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For an equation $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, infinity is a regular singular point only when $p(x)=O(x^{-1})$ and $q(x)=O(x^{-2})$. Here $p=-1$ and $q=-2/x-2/x^2$, so infinity is an irregular singular point.

With varying positive step sizes, the same one-step inequality isSum from $s=1$ to $k$, use $\Vert {\beta }_1-\hat{\beta }{\Vert }_2\le 2R$, and divide by $2A_1$. The Jensen inequality for the weighted average $\tilde{\beta }=A_1^{-1}{\sum }_s{\eta }_s{\beta }_s$ giveswhere $A_1={\sum }_s{\eta }_s$ and $A_2={\sum }_s{\eta }_s^2$.

The stochastic gradients obeyThe nonexpansiveness of metric projection and the subgradient inequality imply, after conditioning on ${\beta }_s$,Summing telescopes. Since both ${\beta }_1$ and $\hat{\beta }$ lie in the radius-$R$ set $C$, their distance is at most $2R$. By the Jensen inequality and convexity of $f$,ChooseThe two terms are then equal, and

Choose $I_s$ independently and uniformly from $\{1,\dots ,n\}$. Starting from any ${\beta }_1\in C$, the projected stochastic gradient descent iteration isConditional on ${\beta }_s$, $\mathbb{E}[G_s\mid {\beta }_s]=\nabla f({\beta }_s)$, so $G_s$ is an unbiased stochastic gradient. The metric projection onto a closed convex set keeps every iterate in $C$. For a fixed step size, return the Cesàro average $\bar{\beta }=k^{-1}{\sum }_{s=1}^k{\beta }_s$.

The empirical risk isFor $g(z)={\log }_2(1+e^{-z})$,so $g$ is a convex function. Each signed margin $y_i{x}_i^T\beta$ is an affine function of $\beta$, and composition with an affine function preserves convexity. Equivalently,is a positive semidefinite matrix. Therefore $f$ is convex.

At time $t$, immediate exercise is worth $(S_t-K{)}^{+}$, while waiting is worth the European call option price with remaining maturity $\tau =T-t$. The conditional version of part (iii) givesUnder the standard Black-Scholes assumption $r\ge 0$, this is at least $S_t-K$, while a call value is nonnegative. Hence $C(t,S_t)\ge (S_t-K{)}^{+}$, and the investor should wait until $T$; with $\sigma >0$ and $\tau >0$, waiting has strictly positive time value.

If negative interest rates are allowed, the unconditional instruction “wait” need not be correct. The exact decision is to exercise whenand otherwise to continue.

Under $\mathbb{Q}$,DefineThe standard normal cumulative distribution function $\Phi$ then givesSubstitution into the discounted expected payoff yields the Black-Scholes formula

For a European call option, the pointwise inequality $(S_T-K{)}^{+}\ge S_T-K$ and risk-neutral pricing give

The model has a unique equivalent martingale measure $\mathbb{Q}$ under whichand the discounted stock is a martingale. The time-zero value of an integrable time-$T$ contingent claim $C$ is therefore

In a Black-Scholes model, the risk-free asset and risky asset satisfywhere $B$ is a Brownian motion, $\sigma >0$, and the coefficients are constant. With $S_0^0=1$,so the risky asset is a geometric Brownian motion.

The Poisson output theorem for a feedback queue, equivalently the Burke theorem applied to the effective queue-length process with service rate $\delta \mu$, shows that successful departures in equilibrium form a Poisson process of rate $\lambda$.

The process of all entries into the queue is different. Its long-run rate isImmediately after any queue-entry event the server is busy, so the conditional probability of another entry in the next interval of length $h$ isthe next entry can be external or can be a service completion followed by feedback. A Poisson process of the displayed long-run rate would instead give $(\lambda /\delta )h+o(h)$. These coefficients are unequal when $\lambda <\delta \mu$, sinceHence the combined arrival process is not Poissonian.

Let $N$ be the number of visits to the server made by one customer. Then $N$ has the geometric distribution with success parameter $\delta$, and its service periods $E_1,E_2,\dots$ are independent exponential random variables of rate $\mu$. For $s\ge 0$, the Laplace transform of the total service time $T=E_1+\cdots +E_N$ isThus

At each service completion, a successful departure reduces the queue length by one and occurs at rate $\delta \mu$ while the queue is nonempty; feedback completions leave its length unchanged. The queue-length process is therefore the birth-death process of an ordinary $M/M/1$ queue with service rate $\delta \mu$. It is positive recurrent exactly when $\lambda <\delta \mu$, and then, with $\rho =\lambda /(\delta \mu )$,

An M/M/1 queue has Poisson process arrivals of rate $\lambda$, independent exponential service times of rate $\mu$, and one server. The two letters $M$ indicate the memoryless property of the interarrival and service-time distributions. Its queue length is a birth-death process with birth rate $\lambda$ and death rate $\mu$ away from zero. When $\rho =\lambda /\mu <1$, its normalized stationary distribution isthe powers ${\rho }^n$ quoted in the question form the corresponding unnormalized invariant measure.

The strong law of large numbers states that if $(X_i{)}_{i\ge 1}$ are independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, thenalmost surely. Under the stronger fourth-moment hypothesis, it has a short proof. Put $\mu =\mathbb{E}{X}_1$, $Y_i=X_i-\mu$, and $S_n={\sum }_{i=1}^n{Y}_i$. Independence and centering make every term in the expansion of $S_n^4$ vanish unless each index occurs at least twice, soFor every $\epsilon >0$, the Markov inequality givesThe probabilities are summable, so the first Borel-Cantelli lemma says that $|S_n|>\epsilon n$ occurs only finitely often almost surely. Apply this simultaneously to $\epsilon =1,1/2,1/3,\dots$ to obtain $S_n/n\to 0$ almost surely, which is the claimed law.

Take the ring torus with minor radius $1$ and major radius $R_n=n+2$. With angular coordinate $\theta$ around the generating circle, its Gaussian curvature isand henceThus the displayed $\mathrm{limsup}$ condition holds, while every $S_n$ has genus one and is not diffeomorphic to the sphere.