambda w_t)$ defines ${\pi }_t$. This reconstructs the wealth equation with $w_t\ge 0$. Localization gives the representation for integrable positive budgets when an $L^2$ assumption is not available.

For smooth strictly concave utilities with the usual marginal-utility range, let $I_u=(u^{\prime }{)}^{-1}$ and $I_U=(U^{\prime }{)}^{-1}$ be the inverse marginal utility functions. The Lagrange-multiplier solution iswhere $y>0$ makes the budget an equality. This is a global argument, not merely formal differentiation: the concave supporting-tangent inequality giveswith the analogous terminal inequality. Taking expectations, integrating, and applying the budget bound makes the difference in objectives nonpositive. Replication of the candidate budget proves attainability and optimality whenever these expectations are well-defined and finite. If the marginal-utility ranges or integrability conditions fail, existence must be established separately rather than inferred from the first-order equations.

A detailed example is constant relative risk aversion utility $u(c)=c^{1-\gamma }/(1-\gamma )$ and $U(w)=\eta w^{1-\gamma }/(1-\gamma )$, with $\gamma >0$, $\gamma \ne 1$, $\eta >0$, and $\delta \ge 0$. PutThe optimal plans from the marginal-utility equations are $c_t^{\ast }=(y{e}^{\delta t}{H}_t{)}^{-1/\gamma }$ and ${\xi }^{\ast }={\eta }^{1/\gamma }(y{e}^{\delta T}{H}_T{)}^{-1/\gamma }$. The normal distribution of $W_t$ gives, for $a=1-1/\gamma$,The budget therefore becomes $y^{-1/\gamma }b(T)=w_0$, determining the multiplier explicitly. Independent Brownian increments in the conditional budget formula give $w_t^{\ast }=c_t^{\ast }b(T-t)$. The diffusion coefficient of $c_t^{\ast }$ is $(\lambda /\gamma )c_t^{\ast }$, so matching the wealth diffusion determines the dollar stock holding. Consequently the finite-horizon power-utility investment and consumption optimum isAll candidate moments are finite on a finite horizon, since powers of a lognormal state-price density have finite moments. The budget and tangent argument above therefore verify optimality in this example, including admissibility and positive wealth.

For another verification of the formulas, the Hamilton-Jacobi-Bellman equation for a value function with discount measured from the current time isSubstitute $V(w,t)=b(T-t{)}^{\gamma }{w}^{1-\gamma }/(1-\gamma )$. The two concave maximizations give the displayed controls, and the remaining equation is $b^{\prime }(\tau )=1-kb(\tau )$ with $b(0)={\eta }^{1/\gamma }$, exactly solved by the formula above. ThusFor $k=0$, interpret the integral as $\tau$, so $b(\tau )=\tau +{\eta }^{1/\gamma }$. Unlike the infinite-horizon Merton consumption-investment problem, finite horizons do not require $k>0$. Consumption vanishes in the terminal-only formulation; then the integral term in $b$ is absent and the optimal stock fraction remains $(\mu -\rho )/(\gamma {\sigma }^2)$. Logarithmic utility is handled directly with inverse marginal utility $1/z$ and the same wealth-to-consumption formula at $\gamma =1$, with $k=\delta$.

Assume $\sigma >0$, $x>0$ and sufficient $C^{2,1}$ regularity to use the Itô formula. Under the physical measure write $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ and let the bank account be $R_t=e^{\rho t}$. If $V_t=f(S_t,t)$ is a self-financing portfolio with ${\Delta }_t$ stock units and ${\beta }_t$ bank-account units, thenThe Itô formula gives $df=(f_t+\mu S_t{f}_x+\frac{1}{2}{\sigma }^2{S}_t^2{f}_{xx})dt+\sigma S_t{f}_{x}d{W}_t$. Equating diffusion terms forces ${\Delta }_t=f_x(S_t,t)$; equating drifts and eliminating ${\beta }_t{R}_t=f-S_t{f}_x$ gives the Black-Scholes equationConversely, if this equation holds, take $\Delta =f_x$ and $\beta =(f-x{f}_x)/R$. The same comparison shows $df=\Delta dS+\beta dR$, so the constructed portfolio is self-financing. This is a local replication equivalence for smooth functions on the state space; global admissibility and a terminal condition specify a unique economically admissible claim price.

With continuous stock-dividend yield $\theta$, holding ${\Delta }_t$ stock units earns $\theta {\Delta }_t{S}_{t}dt$ in addition to their price changes. A self-financing portfolio reinvests that income, soThe same diffusion comparison gives $\Delta =f_x$, and the drift comparison yields the Black-Scholes equation with continuous stock dividendsUnder the risk-neutral measure, the ex-dividend stock consequently has drift $\rho -\theta$, while cash is discounted at $\rho$.

For the last claim, distinguish the terminal payoff function $F$ from a time-dependent price. Put $\tau =t_0-t$. The price with yield $\theta$ is $q(x,t)=e^{-\rho \tau }\mathbb{E}[F(x{e}^{(\rho -\theta -{\sigma }^2/2)\tau +\sigma \sqrt{\tau }Z})]$, with $Z$ a standard normal random variable. In a no-dividend model with interest rate $\rho -\theta$, the stock-transition law is identical but its discount factor is $e^{-(\rho -\theta )\tau }$. Thus, whenever the payoff expectation is finite, the dividend-yield discount shift givesEquivalently substitute this expression in the dividend equation: $q_t=e^{-\theta \tau }(p_t+\theta p)$, and the equation reduces exactly to the no-dividend Black-Scholes equation with rate $\rho -\theta$, with the same terminal data.

For the Brownian first-passage time to a line, set $\lambda =\theta >0$ and choose $h=b+\sqrt{b^2+2\lambda }>0$, so $h^2/2-bh=\lambda$. The Exponential martingale for Brownian motion $M_s=e^{h{W}_s-h^{2}s/2}$, stopped at $T_{a,b}$, obeysThe optional stopping theorem at the bounded time $s\wedge T_{a,b}$ gives expectation one. The contribution from $\{T_{a,b}>s\}$ is at most $e^{ha-\lambda s}$, and tends to zero. Taking $e^{-\lambda \infty }=0$ therefore yieldsLetting $\theta \downarrow 0$ shows that the hitting probability is one for $b\le 0$ and $e^{-2ab}$ for $b>0$; in the latter case the hitting-time law has mass at infinity.

To derive the finite-time linear-boundary Brownian first-passage distribution, put $Z_s=W_s-bs$. This is Brownian motion with drift $\mu =-b$, and the event is $\{{\sup }_{s\le t}{Z}_s\ge a\}$. For driftless Brownian motion, the Brownian reflection principle says that the joint endpoint density on this event equals the normal density $p_t(x)$ when $x\ge a$ and the reflected density $p_t(2a-x)$ when $x<a$, where $p_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$. The constant-drift change of measure weights these densities by $e^{\mu x-{\mu }^{2}t/2}$. Thus the crossing probability isCompleting the square gives respectively $\Phi ((\mu t-a)/\sqrt{t})$ and $e^{2\mu a}\Phi ((-\mu t-a)/\sqrt{t})$. Substituting $\mu =-b$ provesThe density weighting follows directly from the Exponential martingale for Brownian motion, or the Girsanov theorem, on the fixed finite horizon; it does not require eventual hitting.

For the barrier digital call, under the risk-neutral measure in the Black-Scholes model,Assume $\sigma >0$, put $a=\log (c/S_0)/\sigma >0$ and $b=-(\rho -{\sigma }^2/2)/\sigma$. The upper stock barrier is reached exactly when $W_s^Q$ reaches $a+bs$. The payment is at the fixed maturity $t_0$, not at the random hitting time. Hence its price isIf $\sigma =0$, use the deterministic path $S_s=S_0{e}^{\rho s}$ instead: the value is $e^{-\rho t_0}$ if that path reaches $c$ by $t_0$, and zero otherwise. The nondegenerate formula concerns the usual positive-volatility model.